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Satoru Takahashi EC5104/EC5104RDepartment of Economics Mathematical EconomicsNational University of Singapore Semester 1, AY2014/2015

Problem Set 4

due on September 8 (Monday)

Part A

1. Consider the following subspaces of R (with the topologies induced by the usualtopology on R):

R, Q, Z, [0, ), (0, ),

[0, 1], [0, 1), (0, 1), [0, 1] Q, {1, 2, 3}.

(a) Which of them is compact?

(b) Which of them is connected?

2. Show that a set Xwith the discrete topology is compact if and only ifX is finite.

3. (a) Show that a topological spaceXis compact if and only if every family {F}of closed subsets ofXhas the finite intersection property, i.e., whenever everyfinite subfamily{Fk}

nk=1 has a nonempty intersection:

F1 F2 Fn =,

the entire family {F} also has a nonempty intersection:

F =.

(b) Show thatX Rd is compact (i.e., bounded and closed in Rd) if and only if everyweakly decreasing sequence of nonempty closed subsets of Xhas a nonemptyintersection.

4. (a) Let x Rn. Let Y Rn be nonempty and compact. Show that there existsy Ythat minimizes the distance to x, i.e.,x y x yfor anyy Y.

(b) Does this hold ifY is nonempty and closed, but unbounded in Rn?

5. Show that every polynomial equation with odd degree and real coefficients has atleast one real root.

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Part B

1. We say that a topological space (X, ) is Hausdorff if for every x, y Xwithx =y,there exist open sets U, V such that U x, V y, and U V = . Show that

every metric space induces a Hausdorff topology.2. Let Xbe a compact Hausdorff space. Show that a subset Y Xwith the subspace

topology is compact if and only ifY is closed in X.

3. Let X and Y be topological spaces, and f: X Ybe a continuous and bijectivefunction.

(a) Isf1 : Y Xalways continuous?

(b) Show thatf1 is continuous ifX is compact and Y is Hausdorff.

4. Let (X, d) be a metric space. Show that the following are equivalent.

(a) X is compact under the topology induced by d.

(b) (X, d) is complete and totally bounded with respect to d.1

(c) (X, d) is sequentially compact.2

5. In what follows, you will show that (b) does not imply (a) in the previous question ifdifferent distance functions are used for completeness and for total boundedness. Letd(m, n) =|m n| and d(m, n) =|1/m 1/n| for m, n N ={1, 2, . . .}.

(a) Show that bothd and d induce the discrete topology on N.

(b) Show that (N, d) is complete.

(c) Show that (N, d) is totally bounded.

(d) Show thatN is not compact under the discrete topology.

6. A topological space is said to be separable if it admits a countable dense subset.

(a) Show thatRn is separable.

(b) Show that every totally bounded metric space (e.g., every compact metric space)is separable.

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We say that (X, d) is totally bounded if for every >0, there exists a finite subset Y ofXsuch that,for every x X, there exists y Y such that d(x, y)< . Note that ifX Rn and d is equivalent to theEuclidean metric, (X, d) is complete if and only ifXis closed in Rn, and total boundedness is equivalent toboundedness. In more general metric spaces, total boundedness implies boundedness, but not vice versa.For example, any infinite set with the discrete metric is bounded, but not totally bounded.

2We say that (X, d) is sequentially compactif every sequence on Xhas a convergent subsequence withrespect to d.

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7. Recall that [0, 1]N denotes the set of all sequences on [0, 1]. Let

d({xn}, {yn}) = supnN

|xn yn|,

d({xn}, {yn}) =

n=0

2n|xn yn|.

(a) Show that bothd and d are distance functions on [0, 1]N.

(b) Consider a sequence{xkn}on [0, 1]N, i.e., a sequence of sequences on [0, 1]. Show

that {xkn} {x

n} as k in ([0, 1]N, d) if and only ifxkn x

n as k pointwise for every n N, i.e., for every >0 and n N, there exists k,nNsuch that

k > k,n |xkn x

n|< .

(c) Show that {xkn} {x

n} as k in ([0, 1]N, d) if and only ifx

kn x

n as

k uniformly in n, i.e., for every >0, there exists k N such that

k > k n, |xkn x

n|< .

(d) Show that ([0, 1]N, d) is complete but not separable (and hence not compact).

(e) Show that ([0, 1]N, d) is compact.

8. Fix 0 < K < . Let Xbe the set of functions f: [0, 1] [0, 1] that are Lipschitzcontinuous with Lipschitz constant K:

|f(x) f(x

)| K|x x

|with the distance function

d(f, g) = maxx[0,1]

|f(x) g(x)|.

Show that (X, d) is compact.

9. Let R = R {, +}be the set of extended real numbers. We say that a subsetX R is open if all the following conditions hold:

X\ {, +}is open in R in the usual sense;

if X, then there exists K > such that x < Kx X;

if + X, then there exists K Kx X.

Let be the family of all such open sets.

(a) Show that (R, ) is a topological space, i.e., is closed under the operations offinite intersections and infinite unions.

(b) Show that (R, ) is metrizable, i.e., is generated by some distance function.

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(c) Show that (R, ) is compact.

10. (a) Show that a function f: R R is continuous if and only if the graph off

f={(x, y) R2 :y = f(x)}

is connected and closed in R2.

(b) Show that a function f: [0, 1] R is continuous if and only if f is compact.

11. Let (X, d) be a metric space. We say that a functionf: X Ris upper semicontin-uousat xif for every >0, there exists >0 such that

d(x, x)< f(x)< f(x) +.

fis upper semicontinuous if it is upper semicontinuous at every point in X.

(a) Show thatfis upper semicontinuous at xif and only if

lim supn

f(xn) f(x)

for every sequence {xn}that converges to x.

(b) Show that f is upper semicontinuous if and only iff1((, )) = {x X :f(x)< } is open for every R.

(c) Show thatf is upper semicontinuous if and only if the hypograph off

{(x, y) X R :y f(x)}

is closed in X R.

(d) Show that ifXis nonempty and compact, and fis upper semicontinuous, thenfhas a maximum, i.e., there exists x X such that f(x) f(x) for everyx X.

(e) Does an upper semicontinuous function on a nonempty and compact domainhave a minimum?

12. Let (X, d) be a metric space. We say that a preference relation on X is uppersemicontinuous if

{x X :x y}

is closed in X for any y X, lower semicontinuous if{y X :x y}

is closed inXfor anyx X, andcontinuousif it is upper and lower semicontinuous.

(a) Show that is continuous if and only if

{(x, y)X X :x y}

is closed in X X.

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