where we are in your textbook - cpb-us-w2.wpmucdn.com · P goes around each of a, b and c with winding number +2 Q goes around each of a, b and c with winding number +2 P cannot be

where we are in your textbook

I you should’ve read chapter 1, sections 1,3,6

I chapter 1, sections 4,5 are good background reading for realanalysis; rest of chapter 1 is optional

I continue with chapter 2, sections 1,2,3

I will focus on chapter 2, section 6 (compactness)

Tibor Beke metric spaces

answer to problem 1

Let a, b, c be three distinct points in the plane. Can you drawcurves P and Q in the plane, avoiding the points a, b, c , such that

• P goes around each of a, b and c with winding number +2• Q goes around each of a, b and c with winding number +2• P cannot be continuously deformed into Q (without passingthrough one of the points a, b, c).

Yes. Can be done in infinitely many different ways, in fact.(“Different” here means “non-deformable into each other”.We’ll make this precise!)


answer to problem 2(a)(b)

(a) (U ∪ V ) rW = (U rW ) ∪ (V rW ) is equivalent to

(U ∪ V ) ∩W = (U ∩W ) ∪ (V ∩W )

true by virtue of the distributive law for ∩ with respect to ∪.

(b) (U ∩ V ) rW = (U rW ) ∩ (V rW ) is equivalent to

(U ∩ V ) ∩W = (U ∩W ) ∩ (V ∩W )

which follows from properties of ∩.


answer to problem 2(c)

(c) the next two could differ, however . . .

U r (V rW ) = U ∩ V rW = U ∩ V ∩W = U ∩ (V ∪W ) =

U ∩{

(V ∩W ) t (V ∩W ) t (V ∩W )}

=

(U ∩ V ∩W ) t (U ∩ V ∩W ) t (U ∩ V ∩W )

while(U r V ) rW = (U ∩ V ) ∩W = U ∩ V ∩W

Above, t stands for ‘disjoint union’.


problem 4

The ordinary notion of euclidean distance in the plane correspondsto the formula

d(〈x1, x2〉, 〈y1, y2〉) =

√(x1 − y1)2 + (x2 − y2)2

Prove using algebra, not geometry that this distance functionsatisfies the triangle inequality. Concretely, prove the algebraicinequality√

(x1 − z1)2 + (x2 − z2)2 6√(x1 − y1)2 + (x2 − y2)2 +

√(y1 − z1)2 + (y2 − z2)2

When is there an equality?


answer to problem 4 using “on the nose” algebra

Set ui = xi − yi , vi = yi − zi , then ui + vi = xi − zi (i = 1, 2).Need to prove√

(u1 + v1)2 + (u2 + v2)2 6√

u21 + u22 +√

v21 + v22

Square, simplify, divide by 2 to obtain

u1v1 + u2v2 6√

u21 + u22

√v21 + v22

Square, expand, simplify to obtain

2u1v1u2v2 6 u21v22 + u22v

21

0 6 u21v22 − 2u1v1u2v2 + u22v

21

0 6 (u1v2 − u2v1)2

Steps are reversible(!), equality iff the vectors 〈u1, u2〉 and 〈v1, v2〉are parallel, iff the points (x1, x2), (y1, y2), (z1, z2) are collinear.


The above proof generalizes to Rn, though the algebra is trickier.The second stage becomes

n∑i=1

uivi 6

√√√√ n∑i=1

u2i

√√√√ n∑i=1

v2i

( n∑i=1

uivi

)26( n∑

i=1

u2i

)( n∑i=1

v2i

)the beautiful Cauchy-Schwarz inequality (over R). A one-line proofof this inequality is the algebraic identity( n∑

i=1

uivi

)2+

∑16i<j6n

(uivj − ujvi )2 =

( n∑i=1

u2i

)( n∑i=1

v2i

)the n = 2 case of which you discovered on the previous slide.


“textbook” solution to problem 4

We work in Rn. For vectors u = 〈u1, u2, . . . , un〉 andv = 〈v1, v2, . . . , vn〉, inner (a.k.a. ‘scalar’ or ‘dot’) product

u · v :=n∑

i=1

uivi

and norm ‖u‖ :=√

u · u defined as usual.


