Compact Sets. James K. Peterson. September 15, Department of Biological Sciences and Department of Mathematical Sciences Clemson University

Transcription

1 Compact Sets James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University September 15, 2017

2 Outline 1 Closed Sets 2 Compactness 3 Homework

3 Closed Sets Recall a set S is closed if its complement is open. We can characterize closed sets using the closure too. Theorem Let S be a set of real numbers. Then S is a closed set S = S. ( ): We assume S is a closed set. Then S C is open. We wish to show S = S. Now S contains S already, so we must show all boundary points of S are in S. Let s assume x is a boundary point of S which is in S C. Then, since S C is open, x is an interior point and there is an r > 0 so that B r (x) S C. But B r (x) must contains points of S and S C since x is a boundary point. This is impossible. So our assumption that x was in S C is wrong. Hence, S S. This shows S = S.

4 Closed Sets ( ): We assume S = S and show S is closed. To do this, we show S C is open. It is easy to see the complement of the complement gives you back the set you started with, so if S C is open, by definition, (S C ) C = S is closed. We will show any x in S C is an interior point. That will show S C is open. Let s do this by contradiction. To show S C is open, we show that for any x in S C, we can find a positive r so that B r (x) S C. So let s assume we can t find such an r. Then every B r (x) contains points of S and points of S C. Hence, x must be a boundary point of S. Since S = S, this means S S which implies x S. But we assumed x was in S C. So our assumption is wrong and we must be able to find an r > 0 with B r (x) S C. Hence S C is open which means S is closed.

5 Closed Sets Let s look at some set based proofs. Theorem If A and B are open, so is A B. Let p A B. Then p A or p B or both. If p A, p is an interior point because A is open. So there is an r > 0 so that B r (p) A. This says B r (p) A B too. A similar argument works for the case p B. Since p is arbitary, all points in A B are interior points and we have shown A B is open.

6 Closed Sets Theorem (A B) C = A C B C and (A B) C = A C B C. (A B) C A C B C : If p (A B) C, p A B. Thus p A and p B implying p A C and p B C ; i.e. p A C B C. Since p is arbitary, this shows (A B) C A C B C A C B C (A B) C : If p A C B C, p A C and p B C. So p A and p B. Thus p A B telling us p (A B) C. Since p is arbitrary, this shows A C B C (A B) C. The other one is left for you as a homework.

7 Closed Sets Let s look at some set based proofs. Theorem If A and B are closed, so is A B. Let s show A B is closed by showing (A B) C is open. Let p (A B) C = A C B C. Then p A C and p B C. So there is a radius r 1 so that B r1 (p) A C and there is a radius r 2 so that B r2 (p) B C. So if r = min{r 1, r 2 }, B r (p) A C B C. Thus, p is an interior point of (A B) C. Since p is arbitary, all points in (A B) C are interior points and we have shown (A B) C is open.

8 Compactness Definition Let S be a set of real numbers. We say S is sequentially compact or simply compact if every sequence (x n ) in S has at least one subsequence which converges to an element of S. In other words Given (x n ) S, (x nk ) (x n ) and an x S so that x nk x. Theorem A set S is sequentially compact S is closed and bounded. ( ): We assume S is sequentially compact and we show S is both closed and bounded. First, we show S is closed. Let x S. If we show S = S, by the previous theorem, we know S is closed. Since S S, this means we have to show all x S are also in S.

9 Compactness Now if x S, x is a boundary point of S. Hence, for all B 1/n (x), there is a point x n S and a point y n S C. This gives a sequence (x n ) satisfying x n x < 1/n for all n which tells us x n x. If this sequence is the constant sequence, x n = x, then we have x S. But if (x n ) satisfies x n x for all n, we have to argue differently. In this case, since S is sequentially compact, (x n ) has a subsequence (x nk ) which converges to some y S. Since limits of a sequence are unique, this says x = y. Since y is in S, this shows x is in S too. Hence, since x was arbitrarily chosen, we have S S and so S = S and S is closed.

