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More Open Sets and Topological Compactness James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University September 13, 2016 Outline More Set Theory Topological Compactness There are more things we can talk about with sets. So let’s get started. You’ll see that the way we prove propositions in set theory is very different from the way we do this in other parts of this class. All part of your growth! Example Let A and B be open sets. Then A ∪ B is open too. Solution Remember A ∪ B = {c : a ∈ A and / or b ∈ B}. To show this result, we must show an arbitrary c in A ∪ B is an interior point. Now for such a c it is clear c is in A only or it is in B only or it is in both. All that matters here is that c is in one or the other. If c ∈ A, then c is an interior point and there is a Br (c) ⊂ A ⊆ A ∪ B. And if c ∈ B, again c is an interior point and there is a Br (c) ⊂ B ⊆ A ∪ B. So in all cases, c is an interior point and so A ∪ B is open. A countable union is one where the index set is an RII of Z. Usually N is what we use, but it could be such a subset. Example The countable union of open sets is open. Solution Let our RII set be N just for convenience of notation and let the individual open sets here be Oi . We let A = ∪∞ n=1 On = ∪n∈N On denote the countable union. Let x be in the union. Then x is in at least one OQ . Since OQ is open, x is an interior point of OQ . So there is a Br (x) ⊂ OQ . But OQ ⊆ ∪n∈N On . Hence, x is an interior point of the countable union. Since x was arbitrary, the countable union is open. Comment For a countable union of sets, we usually just write ∪n On . Example Unions need not be over countable index sets. A good example is this. Let S = [1, 3) which has an infinite number of points. At each x ∈ S, pick the radius rx > 0 and consider the set A = ∪x∈S Brx . This index set is not an RII so indeed is an example of what is called a uncountable set. To be precise, a countable set Ω is one for which there is a map f which assigns each element in Ω one and only one element of N. A RII is countable because nk is assigned to k in N and this assignment is unique. Hence there is a 1 − 1 and onto map from the RII to N. Example The rational numbers are countable too. We will do the construction for positive fractions only but you should be able to see how to extend it to negative fractions also. We can set up the map like this: Example (Continued): Write down all the fraction in a table: 1/1 2/1 3/1 4/1 .. . 1/2 2/2 3/2 4/2 .. . 1/3 2/3 3/3 4/3 .. . 1/4 2/4 3/4 4/4 .. . 1/5 2/5 3/5 4/5 .. . 1/6 2/6 3/6 4/6 .. . 1/6 2/6 3/6 4/6 .. . ... ... ... ... .. . 1/n 2/n 3/n 4/n .. . ... ... ... ... .. . n/1 .. . n/2 n/3 .. .. . . n/4 .. . n/5 .. . n/6 .. . n/6 .. . ... .. . n/n .. . ... .. . Example (Continued): The mapping is then defined by the numbers in parenthesis chosen in diagonal sweeps: 1/1 (1) 2/1 (3) 3/1 (4) 4/1 (10) .. . n/1 .. . 1/2 (2) 1/3 (6) 2/2 (5) 2/3 (8) 3/2 (9) 3/3 4/2 4/3 .. .. . . n/2 n/3 .. .. . . 1/4 (7) 2/4 3/4 4/4 .. . 1/5 2/5 3/5 4/5 .. . 1/6 2/6 3/6 4/6 .. . 1/6 2/6 3/6 4/6 .. . ... ... ... ... .. . 1/n 2/n 3/n 4/n .. . ... ... ... ... .. . n/4 .. . n/5 .. . n/6 .. . n/6 .. . ... .. . n/n .. . ... .. . Example (Continued): eliminating the fractions, it is easier to see the pattern: (1) (2) (6) (7) (3) (5) (8) (4) (9) (10) (12) .. . (13) .. . (14) .. . (11) (15) . . . .. . ... ... ... ... I So the rationals Q are countable too. We often call this countably infinite to distinguish it from the case where the index set is countably finite like {1, 2, 3, 4, 5}. I So [1, 5] would not be countable as it contains lots of numbers which are not fractions. Example The uncountable union