- Let d=for x,y ∈ ℝ. Show that this is a metric on ℝ.
- Now consider ℝ
^{n}. Let_{∞}≡ max. DefineShow that this is a metric on ℝ

^{n}. In the case of n = 2, describe the ball B. Hint: First show that≤+. - Let Cdenote the space of functions which are continuous on. Define
Verify the following.

≤+. Then use to show that d≡is a metric and that with this metric,is a metric space. - Recall that is compact. This was done in single variable advanced calculus. Also, it is Lemma 2.5.6 above. Thus every open cover has a finite subcover of the set. Also recall that a sequence of numbersis a Cauchy sequence means that for every ε > 0 there exists N such that if m,n > N, then< ε. First show that every Cauchy sequence is bounded. Next, using the compactness of closed intervals, show that every Cauchy sequence has a convergent subsequence. By Theorem 2.2.2, the original Cauchy sequence converges. Thus ℝ with the usual metric just described is complete because every Cauchy sequence converges.
- Using the result of the above problem, show that is a complete metric space. That is, every Cauchy sequence converges. Here d≡
_{∞}. - Suppose you had is a metric space. Now consider the product space
with d

= max. Would this be a metric space? If so, prove that this is the case.Does triangle inequality hold? Hint: For each i,

Now take max of the two ends.

- In the above example, if each is complete, explain whyis also complete.
- Show that Cis a complete metric space. That is, show that ifis a Cauchy sequence, then there exists f ∈ Csuch that lim
_{n→∞}d= lim_{n→∞}= 0. Hint: First, you know thatis a Cauchy sequence for each t. Why? Now let fbe the name of the thing to which f_{n}converges. Recall why the uniform convergence implies t → fis continuous. Give the proof. It was done in single variable advanced calculus. Review and write down proof. Also show that→ 0. - Let X be a nonempty set of points. Say it has infinitely many points. Define d= 1 if x≠y and d= 0 if x = y. Show that this is a metric. Show that inevery point is open and closed. In fact, show that every set is open and every set is closed. Is this a complete metric space? Explain why. Describe the open balls.
- Show that the union of any set of open sets is an open set. Show the intersection of any
set of closed sets is closed. Let A be a nonempty subset of a metric space . Then the closure of A, written as is defined to be the intersection of all closed sets which contain A. Show that = A∪A
^{′}. That is, to find the closure, you just take the set and include all limit points of the set. It was proved in the chapter, but go over it yourself. - Let A
^{′}denote the set of limit points of A, a nonempty subset of a metric space. Show that A^{′}is closed. - A theorem was proved which gave three equivalent descriptions of compactness of a
metric space. One of them said the following: A metric space is compact if and only if it
is complete and totally bounded. Suppose is a complete metric space and K ⊆ X. Thenis also clearly a metric space having the same metric as X. Show thatis compact if and only if it is closed and totally bounded. Note the similarity with the Heine Borel theorem on ℝ. Show that on ℝ, every bounded set is also totally bounded. Thus the earlier Heine Borel theorem for ℝ is obtained.
- Suppose is a compact metric space. Then the Cartesian product is also a metric space. That isis a metric space where d≡ max. Show thatis compact. Recall the Heine Borel theorem for ℝ. Explain why
is compact in ℝ

^{n}with the distance given by d= max. Hint: It suffices to show thatis sequentially compact. Let_{ m=1}^{∞}be a sequence. Then_{m=1}^{∞}is a sequence in X_{i}. Therefore, it has a subsequence_{k1=1}^{∞}which converges to a point x_{1}∈ X_{1}. Now consider_{k1=1}^{∞}the second components. It has a subsequence denoted as k_{2}such that_{k2=1}^{∞}converges to a point x_{2}in X_{2}. Explain why lim_{k2→∞}x_{1}^{k2}= x_{1}. Continue doing this n times. Explain why lim_{kn→∞}x_{l}^{kn}= x_{l}∈ X_{l}for each l. Then explain why this is the same as saying lim_{kn→∞}x^{kn}= x in. - If you have a metric space and a compact subset ofK, suppose that L is a closed subset of K. Explain why L must also be compact. Hint: Go right to the definition. Take an open covering of L and consider this along with the open set L
^{C}to obtain an open covering of K. Now use compactness of K. Use this to explain why every closed and bounded set in ℝ^{n}is compact. Here the distance is given by d≡ max_{1≤i≤n}. - Show that compactness is a topological property in the following sense. If
are both metric spaces and f : X → Y has the property that f is one to one, onto, and continuous, and also f

^{−1}is one to one onto and continuous, then the two metric spaces are compact or not compact together. That is one is compact if and only if the other is. - Consider ℝ the real numbers. Define a distance in the following way.
Show this is a good enough distance and that the open sets which come from this distance are the same as the open sets which come from the usual distance d

=. Explain why this yields that the identity mapping f= x is continuous with continuous inverse as a map fromto. To do this, you show that an open ball taken with respect to one of these is also open with respect to the other. However,is not a complete metric space whileis. Thus, unlike compactness. Completeness is not a topological property. Hint: To show the lack of completeness of, consider x_{n}= n. Show it is a Cauchy sequence with respect to ρ. - A very useful idea in metric space is the following distance function. Let be a metric space and S ⊆ X,S≠∅. Then dist≡ inf. Show that this always satisfies
This is a really neat result.

