Introduction to Geometry and Topology - Matthew Day

7/28/2019 Introduction to Geometry and Topology - Matthew Day

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1 Motivation

Consider the real numbers R. From calculus (or analysis), we know what itmeans for a sequence of points to converge to a point in R, and we knowwhat it means for a function from R to R to be continuous. These ideashave to do with nearness: continuous maps send pairs of nearby points topairs of nearby points and convergent sequences eventually get closer andcloser to their limits.

Part of the motivation behind topology is to formalize and distill theidea of nearness. Explaining this aspect of the structure of R will allowus to generalize this notion of nearness to other mathematical objectswhich do not have the extra structure (geometric structure, order structure,

algebraic structure) that R has.Once we have an abstract model for what non-geometric nearness in

space means, we can use this to understand many spaces other than R,including graphs, Klein bottles, Cantor sets, spaces of functions, and others.We can also then understand operations on these spaces, such as takingproducts, cutting, and gluing.

In analysis, when we want to talk about points that are near a point p,we pick an > 0 and consider the set of all points whose distance to p is lessthan . This is intrinsically geometric. The usual approach in topology is tokeep track of all the possible sets of points that we might want to think ofas being the set of points near p, however, to forget which this set would

correspond to.A neighborhood S of a point p in R is a subset ofR, containing p, such

that S contains all the points in R that are close enough to p. An openset U is a subset ofR that is a neighborhood of every point it contains. Bykeeping track of either open sets or neighborhoods, we can have a notion ofnearness without using any geometry.

Example 1.1. In the real numbers R, the neighborhoods of a point p inR are those subsets ofR which contain an open interval containing p. Theopen sets in R are the subsets ofR that can be written as unions of openinterval, indexed over some possibly infinite index set I:

U =iI

(ai, bi).

The empty set is also open.

We dont consider closed intervals to be open sets because a closed inter-val is not a neighborhood of its endpoints: there are points in R arbitrarily

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close to the endpoint that are not in the closed interval. Having defined

open sets and neighborhoods, it is straightforward to give definitions forcontinuity and convergence for R that make no mention of any > 0.

2 Basics about spaces

2.1 Axioms and definitions

The first goal is to provide a general definition of a topological space, suchthat the real numbers will be an example. We will work with the mostcommon formulation of the definition, which uses open sets.

Definition 2.1. A topological space (X, T) is a set X together with a setT of subsets of X. The elements of T are called open sets and satisfy thefollowing axioms:

1. Arbitrary unions of open sets are open.

2. Finite intersections of open sets are open.

3. The set X and the empty set are both open.

Given a set X, a topology on X is a topological space structure (X, T).

Definition 2.2. Having defined topological spaces, we can define the fol-

lowing vocabulary:

A point in a topological space (X, T) is simply an element of X.

When it is necessary to refer to X as a set (and not as a space), theset X is called the point set of (X, T).

A subset A X is closed if its complement X\A is open in (X, T).

A neighborhood of a point p in a space (X, T) is a set S, p S X,such that there is an open set U with p U S. An open neighborhoodof p is simply an open set containing p.

Having defined neighborhoods, we can now characterize open sets interms of neighborhoods.

Proposition 2.3. A set U is open if and only if for each p U, U is aneighborhood of p.

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Proof. If U is open, then for each p in U, U is a (nonproper) superset of an

open set containing p (that is, U is the open set containing p). So U is aneighborhood of each p in U.

Conversely, suppose that U is a neighborhood of each point p in U. Thenfor each p in U, there is an open set Vp with p Vp U. Let V =

pUVp.

Clearly U V, since any p U is in Vp and therefore in V. However, ifp V then p is in some Vq for some q U; each Vq is a subset of U, so thisp is in U. So V U and therefore U = V. Then U is a union of open setsand is therefore open.

This proposition is the link between open sets and neighborhoods. It canbe used to prove a characterization of topological spaces with neighborhoods

as a primitive notion. This is left as an exercise.Now we show that the standard topology on R fits our definition.

Proposition 2.4. The set of open sets forR specified in Example 1.1 givesa topology onR.

Proof. Well show this set of open sets satisfies the axioms for open sets fromDefinition 2.1. We have defined an open set as a union of open intervals.Since a union of unions of open intervals is a bigger union of open intervals,a union of open sets is open.

Now suppose U =

iI(ai, bi) and V =jJ(ci, di) are open sets. Then

U V = (i

(ai, bi)) (i

(ci, di)) =

iI,jJ

((ai, bi) (cj, dj)).

Since each (ai, bi) (cj, dj) is either an open interval or is empty, we havewritten U V as a union of open intervals and it is open. Inductively anyfinite intersection of open sets is open.

Since R = iN(i, i) and weve declared the empty set to be open, thethird axiom is also satisfied.

Example 2.5. A few more examples of topological spaces:

Let X be any nonempty set. The discrete topology on X is the topo-

logical structure where all subsets of X are open.

Let X be any nonempty set. The indiscrete topology on X is thestructure where only X and are open.

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Let X be any infinite set. Declare the closed subsets of X to be

all the finite sets (so the open subsets of X are the sets with finitecomplement). This defines a topology, called the finite complementtopology on X.

Note that we have defined four different topologies on R. Provided aset X has more than one point, there are many different topologies on X.Therefore it is necessary to specify the full topological structure when intro-ducing a space. However, once weve specified the structure, it is rare thatwe specifically refer to the set of open sets T; instead we simply use theword open to specify its elements. For this reason, its common practiceto refer to a space (X, T) using only the symbol for its point set X.

2.2 Subspaces

Any subset of a topological space inherits the structure of a space.

Definition 2.6. Let (X, T) be a space and let S be a subset of X. Thesubspace topology on S is the one whose open sets are exactly the intersec-tions of open subsets of X with S. Precisely, let S = {U S|U T }. Then(S, S) is the subspace topology on S.

This topology on S is also called the induced topology.

Proposition 2.7. Let X be a space and let S be a subset of X. The sub-space topology on S is a topology on S (that is, it satisfies the axioms for atopological space).

The proof is straightfoward.

Example 2.8. Some examples of subspaces ofR with the standard topol-ogy:

The various kinds of intervals [0, 1], [0, 1), (0, 1], (0, 1).

The set {0} {1/n|n N}.

The rational numbers Q.

The irrational numbers R\Q.

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2.3 Subbases and bases

If you want give someone the definition of a particular topological space, youmust specify the point set and the set of all open sets. One way to specifythe set of all open sets is to let the axioms do the work.

Proposition 2.9. Suppose X is a set and C is a set of subsets of S. LetB0 be the set of all intersections of finitely many elements of C, and letB = B0 {X,}. Let T be the set of all unions of elements of B. Then Tis the set of open sets for a topology on X.

Proof. Clearly unions of elements of T are in T. Let U, V T, with U =

iIUi and V =

jJVj for some Ui, Vj B. Then

U V = (iI

Ui) (jJ

Vj) =

iI,jJ

(Ui Vj).

Since each Ui Vj is clearly in B, it follows that U V is in T and T isclosed under finite intersections. Of course, since X and are in B, theyare in T as well, and T is the set of open sets for a topology on X.

Definition 2.10. Let (X, T) be a topological space. A set B of open setsin X is a basis for (X, T) if the set of all unions of elements of B is T. Aset C of open sets in X is a subbasis for (X, T) if the set of all intersectionsof finitely many elements of X, together with X and , form a basis for(X, T).

So Proposition 2.9 asserts that any set of subsets of a set X is a sub-basis for some topology on X. In particular, one can specify a topology byspecifying a subbasis.

Proposition 2.11. Let X be a set and suppose B is a set of subsets of X.Suppose for any U, V B, the intersection U V can be written as a unionof members of B. Further suppose that each point in X is in some U in B.Then B is a basis for some topology on X.

Proof. Define T to be the set of all possible unions of sets of members ofB. Clearly T satisfies the axiom about unions. For U, V T, we writeU = iIUi and V = jJVj where all the Ui and Vj are in B. Then U Vis

iI,jJ(UiVj). Since each Ui Vj can be written as a union of membersof B, this U V can be written as such a union and inductively, T satisfiesthe axiom about unions. Since each p X is in some set in B, the unionof all members of B is T, and therefore X is in T. Since is the emptyunion, is in T. So T satisfies the axioms for a topology, and B is a basisfor T.

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Example 2.12. Bases and subbases can be used to describe the spaces we

have already seen:

The set of open intervals is a basis for the standard topology on R.

The set of singleton subsets is a basis for the discrete topology on anyset X.

The topology on a set X with the subbasis

{X\{p}|p X}

is the finite complement topology.

