real math analysis please hep me solve problem 10.3.5 10.3 The Bolzano-Weierstrass TheoremOnce we introduced the Nested Interval Property (Axiom 10.1.1), theIntermediate Value Theorem (Theorem 10.2.1) followed pretty readily. Theproof of Extreme Value Theorem (Theorem 10.4.8) takes a bit more work. Firstwe need to show that a function that satisfies the conditions of the EVT isbounded.Theorem 10.3.1. A continuous function defined on a closed, bounded intervalmust be bounded. That is, let f be a continuous function defined on [a,b]. Thenthere exists a positive real number B such that |f (x)| n. (Otherwise n would be a bound for f.) IF the sequence(xn) converged to something in [a, b], say c, then we would have ourcontradiction. Indeed, we would have lim, 00 *n = c. By the continuity of f at cand Theorem 9.2.1 of Chapter 9, we would have lim, 00 f(xn)= f(c). Thiswould say that the sequence (f(xn)) converges, so by Lemma 7.2.7 ofChapter 7, it must be bounded. This would provide our contradiction, as we had|f(xn)| n, for all positive integers n. QED?This would all work well except for one little problem. The way it wasconstructed, there is no reason to expect the sequence (xn) to converge toanything and we can't make such an assumption. That is why we emphasizedthe IF above. Fortunately, this idea can be salvaged. While it is true that thesequence (xn) may not converge, part of it will. We will need the followingdefinition.Definition 10.3.2. Let (nx) be a strictly increasing sequence of positiveintegers; that is, n1

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real math analysis please hep me solve problem 10.3.5 10.3 The Bolzano-Weierstrass TheoremOnce we introduced the Nested Interval Property (Axiom 10.1.1), theIntermediate Value Theorem (Theorem 10.2.1) followed pretty readily. Theproof of Extreme Value Theorem (Theorem 10.4.8) takes a bit more work. Firstwe need to show that a function that satisfies the conditions of the EVT isbounded.Theorem 10.3.1. A continuous function defined on a closed, bounded intervalmust be bounded. That is, let f be a continuous function defined on [a,b]. Thenthere exists a positive real number B such that |f (x)| < B for all x E [a, b].Sketch of Alleged Proof: Let's assume, for contradiction, that there is no suchbound B. This says that for any positive integer n, there must exist x, E [a, b]such that |f(xn)| > n. (Otherwise n would be a bound for f.) IF the sequence(xn) converged to something in [a, b], say c, then we would have ourcontradiction. Indeed, we would have lim, 00 *n = c. By the continuity of f at cand Theorem 9.2.1 of Chapter 9, we would have lim, 00 f(xn)= f(c). Thiswould say that the sequence (f(xn)) converges, so by Lemma 7.2.7 ofChapter 7, it must be bounded. This would provide our contradiction, as we had|f(xn)| > n, for all positive integers n. QED?This would all work well except for one little problem. The way it wasconstructed, there is no reason to expect the sequence (xn) to converge toanything and we can't make such an assumption. That is why we emphasizedthe IF above. Fortunately, this idea can be salvaged. While it is true that thesequence (xn) may not converge, part of it will. We will need the followingdefinition.Definition 10.3.2. Let (nx) be a strictly increasing sequence of positiveintegers; that is, n1 < n2 < n3 < …. If (xn)is a sequence, then= (xn1, Xn2, Xng, ...) is called a {subsequence} of (xn).The idea is that a subsequence of a sequence is a part of the sequence, (xn),which is itself a sequence. However, it is a little more restrictive. We can chooseany term in our sequence to be part of the subsequence, but once we choosethat term, we can't go backwards. This is where the condition n1 < n2 < n3 < ·…comes in. For example, suppose we started our subsequence with the term x100-We could not choose our next term to be æ99. The subscript of the next termwould have to be greater than 100. In fact, the thing about a subsequence is thatit is all in the subscripts; we are really choosing a subsequence (ng) of theseguence of subscripts (n) in (xn). Problem 10.3.5. Suppose lim,→∞ xn = c. Prove that limk→00 Xng = c for anysubsequence (xng) of (xn).Hint.First prove that n. > k.A very important theorem about subsequences was introduced by BernhardBolzano and, later, independently proven by Karl Weierstrass. Basically, thistheorem says that any bounded sequence of real numbers has a convergentsubsequence.

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Problem 10.3.5. Suppose limn→∞ ¤n = c. Prove that lim→ xn = subsequence (ng) of (xn). ▼ Hint. First prove that n. > k. c for any