Solutions to Tutorial 8 (Week 9)

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1 The University of Sydney School of Mathematics and Statistics Solutions to Tutorial 8 (Week 9) MATH3961: Metric Spaces (Advanced) Semester 1, 2018 Web Page: Lecturer: Florica C. Cîrstea Material covered (1) Separation properties of topological spaces (2) Urysohn s Lemma (3) The Tietze Extension Theorem Outcomes This tutorial helps you to (1) work with separation properties of topological spaces; (2) determine an example of a Hausdorff space that is not regular; (3) find equivalent characterisations of normal spaces (via the converse of Urysohn s Lemma); (4) establish compactness and other properties for the Cantor set. Summary of essential material Definition. A property of a topological space is hereditary if, whenever a topological space X has that property, then every subspace of X has the property. Definition. We say that a topological space X is (i) a T 1 space if for every x, y X with x y, there exist open neighbourhoods U of x and V of y with y U and x V ; (ii) a T 2 space or Hausdorff space if for every x, y X with x y, there exist disjoint open neighbourhoods U of x and V of y; (iii) a regular space if for every closed subset E of X and every x X \ E, there exist open disjoint neighbourhoods U of E and V of x; (iv) a T 3 -space if X is both a regular space and a T 1 -space; (v) a normal space if for every pair E and F of disjoint closed subsets of X, there exist disjoint open neighbourhoods U of E and V of F ; (vi) a T 4 -space if X is both a normal space and a T 1 -space. Urysohn s Lemma. Let X be a normal space. Then for any disjoint closed subsets E and F of X, there is a continuous function f : X [0, 1] with f = 0 on E and f = 1 on F. Tietze Extension Theorem. Let X be a normal topological space. Then for any closed subset Y of X and any bounded continuous function f : Y R, there exists a bounded continuous function h : X R such that h = f on Y. Copyright c 2018 The University of Sydney 1

2 Questions to complete during the tutorial 1. Let X be topological space. Prove the following: (a) The set {x} is closed for all x X if and only if X is a T 1 space. Solution: We show that if {x} is closed for any x X, then X is a T 1 - space. Let x, y X with x y. Then, X \ {x} is an open set as the complement of the closed set {x}. Since y X \ {x}, it follows that there exists an open neighbourhood U y of y such that U y X \ {x}, that is, x U y. Similarly, x X \ {y} and there exists an open neighbourhood U x of x such that y U x. We now assume that X is a T 1 -space and show that {x} is a closed set for every x X. Let x X be arbitrary. Then, for any y X \ {x}, there exists an open neighbourhoods U y of y such that x U y. This shows that U y X \ {x} and, hence, X \ {x} is open. Therefore, the set {x} is closed for all x X. (b) If X is a finite T 1 -space, then it is a discrete space. Solution: If X is a finite T 1 -space, then {x} is a closed set for every x X from (a) and any subset U of X is finite. Clearly, the empty set is both open and closed. Moreover, any non-empty subset U of X is closed (as a finite union of closed sets with U = x U {x}) and also open set since U c = X \ U is finite and thus closed. This shows that X is a discrete space. 2. Let X be a topological space equipped with the cofinite topology T. Let τ be another topology on X. Show that the identity map id : (X, τ) (X, T ) is continuous if and only if (X, τ) is a T 1 -space. Solution: Assume that the identity map id : (X, τ) (X, T ) is continuous. Since X \ {x} is open in the cofinite topology for every x X, we have that {x} is closed in the same topology. The continuity of the identity map gives that {x} is closed in (X, τ) for every x X, that is, (X, τ) is a T 1 -space. We now assume that (X, τ) is a T 1 -space. Hence, {x} is closed for every x X. Note that the closed sets in the cofinite topology T of X are exactly the finite subsets of X. Hence, if A is any finite non-empty subset of X, then A = a A {a} is a closed subset in (X, τ) as a finite union of closed sets, proving the continuity of id : (X, τ) (X, T ). 