IfS is a subset of a space X and Bis a basis for X, then {SU|U B}is a basis for S with the subspace topology.

3 Continuity and homeomorphisms

Often in mathematics, one is interested not only in studying mathematicalobjects, but also in studying functions between mathematical objects thatpreserve some of the structure. In topology, we mainly consider continuousfunctions.

Definition 3.1. Suppose X and Y are topological spaces. A functionf: X Y is continuous if for every open subset U of Y, the preimagef1(U) is an open subset of X.

This definition seems a little mysterious at first. In the definitionfor the continuity of a function f: R R, we say f is continuous if forevery point p, it pulls back every small open interval around f(p) to a setcontaining a small open interval around p. This easily translates to sayingthat a continuous function f pulls back every neighborhood of f(p) to aneighborhood ofp. Since an open set is a neighborhood of any of the pointsit contains, continuity should mean that a function pulls back open sets toopen sets.

Continuity means nearby points get sent to nearby points. However closetogether I want points to end up, if they start out close enough together, thefunction sends them to points that close together. however small an openset I want to hit with my function, the set of points that hit it are still anopen set.

Bases make it more convenient to check that a function is continuous.

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Proposition 3.2. Suppose X and Y are topological spaces and Y has a

basis B. Let f: X Y be a function and suppose f1(U) is open for eachbasis element U B. Then f is continuous.

Proof. Let V Y be open. Write V as a union of elements {Ui}iI of B:

V =iI

Ui.

Then:f1(V) = f1(

iI

Ui) =iI

f1(Ui),

which is open.

Some near-trivial observations about continuity:

Proposition 3.3. For any space X, the identity map from X to itselfis continuous.

Compositions of continuous maps are continuous: given continuousmaps between spaces f: X Y and g : Y Z, the composition g f: X Z is continuous.

If S is a subspace of a space X, then the inclusion S X is continu-ous.

If X has the discrete topology, then any function from X to any spaceY is continuous; the discrete topology on X is the only topology on Xwith this property.

If X has the indiscrete topology, then any function from any space Yto X is continuous; the indiscrete topology on X is the only topologyon X with this property.

Definition 3.4. Suppose X and Y are topological spaces. A homeomor-phism from X to Y is an invertible, continuous function with continuousinverse. The spaces X and Y are homoemorphic if a homeomorphism exists

between them.Recall from set theory that a function is invertible if and only if it is one-

to-one and onto (a bijection). Warning: A continuous, invertible functiondoes not necessarily have continuous inverse. For example, consider anybijection from a space with the discrete topology to a space that does nothave the discrete topology.

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The relation of being homeomorphic is an equivalence relation on spaces

that is very strongin some sense, homeomorphism is the right notion ofsameness for spaces. We will proceed to justify this remark.

Suppose (X, T) is a space, Y is a set, and f: X Y is a map of setsthat is one-to-one and onto. Define f(T) to be the set {f(U)|U T }. Call(Y, f(T)) the relabeling of (X, T) by f.

Proposition 3.5. The relabeling (Y, f(T)) is a topological space and therelabeling map f: (X, T) (Y, f(T)) is a homeomorphism.

Proof. For any {Ui}iI elements of T, we have

f(iI

Ui) = iI

f(Ui).

This implies that f(T) satisfies the axiom about unions is satisfied. For anyU, V T, we have

f(U V) = f(U) f(V)

since f is one-to-one. So f(T) satisfies the axiom about intersections. Fi-nally, since f is onto, f(X) = Y and Y f(T). So (Y, f(T)) is a space.

The map f is continuous since f1(f(U)) = U for any U in T. ForU T, the pullback by f1 is f(U), which is clearly in f(T). So f is ahomeomorphism.

When we study a topological space, we are not usually interested inspecific points, or in the names for specific points. The map f is giving anew name to each point of X, but is not otherwise changing the space.

In fact, there is a converse to the observation the relabeling is a homeo-morphism:

Proposition 3.6. Let f: (X, T) (Y, S) be a homeomorphism. ThenS = f(T). In particular, (Y, S) is equal to the relabeling of (X, T) by f.

Proof. Suppose U S. Then since f is continuous, f1(U) T. ThenU = f(f1(U)) is in f(T).

Now suppose U f(T). Then U = f(V) for V T. Then since f

1

is continuous with respect to S, f(V) (the pullback by f1(V)) is in S. SoS = f(T).

To show that two spaces are homeomorphic, usually one has to constructa homeomorphism between them. To show that spaces are not homeomor-phic, one has to show that the spaces differ with respect to some property

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that is preserved under homeomorphism. Since a homeomorphism is a rela-

beling of the points in a space, any property that is defined using only thetopological structure of a space will be preserved under homeomorphism.Properties not preserved under homeomorphisms are not part of the topo-logical structure of a space, but come from some kind of additional structurethat we may have imposed separately: geometric structure, algebraic struc-ture, order structure, names of special points.

Simple examples of topological properties of spaces include:

Existence of finite closed sets.

Existence of infinite closed sets.

Existence of finite open sets.

Much of this course will be concerned with defining topological prop-erties, proving relations between them, and showing that certain examplesexhibit or fail to exhibit these properties.

4 Closures, limit points and boundaries

4.1 Closed sets

Recall that a subset A of a space X is closed if and only if its complementX\A is open.

There is a convenient characterization of the set of closed subsets of atopological space.

Proposition 4.1. The closed sets of a topological space (X, T) satisfy thefollowing axioms:

1. Arbitrary intersections of closed sets are closed.

2. Finite unions of closed sets are closed.

3. The set X and the empty set are both closed.

Conversely, if C is a set of subsets of X satisfying these axioms, then C is

the set of closed sets for some topology on X.

The proof is an exercise using DeMorgans laws on set operations.Closed sets can be used to detect continuity.

Proposition 4.2. A function between spaces f: X Y is continuous ifand only if for every closed A Y, the preimage f1(A) X is closed.

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Proof. The preimages of complementary sets are complementary: if Y =

A U, then X = f1(A) f1(U). So f pulls back closed sets to closedsets if and only if f pulls back open sets to open sets.

Since infinite intersections of closed sets are closed, it is useful to makethe following definition.

Definition 4.3. Let X be a topological space and S a subset of X. Theclosure of S, denoted S, is the intersection of all closed sets containing S.

Example 4.4. In the standard topology on R, the closure of a boundedinterval is the closed interval with the same endpoints.

Some near-trivial observations about closed sets:

Proposition 4.5. Let X be a space and S any subset of X.

The closure S is closed.

S is closed if and only if S = S.

The closure of the closure of S (write it S) is equal to S.

Definition 4.6. A subset S of a space X is dense if its closure is the wholespace: S = X.

Example 4.7. In the standard topology on R, the rationals and the irra-tionals are both dense.

Existence of proper, dense subsets of different sizes (finite or infinite) isan example of a topological property.

4.2 Sequences and limit points

From calculus and analysis, we are used to probing the topological proper-ties ofR using sequences. For completeness, we recall the definition of asequence.

Definition 4.8. A sequence of points in a space X is simply a functionfrom the natural numbers N to X. By convention, the function inputs aredenoted as subscripts, so that we write a sequence as {pi}iN for pi X.

Naturally we must redefine convergence in a purely topological context.

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Definition 4.9. A sequence {pi}iN in a space X converges to a point p X

if for each open neighborhood U of p, there is a number N N (dependingon U) such that for all i > N, we have pi U.

A linguistic convenience: we say a set U contains a tail of the sequence{pi}iN if there is some N N with all pi in U for i > N. So a sequenceconverges to a point if every open neighborhood of the point contains a tailof the sequence.

These definitions make the following important fact nearly a trivialty.

Proposition 4.10. Suppose f: X Y is continuous and {pi}iN is asequence in X converging to p X. Then {f(pi)} converges to f(p).

Proof. Let U be an open neighborhood of f(p). Then f1(U) is an openneighborhood of p. Then f1(U) contains a tail of the sequence {pi}iN.Then U contains a tail of the sequence {f(pi)}iN.

The topological space structure lets us tell when a point is near a set;such a point is called a limit point.

Definition 4.11. Let X be a topological space and let S be a subset ofX. A point p in X is a limit point of S if every open neighborhood U of pcontains a point of S other than p:

U S\{p} = .

Example 4.12. In R with the usual topology, the limit points of an intervalare the endpoints of the interval. Note that it does not matter whether theseendpoints are contained in the interval or not.

Proposition 4.13. For any set S in any topological space X, the closureof S is equal to the union of S with all the limit points of S.