of open sets is open. Solution Let the uncountable index set be Ω. and let the individual open sets here be Oα where α is a member of the index set Ω. We let A = ∪α∈Ω Oα = ∪α Oα . denote the uncountable union. Let x be in the union. Then x is in at least one Oβ . Since Oβ is open, x is an interior point of Oβ . So there is a Br (x) ⊂ Oβ . But Oβ ⊆ ∪α Oα . Hence, x is an interior point of the uncountable union. Since x was arbitrary, the uncountable union is open. Example The intersection of a finite number of open sets is open too. Solution Let A = ∩N n=1 On be a finite intersection of open sets. Let x be in this intersection. Then x is an interior point of all N sets. Hence, we have N circles: Br1 (x) ⊂ O1 through BrN (x) ⊂ ON . If we let r = min{r1 , . . . , rN }, then Br (x) is contained in each of the N sets. Hence A is open. Example The intersection of a countable number of open sets need not be open. Solution Take the sets (1 − 1/n, 4 + 1/n). We claim ∩n (1 − 1/n, 4 + 1/n) = [1, 4], a closed set. To prove these sets are equal, we first show ∩n (1 − 1/n, 4 + 1/n) ⊂ [1, 4] and then [1, 4] ⊆ ∩n (1 − 1/n, 4 + 1/n). (∩n (1 − 1/n, 4 + 1/n) ⊂ [1, 4]): Let x ∈ ∩n (1 − 1/n, 4 + 1/n). Then, x ∈ (1 − 1/n, 4 + 1/n) ⊂ [1, 4] for all n. Thus, x ∈ [1, 4]. ([1, 4] ⊆ ∩n (1 − 1/n, 4 + 1/n)): Let x ∈ [1, 4]. Then x ∈ (1 − 1/n, 4 + 1/n) for all n implying x ∈ ∩n (1 − 1/n, 4 + 1/n). Comment So the property of being open can be violated if a countably infinite number of operations are used. We need lots of small results in this kind of thinking. Here is one: Example ∩N n=1 Bn C C = ∪N n=1 Bn for any sets. Solution C ∩N ⊆ ∪N BCn n=1 Bn n=1 C If x ∈ ∩N , then x 6∈ ∩N n=1 Bn n=1 Bn . That tells us x 6∈ Bn for at least C one k. Hence, x ∈ BCk implying x ∈ ∪N n=1 Bn . C C N ∪N B ⊆ ∩ B . n n=1 n n=1 C C If x ∈ ∪N n=1 Bn , there is an index Q with x ∈ BQ . Thus x 6∈ BQ which C N N tells us x 6∈ ∩n=1 Bn . Hence, x ∈ ∩n=1 Bn . Example The intersection of a finite number of closed sets is closed too. Solution Let A = ∩N n=1 Dn be a finite intersection of closed sets. We now know C C C = ∪N AC = ∩N n=1 Dn n=1 Dn . Since each Dn is closed, Dn is open. We also know the finite union of open sets is open. Then AC is open implying A is closed. Comment C ∞ C We can also prove (∩∞ n=1 Bn ) = ∪n=1 Bn for any sets Bn and prove the intersection of a countable number of closed sets is still closed. Example The union of a countable number of closed sets need not be closed. Solution We claim ∪n [1 + 1/n, 4 − 1/n] = (1, 4), an open set. (∪n (1 + 1/n, 4 − 1/n) ⊂ (1, 4) ): Let x ∈ ∪n [1 + 1/n, 4 − 1/n]. Then, x ∈ [1 + 1/Q, 4 − 1/Q] for some Q. Thus, the distance from x to 1 or 4 is positive. So x ∈ (1, 4). ( (1, 4) ⊆ ∪n [1 + 1/n, 4 − 1/n]): If x ∈ (1, 4), x is an interior point and so there is Br (x) ⊂ (1, 4). That is, 1 < x − r < x < x + r < 4. Let the distance from x − r to 1 be d1 and the distance from x + r to 4 be d2 . Let d = (1/2) min{d1 , d2 }. Choose N so that 1/N < d. Then x ∈ [1 + 1/N, 4 − 1/N] and so x ∈ ∪n [1 + 1/n, 4 − 1/n]. Comment The property of closed can be violated using countably infinite operations. Definition We say a collection of open sets U = {Oα : α ∈ Ω} indexed by an arbitrary set Ω ( so it can be finite, countably infinite or uncountable) is a open cover of a set A if A ⊆ ∪α∈Ω Oα . We say the collection {Oα1 , . . . , OαN } from U , for some natural number N, is a finite subcover or fsc of A if A ⊆ ∪N n=1 Oαn . We are now ready for the big new idea! Definition We say a set A is topologically compact if every open cover of A has a finite subcover. We need to show this idea in our setting < is equivalent to sequential compactness. Theorem A is topologically compact ⇒ A is