Open and Closed Sets

3.2-1: Open and Closed Sets

3.2.2 Let 2 A = (−1)n + : n = 1, 2, 3, ··· and B = {x ∈ : 0 < x < 1}. n Q

Answer the following questions for each set.

a) What are the limit points? A has limit points at −1 and 1. The limit points of B form the closed interval [0, 1]. b) Is the set open? Closed? A is closed but not open, because no neighborhood is contained in A. B is open, because if 0 < x < 1, we can ﬁnd a neighborhood of x that ﬁts inside of (0, 1). B is not closed because it does not contain limit points 0, and 1. c) does the set contain any isolated points? Every point of A except −1 and 1 are isolated. Consider the point 1 1 1 + k (where n = 2k). The closest points in A are 1 + k−1 and 1 2 1 + k+1 . A similar argument could be made for −1 + 2k+1 . B does not contain any isolated points. d) Find the closure of the set. Since A is closed, the closure of A is A. The closure of B is [0, 1].

3.2.3 Decide whether the following sets are open, closed or neither.

a) Q: Q is not open because every neighborhood must contain irrational points. Every point in the set satisﬁes the fact there is no -neighborhood contained in Q. √Q is also not closed because every irrational number (for example 2) is a limit point of the set. b) N: Notice that every point in N is isolated, because if n ∈ N, N ∩ (n − 1/2, n + 1/2) = {n}. Thus, N is not open. N is closed because it has no limit points, and therefore contains all of its limit points. c) {x ∈ R : x 6= 0}: Let x be an element of this set. If x > 0, then the interval (x/2, x + 1) is in the set, because x/2 > 0. Similarly,

1 if x < 0, (x − 1, x/2) is in the set. Thus, the set is open. Notice 1 1 that n is in the set, but ( n ) → 0. Thus 0 is a limit point. Hence, the set is not closed. d) {1 + 1/4 + 1/9 + ··· + 1/n2 : n ∈ N}. Notice that this set contains the terms of an increasing sequence. Thus, all of the terms are isolated and the set cannot be open. However, it cannot be closed either, because the sequence has a limit (π2/6). e) {1 + 1/2 + 1/3 + ··· + 1/n : n ∈ N}. This is also an increasing sequence, and the terms are isolated. Thus it is not open. The sequence does not converge, and therefore it has no limit points. Thus it is closed.

3.2.5 Prove Theorem 3.2.8: A set F ⊆ R is closed if and only if every Cauchy sequence contained in F has a limit that is also an element of F .

Proof. Let F be any subset of R. Suppose F is closed. If (an) is a Cauchy sequence contained in F , then we know that (an) → a for some a. By Theorem 3.2.5, a ∈ F . On the other hand, suppose every Cauchy sequence in F has a limit that is also in F . Let b be a limit point of F . By Theorem 3.2.5, there is a sequence (bn) that converges to it. Since convergent sequences are also Cauchy, we see that b ∈ F . Hence, F is closed.

3.2.7 Given A ⊆ R, let L be the set of all limit points of A. a) Show that the set L is closed. Proof. Suppose a is a limit point of L. Then, there is a sequence (an) → a with an → a. Let > 0 be given. There is an N1 such that if n > N1, |an − a| < /2.

Since each element an is a limit point, we can ﬁnd an element bn ∈ A with |bn −an| < 1/n. Choose N2 so that 1/N2 < /2. Note that this means if n > N2, then 1/n < /2.

Let N = max(N1,N2). If n > N, then 1 |b −a| = |b −a +a −a| ≤ |b −a |+|a −a| < + < + = . n n n n n n n n 2 2 2

Thus, (bn) → a, and so a is a limit point of A, and is in L. Thus L is closed.

2 b) If x is a limit point of A ∪ L, then x is a limit point of A.

Proof. Suppose x is a limit point of A ∪ L. Let (cn) be a sequence converging to x. Either infinitely many elements of the sequence come from A or infinitely many come from L. In the first case, we can find a subsequence of (cn) that lies in A. Since the sequence converges, the subsequence converges to x, and so x is a limit point. In the second case, we can find a subsequence that lies in L, but since L is closed, x ∈ L and is a limit point of A.

Prove Theorem 3.2.12

Proof. From the observation above, the closure of A contains all of its limit points and is therefore closed. Suppose B is a closed subset with A ⊆ B ⊆ A. We will show B = A. Let x ∈ A. Then, x ∈ A or x ∈ L, where L is the set of limit points of A. If x ∈ A, then x ∈ B. If x ∈ L, then there is a sequence (xn) in A that converges to x. However, the sequence is also in B, and since B is closed, it contains the limit. Thus, x ∈ B. This shows A ⊆ B, and so B = A. Thus, A is the smallest closed set that contains A.

3.2.8 Assume A is an open set and B is a closed set. Determine if the following sets are deﬁnitely open, deﬁnitely closed, both or neither.

a) A ∪ B. Definitely closed from Theorem 3.2.12. b) A\B: Definitely open. Let x ∈ A\B. Then x cannot be a limit point of B because B is closed and contains all of its limit points. Thus, there must be an -neighborhood of x that contains no points in B. Since A is open, there is an -neighborhood of x that lies in A. The intersection of these two neighborhoods give us an -neighborhood that lies entirely in A\B. c) (Ac ∪ B)c: Definitely open. Since Ac is closed, and a union of closed sets is closed, Ac ∪ B is closed. Thus its complement is open. d) (A ∩ B) ∪ (Ac ∩ B). Definitely closed. This is just the set B.

3 c e) A ∩ Ac Deﬁnitely open. Since A is open, Ac is closed. Thus, c Ac = Ac. Also, A ⊆ A, so A ⊆ Ac, and the intersection is just c A , which is open.