VIEWS: 20 PAGES: 3 CATEGORY: Legal POSTED ON: 3/15/2010 Public Domain
Math 320 Introduction to Real Analysis II Assignment 2 Solutions Q1. Let {Vα }α∈A be an open cover of K. We have 0 ∈ K so there should exist one α0 in A, such that 0 ∈ Vα0 . Vα0 is an open set so this implies that for some positive , we should have (− , ) ⊂ Vα0 . There are just ﬁnite number of natural numbers n, such that n ≤ 1 . For each such n, there is 1 an αn such that n ∈ Vαn . We take Vα0 , and these ﬁnite collection and we get a ﬁnite subcover of at most 1 + 1 open sets. Therefore K is compact. Q2. We use the previous problem, K was compact and also it had just one limit point which was 0. Q3. We take the set of all open segments (1/n, 1) for n ∈ N. This is a countable open cover of (0, 1) without any ﬁnite subcouver, because among any ﬁnite number of natural number there is a biggest number so if (1/nk , 1) for 1 ≤ k ≤ t is a ﬁnite subcover then by taking the maximum among nk ’s, we have: M := M ax{nk |1 ≤ k ≤ t} , (1/nk , 1) = (1/M, 1) 1≤k≤t As we see this won’t cover the whole (0, 1) interval. √ √ √ √ Q4. Let S = ( 2, 3) ∪ (− 3, − 2), then E = {p ∈ Q|p ∈ S}. Therefore E is bounded in Q, and since Q is dense in R every point of Q is a limit point of Q (here Q is a metric space). Thus E is closed. To prove that E is not compact, we form an open cover of E as follows: {Gα } = {Nr (p)|p ∈ E&(p − r, p + r) ⊂ S} If E is compact then there are ﬁnitely many indices α0 , . . . , αn such that E⊂ G αi 1≤i≤n √ we take p = max1≤i≤n pi , then p is the closest point to 3. But Nr (p) lies √ in E, so [p + r, 3) cannot be covered since Q is dense in R. Therefore E is not compact. It’s easy to see also E is open in Q. 1 Q5. Let X be a separable metric space, and S be a countable dense subset of X. Let for each s ∈ S, {Nr (s)}r∈Q be a collection of open balls with center at s and radius rational numbers. Therefore this is a countable collection of open sets around s. It’s clear that any open set which contains s should contain one of these open balls. Now we take all this collection of open balls for each s in S. To prove this is a base. suppose x is point in X and U is an open set containing x. U is open so There exist an positive such that x ∈ N (x) ⊂ U . S is dense in X so we should have one s0 in S such that s0 ∈ N 4 (x). Now it is clear that x ∈ N 2 (s0 ) ⊂ N (x) ⊂ U (why?). This complete our proof, as we built a base. Q6. Fix δ > 0, and pick x1 ∈ X. Having chosen x1 , . . . , xj ∈ X, choose xj+1 , if possible, so that d(xi , xj+1 ) ≥ δ for i = 1, . . . , j. If this process cannot stop, then consider the set A = {x1 , x2 , . . . , xk }. If p is a limit point of A, then a neighborhood Nδ/3 (p) of p contains a point q = p such that q ∈ A. q = xk for only one k ∈ N . If not, d(xi , xj ) ≤ d(xi , p) + d(xj , p) ≤ δ/3 + δ/3 < δ, and it contradicts the fact that d(xi , xj ) ≥ δ for i = j. Hence, this process must stop after ﬁnite number of steps. Suppose this process stop after k steps, and X is covered by Nδ (x1 ), Nδ (x2 ), . . . , Nδ (xk ) that is, X can therefore be covered by ﬁnite many neighborhoods of radius δ. Take δ = 1/n(n = 1, 2, 3, . . .), and consider the set A of the centers of the corresponding neighborhoods. Fix p ∈ X. Suppose that p is not in A, if Nr (p) be a neighborhood of p . Note that Nr/2 (p) can be covered by ﬁnite many neighborhoods Ns (x1 ), . . . , Ns (xk ) of radius s = 1/n where n = [2/r] + 1 and xi ∈ A for i = 1, . . . , k. Hence, d(x1 , p) ≤ d(x1 , q) + d(q, p) ≤ r/2 + s < r where q ∈ Nr/2 (p) ∩ Ns (x1 ). Therefore, x1 ∈ Nr (p) and x1 = p since p is not in A. Hence, p is a limit point of A if p is not in A, that is, A is a countable dense subset, that is, X is separable. Q7. By Exercise 23, 24 on pg. 45 in Rudin, X is separable. Hence X has a countable base. It follows that every open cover of X has a countable subcover {Gn }, n = 1, 2, . . . . This is because if {Bn } denote the countable basis of X, and A be an open covering of X. For each positive integer n for which it is possible to choose an element An of A containing the basis element Bn . The collection A of the sets An is countable, since it is indexed with a subset J of positive integer numbers. Furthermore it covers X: given point x in X we can choose an element A of A containing x. Since A is open there is a basis element Bn such that x ∈ Bn ⊂ A. Because Bn lies in an 2 element of A, the index n belongs to the set J, so An is deﬁned; since An contains Bn , it contains x. Thus A is a countable subcollection of A that covers x. Suppose X is not compact. Therefore, no ﬁnite subcollection of {Gn } covers X. For each n, deﬁne Fn := X\(G1 ∪ · · · ∪ Gn ). Therefore, Fn = ∅, otherwise G1 ∪ · · · ∪ Gn ⊇ X. Also, we have that F1 ⊇ F2 ⊇ · · · , but since ∞ Gn = X, thus n=1 ∞ n=1 Fn = ∅. For each n, pick an element in Fn and form a set E. Then E is inﬁnite, and so E has a limit point, say x ∈ E . Then x ∈ E ⊆ X would imply that there exists N such that x ∈ GN . But since x is a limit point, then there exists y such that y ∈ GN ∩ E. Since y ∈ GN , then y ∈ FN , FN +1 , . . . But y ∈ E would mean that / y ∈ F1 ∪ F2 ∪ · · · ∪ FN −1 . But |GN ∩ E| ≤ N + 1. Since there are only ﬁnitely many points in GN ∩ E, so x can’t be a limit point of E. Contradiction. 3