Cauchy-Schwarz inequality

Let α, β be arbitrary scalars. Using non-negativity of squares andbilinearity of inner product

0 6 (αu− βv) · (αu− βv) = α2 u · u− 2αβ u · v + β2 v · v

Choose the particular scalars α = ‖v‖ and β = ‖u‖ and substitute:

0 6 ‖v‖2‖u‖2 − 2‖v‖‖u‖u · v + ‖u‖2‖v‖2

‖v‖‖u‖u · v 6 ‖u‖2‖v‖2

u · v 6 ‖u‖‖v‖


Cauchy-Schwarz inequality

u · v 6 ‖u‖‖v‖Using this on −u and v gives

−u · v 6 ‖u‖‖v‖

The two cases can be compressed into the single inequality

|u · v| 6 ‖u‖‖v‖

equivalently

(u · v)2 6 ‖u‖2‖v‖2 = (u · u)(v · v)

‘long-hand’ form

( n∑i=1

uivi)2

6( n∑i=1

u2i)( n∑

i=1

v2i)


triangle inequality

In Rn, we wish to prove

‖x− z‖ 6 ‖x− y‖+ ‖y − z‖

Write u := x− y and v := y − z so u + v = x− z. Want to prove

‖u + v‖ 6 ‖u‖+ ‖v‖

‖u + v‖2 6 (‖u‖+ ‖v‖)2

(u + v) · (u + v) 6 ‖u‖2 + 2‖u‖‖v‖+ ‖v‖2

‖u‖2 + 2 u · v + ‖v‖2 6 ‖u‖2 + 2‖u‖‖v‖+ ‖v‖2

u · v 6 ‖u‖‖v‖

Steps reversible; equality iff equality in Cauchy-Schwarz iff u, vlinearly dependent (with non-negative inner product), iff x, y, zcollinear (in this order).


different metrics, same opens

Recall our three examples of metrics on R2:

d1(〈x1, x2〉, 〈y1, y2〉) =√

(x1 − y1)2 + (x2 − y2)2

d2(〈x1, x2〉, 〈y1, y2〉) = |x1 − y1|+ |x2 − y2|

d3(〈x1, x2〉, 〈y1, y2〉) = max{|x1 − y1|, |x2 − y2|}

Proposition With respect to any one of these metrics, exactly thesame subsets of R2 are open.

Let Bi (x , r) be the ‘open ball’ with center x ∈ R2 and radius r > 0in the metric di (i = 1, 2, 3). Enough to prove that Bi (x , r) is anopen set when considered with respect to the metric dj , i 6= j .Use elementary geometry!


strongly equivalent metrics

Definition Let d1 and d2 be two metrics on the set X .They are said to be strongly equivalent (sometimes, coarselyequivalent) if there exist constants 0 < k < K <∞ (dependingonly on the metrics) such that for all distinct points x , y ∈ X ,

k <d1(x , y)

d2(x , y)< K

This implies thatB1(x , kr) ⊆ B2(x , r)

B2(x ,r

K) ⊆ B1(x , r)

for all x ∈ X and r > 0, where Bi (−,−) means the open ball withrespect to the metric di .


topologically equivalent metrics

Proposition Suppose d1 and d2 are strongly equivalent. ThenU ⊆ X is an open subset in (X , d1) if and only if it is open in(X , d2).

Two metrics d1 and d2 on a set X with the property that theydefine the same open sets are said to be topologically equivalent.

We’ve just seen that strong equivalence is a sufficient (but notnecessary) condition for topological equivalence.


why abstract away from metric spaces?

I the continuity of a map f : (X , dX )→ (Y , dY ) between metricspaces only depends on what the open subsets of X and Y are

I lots of metrics give rise to the same open sets; there is no‘simplest’ or ‘best’ metric in general

I there are spaces on which continuity makes sense that do notarise from a metric (e.g. Frechet spaces; non-Hausdorff spacesin combinatorial and algebraic geometry; infinite products ofmetric spaces; many function spaces)

I there is no natural metric on quotients by equivalencerelations and ‘gluings’ of metric spaces

I we will not abandon metric spaces, really; they will continueto be the main source of examples and motivation for ourdefinitions.


topology on a set of points

Seeing that the notion of continuity can be encapsulated in thealgebra of open sets, we introduce

Definition Let X be a set (to be called ‘the set of points’).A topology on X is a collection of subsets of X (to be called ‘thecollection of open subsets of X ’) satisfying the axioms

I The empty set is open; X itself is open (in X ).