Proof. For a point p X, p is not in S if and only if there is a closed set Awith S A and p / A. There is such an A if and only if there is an openneighborhood U of p with U S = , since we can take U = X\A and viceversa. There is such a U if and only if p is not in S and is not a limit pointof S.

Sequences cannot escape the closure of a set:

Proposition 4.14. Suppose S is a subset of a space X and {pi}iN is asequence of points in S converging to a point q X. Then q is in S.

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Proof. If{pi}iN is eventually constant, then q is equal to pi for large enough

i and q S. If {pi}iN is not eventually constant, infinitely many of the piare not equal to q. Let U be an open neighborhood of q. Then U containsa tail of {pi}iN. Then U contains some point in S (in fact, infinitely manypoints in S), and q is a limit point of S.

Later we will see that the converse to this theorem is not true in allspaces: there are examples of a limit point q of a set S in a space X, butwhere no sequence of elements in S converges to q.

Corollary 4.15. If A is a closed subset of a space X and a sequence ofpoints in A converges to a point q X, then q A.

This is immediate since A = A.

4.3 Topologically derived sets

In addition to the closure and the set of limit points (sometimes called thederived set), there are several ways of using the topological structure toderive new subsets of a space from old ones.

Definition 4.16. Let X be a topological space and S a subset of X.

The interior of S, denoted S, is the union of all open subsets of S.

The exterior of S is the union of a open subsets that are disjoint from

S.

The boundary ofS, denoted S, is the set of points whose every neigh-borhood contains elements of S and elements not in S:

S =

{p X| for all open U X with p U, U S = and U\S = .}

Proposition 4.17. For any subset S of any space X, the exterior, interior,and boundary of S are disjoint, and their union is X.

Proof. If p S, then some open neighborhood of p is a subset of S. So

clearly p is not in S or in the exterior of S. Similarly, ifp is in the exteriorof S, then some open neighborhood of p is disjoint from S, and p cannot bein S.

Now suppose p X and p is not in S. Then some open neighborhoodof p is either disjoint from S or a subset of S. In the first case, p is in theexterior of S, and in the second case, it is in S.

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Proposition 4.18. For any subset S of any space X, we have S = S S

and S = S\S.

Proof. A point p is in the exterior of S if and only if there is an open set Uwith p U and U S = . This is true if and only if there is a closed setA = X\U with p / A and S A. This is true if and only if p is not in S.So therefore S = S S, the complement of the exterior of S.

Then S\S is S.

Proposition 4.19. The following are true:

A subset S of X is open if and only if S = S.

A subset S of X is closed if and only if S S.

A subset S of X is open if and only if S S = .

The proofs are straightforward.There is a rich interplay between these set operations. In fact, just using

the operations of set complementation and set closure, one can derive asmany as 14 different sets from a subset of a space. We will explore this inthe second homework assignment.

4.4 Comparing topologies on the same point set

Definition 4.20. Fix a set X and let S, T be two different topologies onX. Then (X, S) is coarser than (X, T) if S T. In the same situation,(X, T) is finer than (X, S).

To restate: for (X, T) to be finer than (X, S), we need that every openset in S is also open in T, but possibly some other sets are open as well.

The relation coarser is reflexive, transitive relation on the set of alltopologies on a point set. The discrete topology is the finest topology ona set, and the indiscrete topology is the coarsest topology on a set. Thediscrete and indiscrete topologies can both be compared to any topology onthe set. As the following example shows, not all topologies are comparable.

Example 4.21. Let X be a set and let p X. The particular point topologywith respect to p is the topology on X where a set is open if and only ifit contains p. Let p = q X and let T(p) and T(q) be the respectiveparticular point topologies. Then T(p) T(q) and T(q) T(p); neithertopology is coarser than the other.

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Many topologies on a space can be characterized as the coarsest topology

with a particular property.

Example 4.22. Let X be a space and let S X. Then the subspacetopology is the coarsest topology for which the inclusion S X iscontinuous.

The topology of finite complements is the coarsest topology in whichpoints are closed.

The topology generated by a given subbasis B is the coarsest topologyin which every set in B is open.

5 Topologies from other structure

5.1 Order topologies

The standard topology on R is a special case of the following general con-struction:

Definition 5.1. A strict total ordering on a set X is a relation < satisfying:

for all x, y X with x = y, we have x < y or y < x,

for all x,y ,z X with x < y and y < z, we have x < z, and

for all x X, we have x < x.

A set with a strict total ordering is an ordered set.Just like in the real line, we have intervals in any ordered set. Define

(x, y) = {z X|x < z < y}, and define closed and half-closed intervals to bethe union of the open interval with the indicated endpoints. We also havethe intervals (, x) and (x, ) (these are sometimes called open rays).

If X has a strict total ordering


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5.2 Countable and uncountable sets

Recall that a set is countable if it is finite or can be put in bijection with N.A set is uncountable if it is infinite and cannot be put in bijection with N.

Intuitively, countability of sets relates to the amount of informationneeded to specify a member of a set. If there is a universal bound on thenumber of bits needed to specify a member of the set S, then S is finite.If there is no universal bound on the number of bits needed to specify amember of the set S, but every member of S can be specified with a finitenumber of bits, then S is countable. If most members of the set S requirean infinite number of bits to specify, then S is uncountable. (The precedingremarks are not so much rigorous results as they are intuitive guidelines.)

We put together some results from set theory and analysis:Proposition 5.2. Subsets and quotients of countable sets are count-

able.

Unions of countable sets, indexed over a countable set, are countable.

The Cartesian product of a family of finitely many countable sets iscountable.

Power sets of infinite sets are uncountable.

The rational numbers Q are countable, but the real numbers R are

uncountable.

5.3 A pathological example

Definition 5.3. A well ordering on a set X is a strict total order suchthat every nonempty subset of X contains a minimal element. An initialsegment of a well ordered set is a subset that contains all the predecessorsof any element it contains.

The following theorem in set theory depends on the axiom of choice. Wequote it without proof. Halmoss Naive Set Theory is an excellent book andcontains this theorem.

Theorem 5.4 (Well-ordering theorem). Every set is the domain of a wellordering.

From this can deduce the following:

Proposition 5.5. There is an uncountable, well ordered set (X,


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Proof. Suppose (Y,


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Proposition 5.11. The set of all open metric balls in (X, d) form a basis

for a topology on X.

Proof. Let x, y X and let r, s > 0. IfBr(x)Bs(y) = , there is nothing toshow. So consider any z Br(x) Bs(y). Of course, this means d(x, z) < rand d(y, z) < s. Select t > 0 such that:

t < min{r d(x, z), s d(y, z)}.

Suppose that w Bt(z). Then

d(w, x) d(w, z) + d(z, x) < t + d(x, z) < r d(x, z) + d(x, z) = r,

by the triangle inequality and the choice oft. So w is in Br(x). Similarly, anyw Bt(z) is also in Bs(y), and therefore Bt(z) Br(x) Bs(y). So for anyz Br(x) Bs(y), there is a t(z) > 0 such that Bt(z)(z) Br(x) Bs(y).In particular, Br(x) Bs(y) is a neighborhood of each z it contains. SoBr(x) Bs(y) is open.

Definition 5.12. For any metric space (X, d), the topology with basis givenby the set of all open metric balls is the metric topology on (X, d).

The metric topology on R is the same as the order topology. Note thatwe have finally defined a reasonable topology on Rn for n > 1.

Example 5.13. Any set X is a metric space with the distance functiond(x, y) = 1. This metric space topology is the discrete topology.

5.5 The topology on a finite product

The following is the standard way to use the topology on a pair of spaces toput a topology on the Cartesian product of their point sets.

Definition 5.14. Let (X, T) and (Y, S) be spaces. The product topology onX Y is the one with basis

{U V|U T, V U}.

Inductively we can define a topology on Rn by Rn = Rn1 R, where Rhas the standard topology. Luckily, this topology turns out to be the sameas the metric topology. This will probably be a homework exercise.

Proposition 5.15. Some observations:

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For any y Y, the inclusion X X Y by x (x, y) is continuous.

Similarly inclusions from Y to X Y are continuous.

There are canonical coordinate projections p1 : XY X (by(x, y) y) and p2 : X Y Y; these are continuous.

The product topology on X Y is the coarsest topology in which thecoordinate projections are continuous.

The product topology is characterized by the following property: Forany space Z and any continuous mapsf: Z X andg : Z Y, thereis a unique continuous map fg : Z XY such thatp1(fg) = fand p2 (f g) = g.

5.6 Countability properties

As we saw in the example of an uncountable well ordering, there are un-countable spaces that behave very badly. On the other hand, there are un-countable spaces, like R with the standard topology, in which we still havea lot of control. The following definitions help distinguish these situations.

