- Suppose is a sequence contained in a closed set C such that lim
_{n→∞}x_{n}= x. Show that x ∈ C. Hint: Recall that a set is closed if and only if the complement of the set is open. That is if and only if ℝ^{n}∖ C is open. - Show using Problem 1 and Theorem 8.10.12 that every closed and bounded set is sequentially
compact. Hint: If C is such a set, then C ⊆ I
_{0}≡∏_{i=1}^{n}. Now ifis a sequence in C, it must also be a sequence in I_{0}. Apply Problem 1 and Theorem 8.10.12. - Prove the extreme value theorem, a continuous function achieves its maximum and minimum
on any closed and bounded set C, using the result of Problem 2. Hint: Suppose λ =
sup. Then there exists⊆ C such that f→ λ. Now select a convergent subsequence using Problem 2. Do the same for the minimum.
- Let C be a closed and bounded set and suppose f : C → ℝ
^{m}is continuous. Show that f must also be uniformly continuous. This means: For every ε > 0 there exists δ > 0 such that whenever x,y ∈ C and< δ, it follows< ε. This is a good time to review the definition of continuity so you will see the difference. Hint: Suppose it is not so. Then there exists ε > 0 andandsuch that<but≥ ε. Now use Problem 2 to obtain a convergent subsequence. - From Problem 2 every closed and bounded set is sequentially compact. Are these the only sets which are sequentially compact? Explain.
- A set whose elements are open sets C is called an open cover of H if ∪C ⊇ H. In other
words, C is an open cover of H if every point of H is in at least one set of C. Show that if C
is an open cover of a closed and bounded set H then there exists δ > 0 such that whenever
x ∈ H, Bis contained in some set of C. This number δ is called a Lebesgue number. Hint: If there is no Lebesgue number for H, let H ⊆ I = ∏
_{i=1}^{n}. Use the process of chopping the intervals in half to get a sequence of nested intervals, I_{k}contained in I where diam≤ 2^{−k}diamand there is no Lebesgue number for the open cover on H_{k}≡ H ∩I_{k}. Now use the nested interval theorem to get c in all these H_{k}. For some r > 0 it follows Bis contained in some open set of U. But for large k, it must be that H_{k}⊆ Bwhich contradicts the construction. You fill in the details. - A set is compact if for every open cover of the set, there exists a finite subset of the open cover
which also covers the set. Show every closed and bounded set in ℝ
^{p}is compact. Next show that if a set in ℝ^{p}is compact, then it must be closed and bounded. This is called the Heine Borel theorem. Hint: To show closed and bounded is compact, you might use the technique of chopping into small pieces of the above problem. - Suppose S is a nonempty set in ℝ
^{p}. DefineShow that

Hint: Suppose dist

< dist. If these are equal there is nothing to show. Explain why there exists z ∈ S such that< dist+ ε. Now explain whyNow use the triangle inequality and observe that ε is arbitrary.

- Suppose H is a closed set and H ⊆ U ⊆ ℝ
^{p}, an open set. Show there exists a continuous function defined on ℝ^{p}, f such that f⊆, f= 0 if xU and f= 1 if x ∈ H. Hint: Try something likewhere U

^{C}≡ ℝ^{p}∖ U, a closed set. You need to explain why the denominator is never equal to zero. The rest is supplied by Problem 8. This is a special case of a major theorem called Urysohn’s lemma.

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