3. Let (X, d) be a metric space. Let A and K be disjoint subsets of X with A closed and K compact. Show that there exists δ > 0 such that d(a, k) δ for all a A and k K. Solution: We give two different arguments. Method 1. Suppose by contradiction that for every n 1, there exist a n A and k n K such that d(a n, k n ) < 1/n. Since K is compact, we have that K is sequentially compact. Hence, {k n } n 1 has a subsequence {k nj } j 1 converging to some L K. Then, d(a nj, k nj ) < 1/n j 0 as j. Thus, a nj L as j, showing that L Ā = A. Thus, A K is not empty (since it contains L), which contradicts the hypothesis. Method 2. Let f(x) = dist (x, A) for all x X. Then f(x) 0 for all x X and f(x) = 0 if and only if x A = A. Moreover, f is (uniformly) continuous since f(x) f(y) d(x, y) for all x, y X. Since A K =, we have that f(k) > 0. By the compactness of K and continuity of f, we find that f(k) is a compact subset of (0, ) and so inf f(k) = δ > 0, proving that d(k, a) δ for all k K and a A. 4. Show that the following properties are hereditary for a topological space: 2

3 (a) a T 1 -space; Solution: Let (X, τ) be a T 1 -space. Let S X be any non-empty subset of X. The relative topology τ S on S is defined by τ S := {U S : U τ}. For every s S, the singleton {s} is closed in (X, τ), which is a T 1 -space. The set {s} is closed in (S, τ S ) for every s S since {s} = {s} S. Thus, (S, τ S ) is a T 1 -space. (b) a Hausdorff space; Solution: Let (X, τ) be a Hausdorff space and S X be any non-empty subset of X with the relative topology τ S. If x, y S with x y, then there exist disjoint open neighbourhoods U x of x and U y of y. Then, (S, τ S ) is a Hausdorff space since U x S and U y S are disjoint and open in S such that x U x S and y U y S. (c) a regular space. Solution: Let (X, τ) be a regular space and S X be any non-empty subset of X with the relative topology τ S. Let E be closed in S and x S \ E. Then, E = A S for some closed set A in X and x A. Since (X, τ) is regular, there exist open neighbourhoods U of A and V of x such that U V =. Then, U S is open in S and E U S, whereas V S is open in S and contains x such that (U S) (V S) =. Hence, (S, τ S ) is a regular space. 5. Let X be a topological space. Prove the following: (a) If the conclusion of Urysohn s Lemma is valid for X, then X is normal. Solution: Assume that the conclusion of Urysohn s Lemma is valid for X. Then, for every pair E and F of disjoint, non-empty closed subsets of X, there exists a continuous function f : X [0, 1] such that f = 0 on E and f = 1 on F. By the continuity of f, we get that U = f 1 ([0, 1/2)) and V = f 1 ((1/2, 1]) are disjoint open sets in X. Moreover, E U and F V so that U and V are disjoint open neighbourhoods of E and F, respectively. Hence, X is a normal space. (b) If the conclusion of the Tietze Extension Theorem is valid for X, then X is normal. Solution: Assume that the conclusion of the Tietze Extension Theorem is valid for X. Then, for every pair E and F of disjoint, non-empty closed subsets of X, we define f : E F {0, 1} such that f = 0 on E and f = 1 on F. Since E F =, the function f is continuous on the closed set E F. By our assumption, there exists a continuous function h : X [0, 1] such that h = f on E F. Then, as in part (a), we define U = h 1 ([0, 1/2)) and V = h 1 ((1/2, 1]), which are disjoint open subsets of X with E U and F V. Hence, X is normal. 