Definition 5.16. Let X be a space.

X is separable if X contains a countable dense subset.

X is second countable if X has a countable basis. X is first countable if for each p X, there is a countable neigh-

borhood basis: a set Np of open neighborhoods of p, such that everyneighborhood of p contains some member of Np.

Proposition 5.17. Any second countable space is both first countable andseparable.

Proof. Let B be a countable basis for X. For each p X, the set {U B|p U} is a countable neighborhood basis. Let S consist of one point fromeach member of B. Then S is a countable dense subset of X.

Proposition 5.18. Any subspace of a first countable space is first countable,and any subspace of a second countable space is second countable.

Proof idea: We find our new basis or neighborhood bases by intersecting theold ones with the subset.

Proposition 5.19. Finite products of separable spaces are separable.

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Proof idea: The product of the countable dense subsets is a countable dense

subset.

Slightly less trivial:

Proposition 5.20. Any separable metric space is second countable.

Proof. Let X be a metric space with metric d and countable dense subset S.We will show that the set Bof open balls with rational radii around points inS form a countable basis for X. Consider the basic open set Br(x), for somex X and r > 0. Let y Br(x). There is an s > 0 such that Bs(y) Br(x).Let z S B s

3(y) and let t ( s3 ,

2s3 ) Q. Then y Bt(z), and it follows

from the triangle inequality that Bt(z) Bs(y) Br(x). So we have shown

that every point in Br(x) is inside a set in B that is contained in Br(x).Therefore Br(x) is a union of members of B, and therefore B is a basis.Since B is in one-to-one correspondance with the product S ((0, ) Q)of countable sets, B is countable.

Proposition 5.21. If X is first countable, then if x X is a limit point ofa set S X, then there is a sequence in S whose limit is x.

Proof. Let {Ui}iN be a countable neighborhood basis for x. We can assumethat Un+1 Un: if not, we simply replace Un with the finite intersectionn

i=0 Ui. Define a sequence by letting pn be any point in Un S\{x} (thisis nonempty since x is a limit point of S). Since the Ui are nested, thissequence converges to x.

Example 5.22. Countability properties of spaces we have already seen:

Rn with the metric topology is second countable.

The order topology on an uncountable well ordering is not separableor first countable.

A discrete topology is second countable if and only if it is separable ifand only if the point-set is countable. Any discrete topology is triviallyfirst countable. Note that since discrete spaces are metric spaces, there

are metric spaces that are not separable.

Any indiscrete topology is second countable.

An uncountable particular point topology X with particular point phas an uncountable (closed) discrete subspace X\{p}. Therefore X isnot second countable. However, X is separable (with countable dense

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subset {p}) even though X\{p} is not. This X is first countable: {p}

is a neighborhood basis for p and {p, q} is a neighborhood basis forq = p.

Example 5.23. Less trivially: Let be the set of bounded sequences inR, with metric

d({fi}, {gi}) = supi

|fi gi|.

Give the metric topology. Then not separable. Proof: Let S be acountable set and let {(fj)i} be an enumeration ofS, where (fj)1, . . . , (fj)n, . . .is one of the sequences in S. Define a sequence {gi} by:

gi = 0 if |fi| > 1

2 if |fi| 1

Then for each j, we have d({gi}, {(fj)i}) 1, so that the open ball of radius1/2 around gi does not contain any of element of S. The S =

.

Zariski topologies

Definition 5.24. Let F be a field and let n N, n 1. Let F[x1, . . . , xn]be the ring of polynomials over F in n variables. Let X = Fn. The Zariskitopology on X is the topology where the closed sets are exactly the nullsetsof sets of polynomials : A X is closed iff there is S F[x1, . . . , xn] with

A = {(a1, . . . , an) X|f(a1, . . . , an) = 0 for all f S}.

Further, if S is a subset ofX, then the subspace topology on S from theZariski topology is usually also referred to as the Zariski topology. Example:The set of invertible n n matrices over F is a subset of Fn

2

and inheritsa Zariski topology from it.

Proposition 5.25. The Zariski topology is a topology.

Proof. Suppose Ai is the closed set corresponding to the set of polynomialsSi, for i in some index set I. Then A1 A2 is the nullset of the set ofall pairwise products of polynomials in S1 with polynomials in S2. The

intersection

iIAi is the nullset of

iISi.

Proposition 5.26. Some observations about the Zariski topology:

If F is finite then the Zariski topology is discrete.

Ifn = 1 then the Zariski topology is the topology of finite complements.

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Proposition 5.27. If F isR then every nonempty open set in the Zariski

topology is dense.

Proof. Suppose U is a nonempty open set. Let A = X\U. If U = X, thenA = . Then there is nonempty open V A. Let B = X\V. Then Aand B are closed proper subsets of X and A B = X. Then A and Bcorrespond to two sets of polynomials S and T in R[x1, . . . , xn]. Since Aand B are proper, there are f S and g T that are nonzero. Then theproduct f g is a nonzero polynomial. Taking limits of x1, . . . , xn as they goto infinity, the value of f g(x1, . . . , xn) will go to . So there is a pointwhere f g = 0, so A B = X.

Example 5.28. There are subsets ofRn that are closed in the standardtopology but are dense in the Zariski topology. For example, Zn is such aset. However, Z is not Zariski dense in R2. (What is its closure?)

Often when studying structures defined over R or C (such as groups ofmatrices), it is convenient to work simultaneously with the standard topol-ogy and with the Zariski topology. Therefore one often encounters the termZariski dense in mathematical literature; of course, it simply means densein the Zariski topology.

Next time, well talk about separation properties. I briefly mention twoof them:

Definition.A space is T1 if points are closed.

A space X is Hausdorff if for x, y X, x = y, there are disjoint open setsU, V with x U and y V.

The Zariski topology on Rn is an important example of a topologicalspace that is T1 but not Hausdorff.

Separation properties

One measure of the quality of a space is how well the open sets separate

subsets of the space from each other. A series of properties called separationproperties serve this purpose.

Definition 5.29. Two subsets S, T of a space X are separated if S T =S T = .

Definition 5.30. Let X be a space.

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X is T0 if for every pair x, y X, x = y, there is an open set U in X

with exactly one of x, y in U.

X is T1 if for every pair x, y X, x = y, there are open sets U, V withx U, y / U, y V, y / V.

X is T2 if for every pair x, y X, x = y, there are disjoint open setsU, V with x U, y V.

X is T3 if for every closed set A and every x X\A, there are disjointopen sets U, V with A U and x V.

X is T4 if for every pair of disjoint closed sets A and B, there are

disjoint open sets U, V with A U and B V. X is T5 if for every pair of separated sets S and T, there are disjoint

open sets U, V with S U and T V.

X is regular if X is T0 and T3.

X is normal if X is T1 and T4.

X is completely normal if X is T1 and T5.

Hausdorff is a synonym for T2.This is a lot of definitions... I will expect students to know the definitions

of Hausdorff, T1, normal and completely normal, but not the others.The next proposition collects some near-trivial statements.

Proposition 5.31. X is T1 if and only if points are closed.

Hausdorff implies T1 and T1 implies T0.

T5 implies T4, and completely normal implies normal.

Normal implies T3 and Hausdorff.

So completely normal implies any of the other separation properties.

Proposition 5.32. Metric spaces are completely normal.Proof. Let S and T be separated subsets of a metric space (X, d). Then nox S is a limit point of T, so for each x S we can find a real numberr(x) such that Br(x)(x) is disjoint from T. Of course we can also find a realnumber s(y) for each y T such that Bs(y)(y) is disjoint from S. Thenby the triangle inequality,

xA Br(x)/2(x) and

yB Br(y)/2(y) are disjoint

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open sets containing S and T respectively. This proves that metric spaces

are T5.Ifx and y are points in X, then the open balls of radius d(x, y)/2 around

each are disjoint open neighborhoods, showing X is Hausdorff and thereforeT1.

Example 5.33. Since they are metric spaces, Rn with the standardtopology, and any discrete space, will satisfy all separation properties.

Indiscrete spaces are not T0, T1, or Hausdorff, but are vacuously T3,T4, and T5.

The Zariski topology on Rn is T1 but not Hausdorff.

Particular point topologies are T0 but not T1.

Let Y be the indiscrete topology on two points. For any space X, theproduct X Y will not be T0, T1 or Hausdorff, but will be Ti if andonly if X is for i = 3, 4, 5.

Recall that a topological property is hereditary if it is preserved whentaking subspaces.

Proposition 5.34. The following properties are hereditary: being T0, beingT1, being Hausdorff, being T5 (and therefore being completely normal). A

closed subspace of a normal space is normal.

