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Point Set Topology(Last Update : March 2, 2018) myweb.ttu.edu/bban [email protected] Real Analysis Byeong Ho Ban Mathematics and Statistics Texas Tech University Chapter 4. Point Set Topology(Last update : March 2, 2018) 1. If card(X)≥ 2, there is a topology on X that is T0 but not T1. Proof. (Jan 14th 2018) Let x 2 X be given and let (X; T ) be a topological space with the topology given as below. T = fE ⊂ X : E = X or x 62 Eg Then it is topology since ;;X 2 T , and for any subcollection S ⊂ T , [ [ U = X 2 T if X 2 S or x 62 U 2 T if X 62 S U2S U2S \ \ U = X if S = fXg otherwise x 62 U 2 T U2S U2S If y 6= z and y; z 2 X n fxg, then U = X n fx; yg is the open set containing y but not containing z: Thus, (X; T ) is T0 space. However, if we choose y 6= x, there is no open set containing x but not containing y since the open set containing x can only be X. 2. If X is an infinite set, the cofinite topology on X is T1 but not T2, and is first countable iff X is countable. Proof. (Jan 15th 2018) Let the cofinite topology be T . Note that, for any x 2 X, fxgc is open, so fxg is closed in T . Thus, c (X; T ) is T1. However, if x; y 2 X and x 6= y, for any open set U containing x but not y, since U should be finite, any set containing y and disjoint with U should be finite. Thus, there is no open set containing y and disjoint with U. Therefore, (X; T ) is not Hausdorff(T2) space. Suppose that (X; T ) is first countable and let x 2 X. Then there is a countable neighborhood base 1 c fNngn=1 at x. Since (X; T ) is T1, if y 2 fxg , fyg is an open neighborhood of x. So there is n 2 N such that N ⊂ fygc, so y 2 S N c. Thus, fxgc ⊂ S N c. Since each N c is finite, fxgc should be n n2N n n2N n n countable, so X = fxg [ fxgc is countable. Suppose that X is countable and let x 2 X. And let X n fxg = fx g N be defined as below. i i2N N = fX n fx1g ;X n fx1; x2g ;X n fx1; x2; x3g ;X n fx1; x2; x3; x4g ; · · · g Clearly, x 2 E for any E 2 N by construction. Also, if x 2 U and U 2 T , then U c should be finite. Thus, c k k 9k 2 N such that U ⊂ fxigi=1, so X n fxigi=1 ⊂ U, so N is the neighborhood base for T at x. Also, it is clear that N is countable. Therefore, (X; T ) is first countable. 1 2 3. Every metric space is normal. (If A,B are closed sets in the metric space (X; ρ)) consider the sets of points x where ρ(x; A) < ρ(x; B) or ρ(x; A) > ρ(x; B)). Proof. (Jan 14th 2018) Let U = fx 2 X : ρ(x; A) < ρ(x; B)g Q = fx 2 X : ρ(x; A) > ρ(x; B)g Note that A ⊂ U and B ⊂ Q by the definition of metric, and also note that U and Q are disjoint because of the below. U ⊂ fx 2 X : ρ(x; A) ≤ ρ(x; B)g = Qc We need to prove that U and Q are open sets. In order to prove that, it suffice to prove that ρ(x; E) is con- tinuous in x for any closed set E since, if f(x) = ρ(x; A)−ρ(x; B), f is continuous, and so U = f −1((−∞; 0)) and Q = f −1(0; 1) are open. For any x; y 2 X and > 0, without loss of generality, assuming that ρ(x; E) − ρ(y; E) ≥ 0, observe the below. ρ(x; E) − ρ(y; E) = inf fρ(x; z): x 2 Eg − inf fρ(y; z): z 2 Eg = inf fρ(x; z): x 2 Eg + sup {−ρ(y; z): z 2 Eg ≤ inf fρ(x; z): x 2 Eg − ρ(y; z) + (9z 2 E) ≤ ρ(x; z) − ρ(y; z) + ≤ ρ(x; y) + Since is arbitrary, jρ(x; E) − ρ(y; E)j ≤ ρ(x; y) Therefore, ρ(x; E) is Lipschitz function, so it is continuous. 