6. Let K 0 = [0, 1], and construct a nested descending sequence of closed subsets K n be removing middle thirds, i.e., K 1 := K 0 \( 1 3, 2 3 ) = [0, 1 3 ] [ 2 3, 1], K 2 := K 1 \{( 1 9, 2 9 ) ( 7 9, 8 9 )}, and so on. The intersection C = n 1 K n is called the Cantor set. (a) Show that C consists of all real numbers in [0, 1] which have base-3 expansions involving only the digits 0 and 2 (with infinitely recurring 2 s allowed). Solution: Note that the endpoints of the intervals removed are 0, 1 and a subset of the rational numbers m/3 n with 0 < m < 3 n and m not divisible by 3. The left hand endpoints have numerator m 2 mod (3), while the right-hand endpoints (1, 1, 1, 7... ) have numerator m 1 mod (3) and are exactly the elements of [0, 1] which have base-3 expansions involving no 1 s and ending with recurring 2 s. 3

4 (b) Show that C is compact, uncountable, and has empty interior. Solution: Since each K n is closed, the set C is closed and since each K n is compact it follows that C is compact. There are countably many endpoints. Define a function h : [0, 1] C as follows. Let x = 0.a 1 a 2 a 3... be the base-2 expansion of x [0, 1] in non-recurrent form. (Thus 1 = , rather than ) Let h(x) be the real number with 2 base-3 expansion 0.b 1 b 2 b 3..., where b n = 2a n. Then h is injective and maps the interval [0, 1] onto C R, where R is the (countable) set of right-hand endpoints. Hence C has the cardinality of the continuuum: C = 2 ℵ 0. We obtain C from [0, 1] by removing a countable union of intervals, of total length Σ n 1 2 n 1 3 n = 1, which is the length of [0, 1]. Thus C can contain no interval of positive length, and so int C is empty. Alternatively, note that rationals of the form (3m + 1)/3 n with 0 m < 3 n 1 are dense in [0, 1], and no such number is in the set C. Extra questions for further practice 7. Let X := {(x, y) R 2 : y 0 and (x, y) (0, 0)}. Define E = {(x, 0) : x < 0} and F = {(x, 0) : x > 0}. Show that E and F are disjoint closed subsets of X, then find a continuous function f : X [0, 1] such that f = 0 on E and f = 1 on F. Solution: It is clear that E F = since (, 0) (0, ) =. Since X is a metric space with the usual metric from R 2, to show that E is closed in X, it suffices to prove that E is sequentially closed in X. Let {(x n, 0)} n 1 be a sequence in E that converges to (x, y) in X. Hence, we have x n x and y = 0. Moreover, x 0 since (0, 0) X. Since x n < 0 for all n 1 and x 0, we infer that x < 0, that is, (x, y) = (x, 0) E. Thus, E is sequentially closed in X. A similar argument gives that F is sequentially closed in the metric space X and thus F is closed in X. We write each point (x, y) X in polar coordinates: x = r cos θ and y = r sin θ, where r (0, ) and θ [0, π]. The points (x, y) X corresponding to θ = 0 (respectively, θ = π) and any r (0, ) are precisely those in F (respectively, in E). We define f : X [0, 1] by f(x, y) = 1 θ/π [0, 1]. Then, f(x, 0) = 1 if x > 0 and f(x, 0) = 0 if x < 0. The function f is continuous on X, which completes the proof. 8. Show that (R, R l ) is a T 4 -space, where R l is the lower limit topology on R generated by the basis B = {[a, b) : < a < b < }. Solution: Note that any open set in R with the usual topology is open in (R, R l ). Indeed, any non-empty open set U in R with the usual metric is a union of open intervals of the form (a, b) with < a < b < since for every x U, there exists r x > 0 such that (x r x, x+r x ) U so that U = x U (x r x, x+r x ). Now, every interval (a, b) with < a < b < can be written as a union of intervals in the base B that generates the topology R l on R since (a, b) = n n0 [a + 1/n, b), where n 0 1 is a large positive integer such that a + 1/n 0 < b. Thus, any such interval [a, b) is open in (R, R l ), which thus proves that any open set in R with the usual topology is open in (R, R l ). Since for each x R, the set {x} is closed in R with the usual topology, it follows that {x} is closed in (R, R l ). Thus, (R, R l ) is a T 1 -space. To conclude that (R, R l ) is a T 4 -space, it remains to prove that (R, R l ) is normal. Let E and F be a pair of disjoint and (non-empty) closed sets in (R, R l ). Then, since F R\E 4