No proof will be given. It is tricky to give an example of a normal spacewith a non-normal subspace, but we may do it later in the quarter.

The following is perhaps the first nontrivial theorem in point-set topol-ogy.

Theorem 5.35 (Urysohns lemma). A space X is T4 if and only if foreach pair of disjoint closed sets A and B, there is a continuous functionf: X [0, 1] R (with the standard topology, such that f(x) = 0 for allx A, and f(y) = 1 for all y B.

A dyadic rational number is a rational number whose denominator (inreduced form) is a power of two.

Proof. The if direction is easy and is omitted.Let S be the set of dyadic rationals in (0, 1). We will start by inductively

defining two families of open sets U(r) and V(r) for all r S, satisfying

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U(r) V(r) = for all r S,

B U(r) U(r) X\A for all r S, and

for r < r S, we have (X\U(r)) (X\V(r)) = .

Since X is T4, we can select U(1/2), V(1/2) disjoint open sets with A V(1/2) and B U(1/2).

Now let n > 0 and assume we have defined all U(r), V(r) for r S withdenominator 2n or less. Let k N with 0 k < 2n, and set s = k/2n,t = (k + 1)/2n, and r = (s + t)/2 = (2k + 1)/2n+1. If s = 0, we let U(r)and V(r) be disjoint open sets such that A V(r) and X\V(t) U(r). Ift = 1, we let U(r) and V(r) be disjoint open sets such that X\U(s) V(r)

and B U(r). Otherwise, we let U(r) and V(r) be disjoint open sets suchthat X\U(s) V(r) and X\V(t) U(r). This is possible since X is T4,and having made these definitions all sets defined will satisfy the conditions.

Having defined these sets, we now define our function f: X [0, 1].Note that the conditions imply that for r r, we have U(r) U(r). Setf(x) = sup{0} {r S|x U(r)}. Let (a, b) be an open subinterval of (0, 1)and let x f1((a, b)). There are a, b S with f(x) (a, b) (a, b) anda < a (since S is dense in [0, 1]). Then f(U(a)\U(b)) (a, b), U(a)\U(b)is open and contains x. (If y U(a) then f(y) a; if y / U(b) theny / U(b) and y b). In particular, every point in f1((a, b)) is containedin an open subset of f1((a, b)), so f1((a, b)) is open. Similarly, if x

f1

((a, 1]), then there is a

S with f(x) (a

, 1] (a, 1] and a < a

,and x U(a) f1((a, 1]). And of course, if x f1([0, b)), then thereis b S with f(x) [0, b) [0, b), and x X\U(b) f1([0, b)). Thisproves f is continuous. Since B U(r) for all r, we have f(B) {1}; sinceA X\U(r) for all r, we have f(A) {0}.

From Urysohns lemma, we can deduce that for normal spaces, continu-ous functions to [0, 1] separate points.

Proposition 5.36. In a normal space X, for any two points x, y X,x = y, we can find continuous f: X [0, 1] with f(x) = f(y).

Urysohns lemma makes it possible to find a function on a space that

roughly approximates a function on a subset.Lemma 5.37. Let X be normal and let A X be closed. Suppose f: A [1, 1] is continuous. Then there is a continuous functiong : X [1/3, 1/3]such that

supxA

|f(x) g(x)| 2

3.

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Proof. Let B = f1([1/3, 1]), and let C = f1([1, 1/3]). By Urysohns

lemma, there is a function on X that is 1 on B and 0 on C; by rescalingthis function we get the desired function g.

This lemma is key to the following theorem:

Theorem 5.38 (Tietze extension theorem). A space X is T4 if and only iffor every closed subset A and every continuous function f: A R, there isa function g : X R extending f, meaning g|A = f.

Proof sketch. A rigorous proof of this theorem requires quoting some basicresults from analysis, so we give a sketch instead. The if direction iseasy and is omitted. For the only if direction, we compose f with a

homeomorphism R (1, 1) to get a bounded function. Assuming f isbounded, we approximate f by a function f1 defined on X, using the lemma.The difference is presumably nonzero, so we approximate the difference byanother function f2 and then f1 + f2 is a function on X that restricts toa better approximation of f. Repeating this process infinitely many times,the error of the approximations goes to zero exponentially fast, and theapproximations converge to the desired function g (this uses the fact fromanalysis that uniform limits of continuous functions are continuous).

More on metric spaces

Some last comments about separation and countability properties as theypertain to metric spaces. As we mentioned before, metric spaces are sepa-rable if and only if they are second countable. However, taking the balls ofrational radius around each point gives us the following:

Proposition 5.39. Every metric space is first countable.

This means that we can characterize limit points and closures (and there-fore closed sets) in metric spaces in terms of sequences. Further, since metricspaces are Hausdorff, we can use sequences to uniquely specify points:

Proposition 5.40. In a Hausdorff space, any sequence converges to at most

one point.

The proof is an exercise using the definition. In particular, it follows fromthese remarks that any point in a separable metric space can be specifieduniquely by a sequence of points in a fixed countable dense subset (like Qin R).

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6 Connectedness properties

The disjoint union

Definition 6.1. The disjoint union of spaces (X, T), (Y, S) is denoted XY; the point set is the disjoint union XY of the pointsets, and the topologyhas the basis

{U V|U T, V S},

where U and V are considered as subsets of the disjoint copies of X and Yin X Y.

Proposition 6.2. Some observations:

There are canonical inclusions i1 : X X Y and i2 : Y X Y;these are continuous.

The topology on X Y is the finest one that makes the canonicalinclusions continuous.

The disjoint union X Y is characterized by the following property:For any space Z and any pair of continuous maps f: X Z andg : Y Z, there is a unique continuous function f g : X Y Zthat with (f g) i1 = f and (f g) i2 = g.

Connectedness properties

Definition 6.3. A separation of a space X is a pair of disjoint open nonemptysubsets U, V with U V = X.

A space is connected ifX has no separation. Otherwise it is disconnected.A subset S of a space X is connected if it is when considered as a subspace.

Proposition 6.4. R with the standard topology is connected.

Proof. IfR = U V were a separation, then without loss of generality therewould be some x R\U with U (, x) = . Then by replacing U withU (, x) and V with V (U (x, ), we may assume that U is boundedabove. Let y be the least upper bound for U. Certainly y is not in U (if so,

an open interval around y is contained in U and y is not an upper boundfor U). If y V, then an open interval around y is contained in V, andV U = . Then v / U V, a contradiction.

Example 6.5. Any disjoint union of nonempty spaces is disconnected.If X is a space and S, T X are nonempty separated sets, then the

subspace topology on S T is disconnected.

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Note that the only properties ofR that we used in proving connectedness

were the least upper bound property, and the fact that an open intervalaround any point p contains points strictly less than and strictly greaterthan p. An ordered set with these properties is called a linear continuum,and it follows that all linear continua are connected in the order topology.

Proposition 6.6. If S X is connected and T X with S T S, thenT is connected.

Proof. Suppose T = U V is a separation of T. Any point in T is in Sor a limit point of S. If x U is a limit point of S, then since U is open,US = . Similarly, VS = . So S = (US)(VS) is a separation.

Proposition 6.7. The image of a connected space under a continuous func-tion is connected.

Proof. Suppose f: X Y is a continuous function. Iff(X) is disconnectedwith a separation f(X) = UV, then X = f1(U)f1(V) is a separation.

Example 6.8. Since any open interval is homeomorphic to R, any openinterval is connected. Since any interval is a subset of the closure of anopen interval, any interval is connected. Alternatively, we can see that anyinterval is connected because any interval is a continuous image of R.

Since intervals are connected, we can test for connectedness using mapsof intervals.

Definition 6.9. In a space X with x, y X, a path from x to y is acontinuous map f: [0, 1] X with f(0) = x and f(1) = y. A path from xto y is an arc if it is one-to-one.

A space is path-connected if any two points can be joined by a path. Aspace is arc-connected if any two distinct points can be joined by an arc.

No proof is given for the following.

Proposition 6.10. The image of a path-connected space under a continuous

map is path-connected, and the image of an arc-connected space under aninjective, continuous map is arc-connected.

Proposition 6.11. Arc-connected implies path-connected and path-connectedimplies connected.

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Proof. Every arc is a path, so arc-connected implies path-connected.

Suppose X is a space and X = U V is a separation. Let is a pathfrom a point in U to a point in V and let A = ([0, 1]). Then A is the imageof a connected space and is connected. But A = (A U) (A V) is aseparation; this is a contradiction.

Example 6.12. The indiscrete topology on a countable set is path-connectedbut not arc-connected.