4. Let X = R, and let T be the family of all subsets of R of the form U [ (V \ Q) where U,V are open in the usual sense. Then T is a topology that is Hausdorff but not regular. (In veiw of Exercise 3, this shows that a topology stronger than a normal topology need not be normal or even regular) Proof. (Jan 15th 2018)(Regular part is Incomplete) Note that R and ; are open set in usual sense. So R = R [ (;\ Q) and ; = ;[ (;\ Q) are in T . Also, if S = fQigi2A ⊂ T where A is any index set, any element Qi in S is Ui [ (Vi \ Q) where each Ui and Vi are open in usual sense. Then, ! ! ! [ [ [ [ Qi = Ui [ (Vi \ Q) = Ui [ Vi \ Q 2 T i2A i2A i2A i2A S S since i2A Ui and i2A Vi are open in usual sense. n Also, observe that, for any finite subcollection fQigi=1 ⊂ T , n n ! n ! ! \ \ \ Qi = Ui [ Vi \ Q 2 T i=1 i=1 i=1 Tn Tn since ( i=1 Ui) and ( i=1 Vi) are open in usual sense. Thus, T is a Topology on R 3 If x; y 2 R and x 6= y, since every metric space is normal, so Hausdorff space, there are two disjoint open sets(in usual sense) Q and U such that x 2 Q and y 2 U. And observe the below. Q = Q [ (;\ Q) U = U [ (;\ Q) Thus, Q; U 2 T , so (X; T ) is Hausdorff space. However, (X; T ) is not a regular space because of following counter example. c Note that Qc = (;[ (R \ Q)) is a closed set and let Q = U [ (V \ Q) be any open set containing Qc. c Then Q ⊂ U 5. Every separable metric space is second countable. Proof. (Jan 15th 2018) 1 Let X be a separable metric space and let fxngn=1 be the countable dense subset of X. Let B(xn) be defined as below. B(xn) = fB(xn; r): r 2 Qg where B(xn; r) is an open ball centered at xn and having radius r. Note that B(xn) is countable. Now, let B be defined as below. 1 [ B = B(xn) n=1 Also, note that B is countable since it is countable union of countable collections. Now, we will show that B is a base for X. 1 Let x 2 X be given. If x 2 fxngn=1, B(x) is clearly the neighborhood base at x. Otherwise, for any r 2 Q, there exists n 2 N such that x 2 B(xn; r). Thus, B(x) = fB(xn; r): r 2 Q; n 2 N; x 2 B(xn; r)g is the neighborhood base at x, since x 2 E 8E 2 B by the construction . Let x 2 U and U be open in the metric space. Since U is open, for any xi 2 fxngn2 there is r 2 Q such that B(xi; r) ⊂ U. Since x 2 U 1 N and fxngn=1 is dense in X, there is n 2 N such that x 2 B(xn; r) ⊂ U. Thus, B is the countable base for X and so it is second countable. 6. Let E = f(a; b]: −∞ < a < b < 1g a:E is a base for a topology T on R in which the members of E are both open and closed. b:T is first countable but not second countable. c:Q is dense in R with respect to T .(Thus the converse of Proposition 4.5 is false.) Proof. (Jan 15th 2018) (a) Let x 2 R be given. Then note that x 2 (x − 1; x] 2 E. Also, if (a; b]; (c; d] 2 E and 9x 2 (a; b] \ (c; d] = (c; d] (without loss of generality let c < b), then x 2 (c; d] 2 E. Thus, E is a base for a topology. Now, recall that every non-empty member of T is a union of the members of E. Thus, since any member in E is a union of itself, every member in E is open. Also, for any (a; b] 2 E,observe the below. 1 [ (a; b]c = (b + n − 1; b + n] [ (a − n; a − n + 1] 2 T n=1 4 Therefore, every member of E is both open and closed. (b) Let x 2 R be given. Let B(x) be given as below. 1 B(x) = (x − ; x]: n 2 n N 1 Then, for all n 2 N, x 2 (x − n ; x]. Now observe that if x 2 E 2 T , there is a subcollection S ⊂ T such S 1 that E = (a;b]2S (a; b] and x 2 (a; b] for some (a; b] 2 S. Let x − a = d then there is n 2 N such that n < d 1 so (x − n ; x] ⊂ (a; b]. Lastly, note that B(x) is countable.
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