closed and bounded. Proof (⇒): Let’s assume A is topologically compact. If we take the union of all Br (x) for a fixed radius r for all x in S. this is an open cover. Since A is topologically compact, this cover has a fsc which contains A. We can express the fsc as V = {Br (x1 ), . . . , Br (xN )} for some N. Now let x ∈ A. Then x ∈ Br (xj ) for some 1 ≤ j ≤ N. Thus, |x − xj | < r ⇒ |x| < r + |xj | using the backwards triangle inequality. And in turn, r + |xj | ≤ max{r + |x1 |, . . . , r + |xN |} = B. So |x| ≤ B for all x ∈ A implying A is bounded. To show A is closed, we will show AC is open. Let x ∈ AC . Given any y ∈ A, since x 6= y , the two points are separated by a distance d xy . Thus, Bd xy /2 (y ) and Bd xy /2 (x) have no points in common. Proof The collection of all the circles Bd xy /2 (y ) for y ∈ A gives an open cover U of A which must have a fsc, V = {Bd1xy (y1 ), . . . , BdNxy (yN )}, so that xy A ⊆ ∪N n=1 Bdn (yn ). xy xy Let W = ∩N n=1 Bdn (x). The circles Bdn (x) are all separate ( we say disjoint ) from Bdnxy (y ). Hence, points in the intersection are outside the union of the fsc V . Then W is open circle as it is a finite intersection of open circles and it is outside of the union over the fsc V which contains A. Since each of these sets is contained in AC , we see W ⊂ AC implying x is an interior point of AC . Since x was arbitrary, we see AC is open. Hence A is closed. Theorem The closed and bounded finite interval [a, b], a < b, is topologically compact. Proof This proof is similar to how we proved the Bolzano Weierstrass Theorem (BWT). So look at the similarities! Assume this is false and let J0 = [a, b]. Label this interval as J0 = [α0 , β0 ]; i.e. α0 = a and β0 = b. The length of J0 is b − a = B which we will call `0 . Assume there is an open cover U which does not have a fsc. Now divide J1 into two equal pieces b−a {x0 = α0 , x1 = α0 + 1 b−a 2 , x2 = a + 2 2 = b}. We have [α0 , β0 ] = [x0 , x1 ] ∪ [x1 , x2 ] At least one of these intervals can not be covered by a fsc as otherwise the entire interval [α0 , β0 ] has a fsc which we have assumed is not true. Call this interval J1 and note the length of J1 is `1 = (1/2)`0 = B/2. Proof Label this interval as J1 = [α1 , β1 ]. We can continue in this way by induction. Assume we have constructed the interval Jn = [αn , βn ] in this fashion with length `n = (1/2)`n−1 = B/2n and there is no fsc of Jn . Now divide Jn into 2 equal pieces as usual {x0 = αn , x1 = αn + 1 `n , x2 = αn + 2 `n = βn }. These points determine 2 subintervals and at least one of them can not be covered by a fsc. Call this interval Jn+1 = [αn+1 , βn+1 ] which has length `n+1 = (1/2)`n like required. We can see each Jn+1 ⊂ Jn by the way we have chosen the intervals and their lengths satisfy `n → 0. Using the same sort of arguments that we used in the proof of the BWT for Sequences, we see there is a unique point z which is the sup(αn ) = inf(βn ). Another way to say this in our new set language is that ∩∞ n=1 [αn , βn ] = {z}! Proof Since z in in J1 , there is a O1 in U with z ∈ O1 because U covers A. Also, since O1 is open, there is a circle Br1 (z) ⊂ O1 . Now choose K so that `K < `1 . Then, Jk is contained in O1 for all k > K . This says O1 is a fsc of Jk which contradicts our construction process. Hence, our assumption that there is an open cover with no fsc is wrong and we have [a, b] is topologically compact. Homework 12 12.1 If A and B are nonempty sets, show all the details of the proof for (A ∪ B)C = AC ∩ B C . 12.2 Find an open cover with no fsc for the set (1, 3). Explain why this shows (1, 3) is not topologically compact. 12.3 If A and B are nonempty sets, show all the details of the proof for (A ∩ B)C = AC ∪ B C .