The topologists sine curve is connected but not path-connected. Let Sbe the graph of f: (0, 1] R, by f(x) = sin(1/x), a subset ofR2. Thetopologists sine curve is the closure of S, which is S 0 [1, 1]. S is acontinuous image of a path-connected space and is therefore connected, and

S is therefore connected. However, there is no path from a point in S to apoint in {0}[1, 1]. Suppose were such a path. Without loss of generality,([0, 1)) S. Since the projection of to the x-axis is continuous, hits ally-values in [1, 1] on a sufficiently small interval (a, 1]. Let s [1, 1] with(0, s) = (1). We can find a sequence {pi}i in [0, 1] with {(pi)}i convergingto (0, s). Then is not continuous, a contradiction.

Types of components

Definition 6.13. Let X be a space. Define a relation c on X by a c b ifthere is some connected S X with a, b S. The equivalence classes of X

under c are the connected components (or just components) of X.Define a relation q on X by a q b if a and b are on the same side ofevery separation: for every separation X = U V with a U, we have balso in U. The equivalence classes of X under q are the quasicomponentsof X.

The existence of paths from a to b defines an equivalence relation; thepath components of X are the equivalence classes under this relation.

Theres a problem with defining arc components, specifically, the ex-istence of an arc between two points is not in general a transitive relation.

Example 6.14. Let Y be the two-point space 0, 1 with the indiscrete topol-

ogy. Let X0 = Y [0, ), and let X = {(1, 0)} {0} [0, ). Then Xis [0, ) with the standard topology, but with 0 doubled. There is an arcfrom (0, 1) to any point in X, but no arc from (1, 0) to (0, 0).

Then X is path connected but not arc connected, and the notion of arccomponent doesnt make sense for X.

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Proposition 6.15. Let X be a space and x X. Then the path compo-

nent of x is a subset of the connected component of x, and the connectedcomponent of x is a subset of the quasicomponent of x.

Proof. If there is a path from x to y, then the image of this path is aconnected subspace of X containing x and y. This shows the first inclusion.If S is a connected subspace of X containing x and y, then any separationX = U V will have to have S on one side, say S U. Otherwise theseparation will induce a separation of S. So x and y are both in U.

Proposition 6.16. IfX has only one quasicomponent, thenX is connected.

The proof is easy and is omitted.

Proposition 6.17. The quasicomponent of x is the intersection of all setscontaining x that are both closed and open.

Proof. Suppose y is in the quasicomponent of x. Then for each separationX = U V with x U, we have y U. If U is closed and open with x U,then either U = X or X = U (X\U) is a separation. In either case, y U,so y is in the intersection of all such sets.

Suppose y is not in the quasicomponent of x. Then there is some sepa-ration X = U V with x U and y V. Then U is closed and open andy / U, so y is not in the intersection of all such sets.

Example 6.18. For n N, n > 0, let Rn be the rectangle in R2 withcorners (1 1/n,n), (1 1/n, n), (1 + 1/n,n) and (1 + 1/n, n), letL1 = {1} R and L2 = {1} R. Let S = L1 L2

n Rn. Let S = U V

be a separation of S. Suppose L1 U. Then for large n, Rn U, andtherefore L2 U. However, any given Rn is can be separated from L1 L2.So L1 L2 is a quasicomponent of S. But clearly, L1 and L2 are differentconnected components of S.

Proposition 6.19. Quasicomponents are closed and connected componentsare closed.

The proofs are easy and are omitted.Example 6.20. The path components of the topologists sine curve S aboveare S and {0} [1, 1], but S is connected. Note that S is not closed; sogenerally, path components are not necessarily closed.

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Local connectedness properties and disconnectedness

Definition 6.21. A space X is locally connected if for every point x X andevery open neighborhood U of x, there is a connected open neighborhoodV of x with V U.

A space X is locally path connected if for every point x X and everyopen neighborhood U of x, there is a path-connected open neighborhood Vof x with V U.

The following needs no proof.

Proposition 6.22. A locally path connected space is locally connected.

Example 6.23. The topologists sine curve S is connected but not locallypath connected: every sufficiently small neighborhood of (0, 0) has infinitelymany path components.

You should be able to find an example of a locally connected space thatis not connected.

Lemma 6.24. In a space X, suppose S and T are connected sets withnonempty intersection. Then S T is connected. Similarly, if S and T arepath connected sets with nonempty intersection then ST is path connected.

Proof is easy and is omitted.

Proposition 6.25. If X is locally connected, then each connected compo-nent of X is open; therefore the quasicomponents of X coincide with theconnected components.

Proof. If S is a connected component of X and S is not open, then there isa point x S that is a limit point of X\S. If X is locally connected, thenX has a connected open neighborhood U. Since S and U are connected andS U is nonempty, S U is connected, and therefore U S. but then x isnot a limit point of X\S. So S is open.

Then S is open and closed, but since S is connected no proper subset ofS is both open and closed. So S is a quasicomponent.

Proposition 6.26. IfX is locally path connected, then the quasicomponentsof X coincide with the path components.

Proof. Suppose S is a path component of X. Suppose x is in the closure ofS. Since X is locally path connected, x has a path connected neighborhoodU. Then S U is nonempty, and therefore S U is path connected, andU S. So S S and S is closed. Further, such an x cannot be a limit

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point of X\S, and therefore S is open. Since S is connected and open and

closed, S is a quasicomponent of X.

Definition 6.27. A space is totally disconnected if its connected compo-nents are all points.

Example 6.28. A space is both totally disconnected and locally connectedif and only if it is discrete.

The subspaces Q and R\Q ofR with the standard topology are totallydisconnected.

Any subspace of a totally disconnected space is totally disconnected.

7 QuotientsIf f: X S is a surjective function from a space X to a set S, then we canuse f to define a topology on S.

Definition 7.1. Let X be a space, S a set, and f: X S a surjectivefunction. The quotient topology on S (with respect to f: X S) is thetopology where the open sets of S are exactly the sets U S such that thepreimage f1(U) is open.

It is routine to verify that this is a topology. Note that ifV X is open,this definition does not imply that f(V) is open.

Example 7.2. Let X = [0, 1] with the standard topology and let S = [0, 1).Define f: X S by f(x) = x for x [0, 1) and f(1) = 0. The quotienttopology on S has a basis of sets of the form (a, b) [0, 1) and [0, b)(a, 1) [0, 1]. This topology makes S homeomorphic to the unit circle in R2 in thestandard topology.

Since we are defining a topology by collapsing together points of thespace, to define the topology as generally as possible we want lose as littleinformation as possible when forming the quotient. Losing as little informa-tion as possible corresponds to taking as fine a topology as possible.

Proposition 7.3. The quotient topology on S with respect to f: X S is

the finest topology on S such that f: X S is continuous.Proof. It is immediate from the definition that f is continuous with respectto the quotient topology. Now let S be another topology on S that makes fcontinuous. Suppose U S. Then f1(U) is open in X, and by definitionU is also open in the quotient topology. So S is coarser than the quotienttopology.

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For every quotient space S with respect to f: X S as above, the

function f is a continuous surjection with respect to the quotient topol-ogy. However, not every continuous surjection f: X Y of spaces can berealized this way.

Definition 7.4. Let X and Y be spaces. A quotient map f: X Y is acontinuous surjection such that the topology on Y is equal to the quotienttopology on Y with respect to the set map f: X Y. Equivalently, acontinuous surjection f: X Y is a quotient map if and only if the topologyon Y is the finest topology making f continuous.

Example 7.5. Let f: X Y be a bijection and let X have the discretetopology and let Y have the indiscrete topology. Then f is continuous butnot a quotient map.

There are a couple of easy criteria to recognize quotient maps.

Definition 7.6. Let X and Y be spaces. A function f: X Y is open iffor every open U X, we have f(U) open in Y. Similarly, f is closed if forevery closed A X, we have f(A) closed in Y.

Note that a continuous bijection is a homeomorphism if and only if it isclosed if and only if it is open.

Proposition 7.7. Every open continuous surjection is a quotient map, and

every closed continuous surjection is a quotient map.

Proof. Suppose f: X Y is an open continuous surjection. Suppose U Y. Because f is continuous, f1(U) is open if U is. Because f is open,U = f(f1(U)) is open iff1(U) is. Then the topology on Y is the quotienttopology, and f is a quotient map.

If f is closed and U Y, then U = Y\f(X\f1(U)) is open if U is. Sofor the same reasons as above, a closed continuous surjection is a quotientmap.