10 Compactness Next, we show S is bounded by contradiction. Let s assume it is not bounded. Then given any positive integer n, there is an x n in S so that x n > n. This defines a subsequence (x n ) in S. Since S is sequentially compact, (x n ) has a convergent subsequence (x nk ) which converges to an element y in S. But since this subsequence converges, this tells us (x nk ) is bounded; i.e. there is a positive number B so that x nk B for all n k. But we also know x nk > n k and so n k < x nk B for all n k. This is impossible as n k and B is a finite number. Thus, our assumption that S was unbounded is wrong and so S must be bounded.

11 Compactness ( ): We assume S is closed and bounded and we want to show S is sequentially compact. Let (x n ) be any sequence in S. Since S is a bounded set, by the Bolzano Weierstrass Theorem for Sequences, since (x n ) is bounded, there is at least one subsequence (x nk ) of (x n ) which converges to a value y. If this sequence is a constant sequence, then x nk = y always which tells us y is in S too. On the other hand, if it is not a constant sequence, this tells us y is an accumulation point of S. Now if y was not in S, then y would be in S C and then since x nk y, every B r (y) would have to contain points of S and points of S C. This says y has to be a boundary point. But since S is closed, S contains all its boundary points. Thus the assumption y S C can t be right and we know y S. This shows the arbitrary sequence (x n ) in S has a subsequence which converges to an element of S which shows S is sequentially compact.

12 Compactness Example Let S = {2} (3, 4). Note S is not closed as 3 and 4 are boundary points not in S. Hence, S is not sequentially compact either. Example Let S = {2} [5, 7]. Then S is closed as it contains all of its boundary points and so S is also sequentially compact. Example Let S = [1, ). Then S is not bounded so it is not sequentially compact.

13 Compactness Theorem If S is sequentially compact, then any sequence (x n ) in S which converges, converges to a point in S. Since S is sequentially compact, such a sequence does have a convergent subsequence which converges to a point in S. Since the limit of the subsequence must be the same as the limit of this convergent sequence, this tells us the limit of the sequence must be in S.

14 Compactness Example Let S be a nonempty and bounded set of numbers. Then α = inf(s) and β = sup(s) are both finite numbers. Given any r n = 1/n, the Supremum Tolerance Lemma tells us there is a x n S so that β 1/n < x n β for all n. Thus, rearranging this inequality, we have 1/n < x n β 0 < 1/n which says x n β < 1/n for all n. This tells us x n β. We can do a similar argument for the infimum α and so there is a sequence (y n ) in S so that y n α. Now in general, we don t know if α or β are the minimum and maximum of S, respectively. But if we also knew S was sequentially compact, we can say more. By the Theorem above, since x n β and (x n ) S, we know β S. Hence β is a maximum. A similar argument shows α is a minimum.

15 Compactness Definition Let S be a nonempty and bounded set of numbers. Then α = inf(s) and β = sup(s) are both finite numbers. A sequence (y n ) S which converges to inf(s) is called a minimizing sequence and a sequence (x n ) S which converges to sup(s) is called a maximizing sequence. What we want are conditions that force minimizing and maximizing sequences to converge to points inside S; i.e. make sure the minimum and maximum of S exist. One way to make this happen is to prove S is sequentially compact.

16 Homework Homework Let S = {3} (0, 3). Explain why S is or is not sequentially compact Let S = (, 3]. Explain why S is or is not sequentially compact Let S = (x n ) = (sin(n)) n 1. Since (x n ) [ 1, 1], why do you know that (x n ) has a subsequence which converges to a point x in [ 1, 1]? Can you see why it is hard to decide if S is a closed set? This question is just to show you deciding if a set is closed or open can be difficult Prove for any sets A and B, (A B) C = A C B C If A and B are open, prove A B is open also.