I If each Ui is open, i ∈ I (arbitrary collection!), so is theirunion

⋃i∈I Ui .

I If each Ui is open, i = 1, 2, . . . , n (finite collection!), so istheir intersection

⋂ni=1 Ui .


topological spaces

A topological space is a set equipped with a topology.

Any metric space is automatically a topological space: given themetric, define the notion of open set with the help of ε-balls asbefore. The open subsets of any metric space satisfy the axiomsfor a topology.

Example The following is a topological space:

I X = {?, •}I the open subsets are the empty set ∅, X itself and {•}.

This topology does not come from any metric on the set X .


interior

Let X be a topological space and W an arbitrary subset of X .

The point x ∈W is an interior point of W if there exists an openset U such that x ∈ U ⊆W . The set of interior points of W is theinterior of W , denoted int(W ).

I int(W ) is always an open set

I int(W ) is the largest open set contained in W (the union ofall open sets contained in W )

I a set is open if and only if it equals its interior.


closed sets; closure

Let X be a topological space. A subset of X is closed (bydefinition) if its complement is open. ∅ and X are closed; arbitraryintersection of closed sets is closed; finite union of closed sets isclosed.

Let W be an arbitrary subset of X . The point x ∈ X is a point ofclosure of W if every open set containing x intersects Wnon-trivially. The set of points of closure of W is the closure ofW , denoted W .

I W is always a closed set

I W is the smallest closed set containing W (the intersection ofall closed sets containing W )

I a set is closed if and only if it equals its closure.


For any topological space X and any subset S ⊆ X ,

int(S) t X r S = X

S t int(X r S) = X

(Recall that ‘t’ stands for ‘disjoint union’.)


some terminology

Let X be a topological space, x ∈ X a point. An openneighborhood U of x is any open subset of X containing x .A neighborhood S of x is any subset of X that contains an openneighborhood of x . (So, x ∈ U ⊆ S ⊆ X and U is open in X .)

Remark The meaning of ‘neighborhood’ varies slightly fromtextbook to textbook, but ‘open neighborhood’ means the same toall authors.


continuity; homeomorphism

Let f : X → Y be a function from one topological space toanother. f is continuous (over its entire domain) if for every opensubset U of Y , its preimage f −1(U) is an open subset of X .

The composite of two continuous maps is continuous.

f : X → Y is a homeomorphism if it is a continuous bijection witha continuous inverse.


compactness of the closed unit interval (version one)

Theorem Let (ai , bi ), i ∈ I , be a family of open intervals suchthat

[0, 1] ⊂⋃i∈I

(ai , bi ) .

Then there exist finitely many of these open intervals, say(ai1 , bi2), . . . , (ain , bin), so that already

[0, 1] ⊂n⋃

k=1

(aik , bik ) .

Slogan: “ Every cover of [0, 1] by open intervals has a finitesubcover ”


Homework Problem 12

(a) List all possible topologies on the 3-element set {a, b, c}.(b) How many different homeomorphism types of 3-point spacesare there?

Due Wed, Feb 12, in class.


Homework Problem 13

Your textbook, Chapter 2, Section 1, Exercise 2 on page 63.



Homework Problem 14∗ (optional)


Due (optionally) Wed, Feb 19, in class.


Homework Problem 15




Homework Problem 16

Let (X , dX ) and (Y , dY ) be metric spaces. On the cartesianproduct X × Y , define

dX×Y(〈x1, y1〉, 〈x2, y2〉

)= dX (x1, x2) + dY (y1, y2)

(a) Show that dX×Y is a metric on the set X × Y .(b) Show that the projection maps X × Y → X and X × Y → Yare continuous in this metric.(c)∗ (optional) Let (Z , dZ ) be an arbitrary metric space andf : Z → X and g : Z → Y two functions. Show that the maph : Z → X × Y that takes z ∈ Z to 〈f (z), g(z)〉 is continuous ifand only if both of the coordinate functions f and g arecontinuous.



Documents

where we are in your textbook - cpb-us-w2.wpmucdn.com · P goes around each of a, b and c with winding number +2 Q goes around each of a, b and c with winding number +2 P cannot be