- 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. That is, every sequence has a convergent subsequence. (We will go over it in here as well.) 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. It is shown later that if this is true, 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. - 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 is might not be so if K is not compact.
- You know that if f : X → X for X a complete metric space, then if
d< rdit follows that f has a unique fixed point theorem. Let f : ℝ → ℝ be given by
Show that

<, but f has no fixed point. - If is a metric space, show that there is a bounded metric ρ such that the open sets forare the same as those for.
- Let be a metric space where d is a bounded metric. Let C denote the collection of closed subsets of X. For A,B ∈C, define
where for a set S,

Show x → dist

is continuous and that therefore, S_{δ}is a closed set containing S. Also show that ρ is a metric on C. This is called the Hausdorff metric. - ↑Suppose is a compact metric space. Showis a complete metric space. Hint: Show first that if W
_{n}↓ W where W_{n}is closed, then ρ→ 0. Now letbe a Cauchy sequence in C. Then if ε > 0 there exists N such that when m,n ≥ N, then ρ< ε. Therefore, for each n ≥ N,Let A ≡∩

_{n=1}^{∞}∪_{k=n}^{∞}A_{k}. By the first part, there exists N_{1}> N such that for n ≥ N_{1},Therefore, for such n, A

_{ε}⊇ W_{n}⊇ A_{n}and_{ε}⊇_{ε}⊇ A because - ↑ Let X be a compact metric space. Show is compact. Hint: Let D
_{n}be a 2^{−n}net for X. Let K_{n}denote finite unions of sets of the form Bwhere p ∈D_{n}. Show K_{n}is a 2^{−(n− 1) }net for.

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