The real utility of quotient spaces is that we can define surjective set

maps that identify any sets of points that we wish. The quotient space isthen the closest space to the original space in which these sets of points areidentified. Identifying sets of points corresponds to the intuitive notions ofgluing and collapsing.

Definition 7.8. Let X be a space and let be an equivalence relation onS. The quotient of X with respect to is the quotient (in the sense of

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functions) with respect to the surjective function X X/, where X/ is

the set of -equivalence classes. Often the quotient space is also denotedX/.

Example 7.9. Consider the interval [0, 1] in the subspace topology. We geta space homeomorphic to the circle (from the previous example) by takingthe quotient with respect to the equivalence relation with x y if and onlyif x = y or {x, y} = {0, 1}.

Perhaps less trivially, the quotient ofR by the equivalence relation x x + k for all k Z is also homeomorphic to the circle.

Example 7.10. Let X = [0, 1] [0, 1] in the standard topology. Definean equivalence relation by (a, b) (c, d) if and only if (a, b) = (c, d) or (a = c and {b, d} = {0, 1} ) or (b = d and {a, c} = {0, 1}) or (a, b), (c, d) {(0, 0), (0, 1), (1, 0), (1, 1)}. Then X is a torus; X is homeomorphic to thesubspace ofR3 in the standard topology given by surface of revolution of acircle not intersecting the axis of revolution (the surface of a donut).

Example 7.11. In general, the quotients of X with respect to the definingrelations for quasicomponents, connected components, and path componentscan be interesting spaces. If X is locally connected, then the connectedcomponent and quasicomponent quotients are discrete. If X is locally pathconnected, all these quotients are discrete.

Example 7.12. The quotient of the topologists sine curve S by pathcomponents is homeomorphic to the Sierpinski space {0, 1} with open sets, {0}, {0, 1}.

Example 7.13. The quotient of the space X of nested rectangles by quasi-components is homeomorphic to the subspace {0} {1/(n + 1)|n N} Rin the standard topology. The quasicomponent of the two vertical linesmaps to 0 in any such homeomorphism. The quotient of X by connectedcomponents is the same space but with the point 0 doubled.

The properties preserved under taking quotients (in general) are notmuch better than the properties preserved under the images of general con-

tinuous surjections. Quotients of connected spaces are connected. Separa-tion properties are not necessarily preserved.

Example 7.14. Let be any irrational number. Define an equivalencerelation on R by declaring x x + k + m for all x R and k, m Z. LetX be R/ where R has the standard topology, and let p : R X denotethe quotient map. Note that since each equivalence class is countable, X

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is uncountable (otherwise would decompose R into a countable union of

countable sets). Suppose x, y R with p(x) = p(y). Let U X be an openneighborhood of p(x). Then f1(U) is an open set. For some a, b R, wehave x (a, b) f1(U). Since is irrational, the set {k + m|k, m Z}is dense in R and we can find k, m Z with y + k + m (a, b). Of course,this means y f1(U), and therefore that p(y) U. We have shown thatany nontrivial open subset of X is all of X. Therefore we have exhibitedthe uncountable indiscrete topology as a quotient of the standard topologyon R. In particular, without further hypotheses on the equivalence relationor the map, a quotient of a completely normal space need not even be T0.

8 Compactness and related propertiesOpen covers and the Lindelof property

Definition 8.1. Let X be a space and let S be a subset of X. An opencover of S is a collection {Ui}iI (for some index set I) of open subsets ofX, such that S

iIUi. A subcover of {Ui}iI is {Uj}jJ for some subset

J of the index set I, such that we still have S jJUj. An open cover

{Vj}jJ is a refinement of the open cover {Ui}iI if for every j J, thereis some i I with Vj Ui.

By the cardinality of a cover, we mean the cardinality of its index set.

One of the properties of spaces related to covers is also related the count-ability properties we discussed earlier.

Definition 8.2. A space X is Lindelof if every open cover of X has acountable subcover.

Proposition 8.3. Every second countable space is Lindelof.

Proof. Suppose {Ui}iI is an open cover of the second countable space X.Let {Vn}nN be a countable basis for X. Let S N be the subset of suchthat n S if and only if there is some i I with Vn Ui. Each Ui can beexpressed as a union of elements of {Vn}nS, so {Vn}nS is an open cover

of X and a countable refinement of {Ui}iI. For each n S, select somei(n) I such that Vn Ui(n). Then {Ui(n)}nS is a countable subcover.

Proposition 8.4. A metric space is Lindelof if and only if it is separable.

Proof. Since we know separable metric spaces are second countable andtherefore Lindelof, we only need to show that a Lindelof metric space X is

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separable. For each n N\{0}, the set of all open balls of radius 1/n is an

open cover ofX; let x(n, m) X be points such that {B1/n(x(n, m))}mN isa countable subcover. Then the set {x(n, m)|m, n N, n = 0} is a countabledense subset: ifBr(x) is an open ball, let n be large enough that 1/n < r/2;if no x(n, m) is in Br(x), then x is not in any B1/n(x(n, m)) and the set ofsuch balls could not be a cover.

Proposition 8.5. The property of being Lindelof is preserved under takingcontinuous images and closed subspaces.

The proofs are straightforward.

CompactnessCompactness is an important property of certain spaces that is difficult todescribe intuitively. It is a sort of intrinsic boundedness of spacesa wayof characterizing a space as bounded without any reference to distance.

Definition 8.6. A subset S of a space X is compact if every open cover ofS has a finite subcover. The space X is compact if every open cover of Xhas a finite subcover.

Proposition 8.7. A subset S of a space X is compact if and only if S is acompact space in the subspace topology.

The proof is straightforward. Note that any finite space is compact, anda discrete space is compact if and only if it is finite. We need a nontrivialexample of a compact space.

Proposition 8.8. Any closed interval is a compact subset ofR in the stan-dard topology.

Proof. We prove the proposition for [0, 1] and the general statement followseasily. Suppose C is an open cover of [0, 1] in R. Let S [0, 1] be the setwith x S if and only if there is a finite subcollection of C covering [0, x].Of course, 0 S, so S is nonempty. Let y be the least upper bound of S.If y / S, there there is U C with y U. There is y U with y < y and

(y, y) U, and there is a finite subcollection C of C covering [0, y]. Butthen C {U} is a finite cover of [0, y], a contradiction. If y S and y < 1,then there is a finite subset C of C and a U C with y U. Since U is anopen set, there is y U with y > y, contradicting the supposition that yis the least upper bound of S. So we must have y = 1, meaning that C hasa finite subcover covering [0, 1].

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The previous proposition and the following observation give many other

examples of compact spaces.

Proposition 8.9. Closed subspaces of compact spaces are compact.

The proof is straightforward.Compact spaces are much better behaved when we add the hypothesis

that they be Hausdorff.

Proposition 8.10. Compact subspaces of Hausdorff spaces are closed.

Proof. Let S be a compact subset of a Hausdorff space X. Let x X\S.For each y S, there are open disjoint open sets Uy and Vy with y Uy

and x Vy. Since {Uy}yS is an open cover of S, we have a finite subcover{Uy1 , . . . , U yn}. Then

ni=1 Vyi is an open neighborhood of x that is disjoint

from S. This shows X\S is open and S is closed.

At this point we can easily prove the Heine-Borel theorem.

Theorem 8.11. A subset S ofR in the standard topology is compact if andonly if it is closed and bounded.

Proof. Suppose S is closed and bounded. Then S is a subset of a closedinterval [a, b]. Since [a, b] is compact and S is a closed subset of [a, b], S iscompact.

Now suppose S is compact. Then since R is Hausdorff, S is closed. Theopen cover {(k, k + 2)}kZ of S has a finite subcover, from which we deducethat S is bounded.

Another benefit of assuming spaces are Hausdorff:

Proposition 8.12. Every compact Hausdorff space is normal.

Proof. Suppose A and B are closed subsets of a compact Hausdorff space X.Then A and B are compact. Fix a A. For b B, find disjoint open Ub andVb with a Ub and v Vb. Of course, a finite subcollection {Vb1 , . . . , V bn}covers B, and taking the union of these Vbi and the intersection of these Ubi

gives us disjoint open sets Ua and Va with a Ua and B Ub. Then a finitesubcollection {Ua1 , . . . , U an} of the Ua covers A; and the union of these Uaiand the intersection of these Vai give disjoint open sets that respectivelycontain A and B.

The following proposition has a straightforward proof that is omitted.

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Proposition 8.13. Continuous images of compact spaces are compact.

This then implies one of the most useful facts about compact spaces:

Proposition 8.14. Let X be a compact space and f: X R a continuousfunction. Then f is bounded and there are points x, y X realizing theextreme values of f: for all z X, we have f(x) f(z) f(y).