- If K is a compact subset of and yK, show that there always exists x ∈ K such that d= dist. Give an example in ℝ to show that this might not be so if K is not compact.
- If S is a nonempty set, the diameter of S denoted as diamis defined as follows.
Suppose

is a complete metric space and you have a nested sequence of closed sets whose diameters converge to 0. That is, each A_{n}is closed,A_{n}⊇ A_{n+1}and lim_{n→∞}diam= 0. Show that there is exactly one point p contained in the intersection of all these sets A_{n}. Give an example which shows that if the condition on the diameters does not hold, then maybe there is no point in the intersection of these sets. - Two metric spaces ,are homeomorphic if there exists a continuous function f : X → Y which is one to one onto, and whose inverse is also continuous one to one and onto. Show that the intervalis not homeomorphic to the unit circle. Hint: Recall that the continuous image of a connected set is connected, Theorem 2.11.3. However, if you remove a point fromit is no longer connected but removing a single point from the circle results in a connected set.
- Using the same methods in the above problem, show that the unit circle is not
homeomorphic to the unit sphere and the unit circle is not homeomorphic to a figure eight.
- The rational numbers ℚ and the natural numbers ℕ have the property that there is a
one to one and onto map from ℕ to ℚ. This is a simple consequence of the Schroeder
Bernstein theorem presented earlier. Both of these are also metric spaces with respect to
the usual metric on ℝ. Are they homeomorphic? Hint: Suppose they were.
Then in ℚ consider , all the rationals between 1 and 2 excluding 1 and 2. This is not a closed set because 2 is a limit point of the set which is not in it. Now if you have f a homeomorphism, consider f. Is this set closed?
- If you have an open set O in ℝ, show that O is the countable union of disjoint open intervals. Hint: Consider the connected components.
- Addition and multiplication on ℝ can be considered mappings from ℝ × ℝ to ℝ as
follows. +≡ x + y,⋅≡ xy. Here the metric on ℝ × ℝ can be taken as d≡ max. Show these operations are continuous functions.
- Suppose K is a compact subset of a metric space and there is an open cover C of K. Show that there exists a single positive δ > 0 such that if x ∈ K,Bis contained in some set of C. This number is called a Lebesgue number. Do this directly without using the equivalence of compactness and sequential compactness.
- If K is a sequentially compact subset of a metric space , and there is an open cover C of K, show that there exists a single positive δ > 0 such that if x ∈ K,Bis contained in some set of C. This number is called a Lebesgue number. Do this directly without using the equivalence of compactness and sequential compactness.
- Using the above problem, show directly that if K is a sequentially compact set in a
metric space , then it is compact.
- Show directly that compactness implies sequential compactness in a metric space
. Thus with the above, this will show that the two definitions of compactness are the same for a metric space.
- Let f : D → ℝ be a function. This function is said to be lower
semicontinuous
^{1}at x ∈ D if for any sequence⊆ D which converges to x it followsSuppose D is sequentially compact and f is lower semicontinuous at every point of D. Show that then f achieves its minimum on D. Here D is some metric space. Let f : D → ℝ be a function. This function is said to be upper semicontinuous at x ∈ D if for any sequence

⊆ D which converges to x it followsSuppose D is sequentially compact and f is upper semicontinuous at every point of D. Show that then f achieves its maximum on D.

- Show that a real valued function defined on a metric space D is continuous if and only if it is both upper and lower semicontinuous.
- Give an example of a lower semicontinuous function defined on ℝ which is not continuous and an example of an upper semicontinuous function which is not continuous.
- More generally, one considers functions which have values in . Such a function f is upper semicontinuous if, whenever x
_{n}→ x,and lower semicontinuous if whenever x

_{n}→ x,Suppose

is a collection of continuous real valued functions defined on a metric space. LetShow F is an upper semicontinuous function. Next let

Show G is a lower semicontinuous function.

- The result of this problem is due to Hausdorff. It says that if you have any lower
semicontinuous real valued function defined on a metric space , then it is the limit of an increasing sequence of continuous functions. Here is an outline. You complete the details.
- First suppose f≥ 0 for all x. Define
Then f

≥ f_{n}and f_{n}is increasing in n. Also each f_{n}is continuous becauseThus

Why? It follows that

Why?

- Let h= lim
_{n→∞}f_{n}. Then h≤ f. Why? Now for each ε > 0, and fixed x, there exists z_{n}such thatWhy? Therefore, z

_{n}→ x. Why? - Then
Why? Therefore, h

≥ fand so they are equal. Why? - Now consider f : X →and is lower semicontinuous as just explained. Consider+ arctan f≡ g. Since f has real values,
Then g

is also lower semicontinuous having values in. Why? By what was just shown, there exists g_{n}↑ gwhere each g_{n}is continuous. Consider f_{n}≡ tan. Then f_{n}is continuous and increases to f.

- First suppose f
- Generalize the above problem to the case where f is an upper semicontinuous real
valued function. That is,
whenever x

_{n}→ x. Show there are continuous functionssuch that f_{n}↓ f. Hint To save trouble, maybe show that f is upper semicontinuous if and only if −f is lower semicontinuous. Then maybe you could just use the above problem. - What if f is lower (upper) semicontinuous with values in ? In this case, you consideras a metric space as follows:
Then you can generalize the above problems to show that if f is lower semicontinuous with values into

then it is the increasing limit of continuous functions with values in. Note that in this case a function identically equal to ∞ would be continuous so this is a rather odd sort of thing, a little different from what we normally like to consider. Check the details and explain why in this setting, the lower semicontinuous functions are exactly pointwise limits of increasing sequences of continuous functions and the upper semicontinuous functions are exactly pointwise limits of decreasing sequences of continuous functions.