5 and R \ E is open in (R, R l ), it follows that for each x F, there exists U x R l with x U x R \ E. Since B is a base for the topology R l, there exists [a, b) B containing x and [a, b) U x. In particular, there exists r x > 0 such that [x, x + r x ) U x R \ E. With the same reasoning applied to E instead of F, we find that for each y E, there exists r y > 0 such that [y, y + r y ) R \ F. Then, by defining U F := x F [x, x + r x /2) and U E := y E [y, y + r y /2), we see that U F and U E are open in (R, R l ) with F U F, E U E and U F U E =. To show that U F U E =, we proceed by contradiction. If there exists z U F U E, then z [x, r x /2) for some x F and z [y, y + r y /2) for some y E. Hence, we have x z < x + r x /2 and y z < y + r y /2, which is a contradiction with y [x, x + r x ) and x [y, y + r y ). Hence, E and F are separated by the open (in R l ) neighbourhoods U E and U F. This shows that (R, R l ) is normal. 9. Let B be the collection of subsets of R of the form (a, b), together with those of the form (a, b) Q, where a, b R with a < b. Prove the following: (a) The collection B is a base of open sets for a topology T on R. Solution: Firstly, for every x R, we have x (x 1, x + 1) B. Secondly, for any two sets in B, their intersection is either empty or belongs to B. This shows that B is a base of open sets for a topology T on R. (b) (R, T ) is a Hausdorff space. Solution: Any open set in R with the usual metric is open in (R, T ) since the collection B of subsets of R of the form (a, b) with < a < b < is a base for the usual topology on R and B B. Clearly, R with the usual topology is Hausdorff (for any x, y R with x y, there exist disjoint open neighbourhoods U x of x and U y of y) so that (R, T ) is Hausdorff. (c) R \ Q is T -closed. Solution: We see that Q is a union of intervals of the form (a, b) Q with a, b R and a < b using that B is a base for R with the usual topology. Therefore, Q is open in (R, T ), which means that R \ Q is closed in (R, T ). (d) If f : (R, T ) R is a continuous function with f = 0 on R \ Q, then f = 0 on R. Solution: Let f : (R, T ) R be a continuous function with f = 0 on R \ Q. Assume by contradiction that there exists p Q such that f(p) = α 0. Without loss of generality, we can assume α > 0 (otherwise, apply the proof for f). Let 0 < β 1 < β 2 < α. Then, by the continuity of f, we have f 1 (β 2, ) is open in (R, T ) and p f 1 (β 2, ). Hence, there exists a, b R with a < b such that p (a, b) Q f 1 (β 2, ), that is f(x) > β 2 for all x (a, b) Q. Now, let q (R \ Q) (a, b) (which is possible by the density of R \ Q in R). Then, since f(q) = 0, by the continuity of f, we have f 1 (, β 1 ) is open in (R, T ) and q f 1 (, β 1 ) using that 0 (, β 1 ). Thus, there exists a set V q in B such that q V q f 1 (, β 1 ). It means that V q = (c, d) for some c, d R with c < d (the other possibility V q = (c, d) Q is excluded since q Q). It follows that f(x) < β 1 for all x (c, d). Since q (a, b) (c, d), we infer that there exists a rational number z in (c, d) (a, b). Then f(z) > β 2 and also f(z) < β 1, which is a contradiction with β 1 < β 2. This shows that f = 0 on Q (and thus on R). (e) (R, T ) is not regular. Solution: We show that R\Q (which is closed in (R, T )) and {0} R\Q cannot be separated by disjoint sets that are open in (R, T ). Indeed, for any set U open 5

6 in (R, T ) containing 0, there exists ε > 0 such that ( ε, ε) Q U (using that B is a base for the topology T ). Now, for any set V open in (R, T ) containing an irrational number in ( ε, ε) must also contain a non-empty open interval, which will then meet ( ε, ε) Q. Hence, U and V cannot be disjoint, showing that (R, T ) is not regular. 6