Proof. We know f(X) is a compact subset ofR and is therefore closed andbounded. Let a be the least upper bound of f(X). Of course a is in theclosure f(X) (either a f(x) or else a is certainly a limit point of f(x)), sosince f(X) is closed, a f(X). Similarly f(X) contains its greatest lowerbound, and therefore points in x map to both of these extremal values. Of

course, this means that there are points in X mapping to these values.

Another special property of maps from compact spaces:

Proposition 8.15. Let X be compact and let Y be Hausdorff. Then anycontinuous map f: X Y is closed. Further, f: X f(X) is a quotientmap. Finally, if f: X f(X) is injective, it is a homeomorphism.

Proof. Let A X be closed. Then A is compact. Then the image f(A)is compact. Since Y is Hausdorff, f(A) is closed. The map f: X f(X)is a closed surjection, so it is a quotient map. Bijective quotient maps arehomeomorphisms, so the last statement also follows.

Definition 8.16. A surjective, closed, continuous map of spaces f: X Yis perfect if for each y Y, the set f1({y}) is compact in X.

Lemma 8.17. Suppose f: X Y is a perfect map. Let y Y and let Ube an open subset of X containing f1({y}). Then there is an open subsetV of Y with f1({y}) f1(V) U.

Proof. The complement X\U is a closed set that does not intersect f1({y}).Then since f is closed, f(X\U) is closed and does not contain y. LetV = Y\f(X\U). Then y V, so f1({y}) f1(V). Of course f1(V) isopen, and it is not hard to see that f1(V) U.

Proposition 8.18. Suppose f: X Y is a perfect map and Y is compact.Then X is compact.

Proof. Suppose C is an open cover of Y. For each y Y, the set f1({y})is compact, and so there is a finite subset Cy of C that covers f

1({y}). LetUy =

Cy. By the lemma, for each y, there is an open set Vy Y with

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f1({y}) f1(Vy) Uy. Then the collection {Vy}yY is an open cover of

Y and has a finite subcover {Vy1 , . . . , V y1}. Then the collection Cy1 C ynis a finite subcover of C covering Y.

Corollary 8.19. A product of two compact spaces is compact.

Proof. Let X and Y be compact spaces and consider the coordinate projec-tion p : X Y Y. Of course p is continuous and surjective, and for anyy Y the preimage p1(y) is homeomorphic to X and is therefore compact.So to show that p is a perfect map, we just need to show that it is closed.Let A X Y be closed. Suppose y Y is a limit point of p(A) and sup-pose for contradiction that y / p(A). Then X Y\A is open and contains

p1(y) as a subset. Since X Y\A is open, we can write it as a union ofbasis elements of the form Ui Vi for Ui open in X and Vi open in Y. Thiscollection of basis elements then covers p1(y); since p1(y) is compact, wecan find a finite subcover U1 V1, . . . , U n Vn. We can assume each Ui Viintersects p1(Y), so that V = V1 Vn is an open neighborhood of y inY. Then p1(V) contains p1(y) but is a subset of X Y\A. This impliesthat y V but V p(A) = , contradicting our hypothese that y was alimit point ofp(A). So p is closed and therefore perfect, and therefore X Yis compact.

The following needs no further proof.

Corollary 8.20. A subset ofRn

is compact if and only if it is closed andbounded.

There is also a characterization of compactness in terms of intersectionsof closed sets.

Definition 8.21. A collection of sets C has the finite intersection propertyif for every finite subcollection C, we have

C = .

Proposition 8.22. A space X is compact if and only if for every collectionC of closed sets with the finite intersection property, we have that the totalintesection C is nonempty.Proof. Given a collection of closed sets C, we can define a collection of opensets by taking the complements: B = {X\A|A C}. Note that

C =

if and only if

B = X. If a space X is compact and C is a set of closedsets with

C = , then its collection of complements B covers X and has a

finite subcover B. The collection of complements C ofB is a finite subset of

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C with C = . So ifC has the finite intersection property, then certainlyC = .

Now suppose that every collection of closed sets with the finite inter-section property has nonempty intersection. Let B be an open cover of X.Then the collection of complements C has empty intersection, and thereforehas a finite subcollection C with empty intersection. Then the collection ofcomplements of C is a finite subcover of B.

Local compactness

Definition 8.23. A space X is locally compact at a point p X if p has acompact neighborhood, and X is locally compact if it is locally compact at

every point.

Definition 8.24. Given spaces X and Y, a continuous map f: X Y isproper if for every compact A Y, we have f1(A) compact in X.

Proposition 8.25. If X is a space, Y is a locally compact Hausdorff spaceand f: X Y is proper and injective, then f is a homeomorphism onto itsimage (meaning f: X f(X) is a homeomorphism).

Proof. Let f: X Y be as above and let A X be closed. Suppose y Yis a limit point of f(A). Let B Y be a compact neighborhood of y. LetC = A f1(B) X. Then y is a limit point of f(C): ifU Y is an open

neighborhood ofY, it contains a smaller open neighborhood V with V B,and V contains a point of f(A)\{y} which is therefore in f(C). Since fis proper, f1(B) is compact, and since A is closed, C = A f1(B) iscompact. Since Y is Hausdorff, f(C) is compact, and therefore closed. Soy f(C), and therefore y f(A), and f(A) is closed. Since f sends closedsubsets of X to closed subsets of f(X), f is closed. Since it is injective,f: X f(X) is a closed bijection and is therefore a homeomorphism.

Definition 8.26. Let X be a Hausdorff space. The one-point compactifica-tion of X is the space Y = X {y0}, where the open sets are:

U X Y for any open U X, and

{y0} X\A for any compact subset A of X.

Proposition 8.27. Let X be a Hausdorff space and let Y be its one-pointcompactification. Then Y is a compact space, and is Hausdorff if and onlyif X is locally compact.

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Proof. Let y0 denote the point in Y\X. First we show that Y is indeed a

space. Of course the intersection of two open subsets of X is open in Y. IfU = {y0} X\A and V U are open, then U V = U\A, which is an opensubset of X and is therefore open in Y. If instead V = {y0} X\B, thenU V = {y0} X\(A B), which is open because A B is compact. If{Ai}iI is a collection of compact subsets of X, then since X is Hausdorff itis closed, and

iIAi is closed and is compact because it is a closed subset of

the compact set Aj for some fixed j I. Then the union of{{y0} X\Ai}iIis {y0} X\(

iIAi), which is open. If we take a union of a collection of

open sets, some of which contain y0 and some of which do not, then wecan take the union by separately taking the unions of those containing y0and then those that do not, and then taking the union of these two sets. Of

course, the two sub-unions are both open. IfU is open in X and {y0} X\Ais open in Y, then their union is {y0} X\(A\U). Since A\U is a closedsubset of a compact subset A of X, it is compact, and unions of open setsin Y are open. Of course is open in Y. The entire space Y is open in Ybecause is a compact subset of X.

Next we show that Y is compact. Suppose C is an open covering of Y.Then there is some U C with y0 U, and some compact A X withU = {y0} X\A. Then C\{U} is an open covering of A, and has a finitesubcover C. Then {U} C is a finite subcover of C covering Y.

If x, y X, then there are disjoint open neighborhoods of x and y inY because X is Hausdorff. If x X and then there are disjoint open

neighborhoods ofx and y0 in Y if and only if X is locally compact: since Xis locally compact, we have some A, U X with x U A with U openand A compact; then U and {y0} X\A are disjoint open neighborhoodsrespectively containing x and y0. Conversely, if x and y0 have disjoint openneighborhoods U and V, then V = {y0} X\A for some compact A anddisjointness implies U A (so that X is locally compact at X).

Proposition 8.28. Suppose X and Y are locally compact Hausdorff spaceswith respective one-point compactificationsX and Y, withX\X = {x0} andY\Y = {y0}. Then a continuous map f: X Y extends to a continuousmap f: X Y with f(x0) = y0 if and only if it is proper.

Proof. First we suppose that f is the restriction off. Let A Y be compact.Then U = Y\A is open in Y. Then f1(U) is open in X. Since f(x0) = y0,we have x0 f1(U). Note that f1(U) = {x0} X\f1(A). In particular,this implies f1(A) is compact and therefore f is proper.

Now suppose f is proper and extend it to f as described. Let U Ybe open. If U Y, then f1(U) = f1(U) is open. If y0 U, then

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A = Y\U is a compact subset of Y. Then f1(A) is compact since f is

proper, and therefore f1(U) = {x0} X\f1(A) is open in X. Thereforef is continuous.

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Introduction to Geometry and Topology - Matthew Day