Topology Part of the Qualify Exams of Department of Mathematics, Texas A&M University Prepared by Zhang, Zecheng

Topology Part of the Qualify Exams of Department of Mathematics, Texas A&M University Prepared by Zhang, Zecheng Remark 0.1. This is a solution Manuel to the topology questions of the Topology Geometry Qualify Exam of Department of Mathematics, Texas A&M University. The exams cover only point and set topology. It is easy to find the syllabus of the exams in the homepage of the mathematics department. This is not a complete solution Manuel but contains most of the questions of the exams staring from August 2008 to January 2017. For the questions which are not included in this Manuel, you can try to find them in the Math 636 assignments which are also in this web page. I also add many interesting questions which are not in the exams. I believe there are mistakes in my solutions; so please let me know if you find any mistake in the proof. I appreciate it. Zecheng Zhang August 2017 1. (J2017) Show that if F is a locally finite family of closed sets then W = C F C is closed. Proof. We can show X W is open. Let x X W and U be the open neighborhood of x such that only {C i } n i=1, finitely many of members of F intersects U non-trivially. Let us define U i = U C i for all i = 1,..., n and V = n i=1 U i. Clearly x U i and U i is open for all i; it follows that x V and V is open. Moreover, {W n i=1 C i} U i = ; this implies that V W =. It follows that X W is open. 2. (J2017) Every compact Hausdorff is regular. Proof. Suppose X is a compact Hausdorff topological space. Let x X and A X be closed such that x A. Since X is Hausdorff, for any y i A, there exist U i and V i open subset of X such that y i U i and x V i satisfying U i V i =. Clearly i U i is an open cover of A; by the compactness of A inherited from X, we can find a finite subcover U = n i=1 U i; let {V i } be the co-responding open sets containing x and V = n i=1 V i. Clearly V is open and V U = ; this means that X is regular. 3. (J2017) Let X and Y be topological spaces and Y be connected. Suppose f : X Y is continuous and surjective such that f 1 (y) is connected for any y Y. (a) Show that if f is a quotient map, then X is connected. Proof. Assume X is NOT connected; i.e., there exist open disjoint subsets U and V such that X = U V. For any y Y, W = f 1 (y) must be completely contained either in U or V ; otherwise, W = {W U} {W V }; clearly {W U} and {W V } are disjoint and open in subspace topology of W ; this implies W is disconnected, which contradicts with the assumption. It follows that U and V are saturated; and f(u) and f(v ) are open in Y. Clearly, f(u) f(v ) = and since f is sutjective, we have f(u) f(v ) = Y. This implies that Y is NOT connected, which is the contradiction; hence, X is connected. Remark. Remarks to the above proof. i. Why is U saturated? We know that A X is saturated if and only if A = {x X, f(x) = f(a), for some a X}. Now, f 1 (y) is completely contained in U, i.e., {x X, f(x) = y, for some y Y } U. This means U is saturated. Actually we have, if A = some y Y f 1 (y), then A is saturated. ii. Why is f(u) open? We have the theorem: Let f : X Y be surjective, then f is quotient if and only if τ Y = {f(u), U is saturated} (b) Provide an example in which f is not quotient and X is not connected. Proof. I do not know. 4. (a2016) Show that a topological space X is discrete if and only if every function f : X Y is continuous for every Y. 1

Proof. Assume X is discrete. f 1 (U) = some y {y} for U open in Y. Since singleton is open in the discrete space X, f 1 (U) is open and hence f is continuous. Conversely, since arbitrary function f is continuous for any Y, let Y = X equipped with discrete topology and f is the identity map. By assumption, f is continuous; it follows that {x} = f 1 ({x}) is open; this implies that X is discrete. 5. (J2016) Show that a bijection f : X Y is a homeomorphism if and only if f(a) = f(a) for every A X. Proof. Let us first show f is continuous under the condition f(a) = f(a). D = f 1 (A); we need to show D is closed. Clearly we have Let A Y be closed and f(d) f(d) A = A. Now take the inverse, we have D D, which gives us that D = D. Now let us show f 1 is continuous; that is we need to show f is closed, i.e., for any A X closed, f(a) is closed; that is we need to show f(a) f(a) = f(a); but by assumption we have f(a) f(a); hence f(a) is closed. Conversely, assume f is continuous. It follows that f 1 (f(a)) is closed and contains A. Since smallest closed set containing A, we have Ā f 1 (f(a)); this implies that f(ā) f(a). Ā is the Now let us assume f 1 is continuous, i.e., f is closed. Let A be any set in X; we have A Ā; it follows that f(a) f(ā). By the assumption, f(ā) is closed; hence f(a) f(ā) since f(a) is the smallest close set containing f(a). 6. (J2016) Let X be a topological space and Y X. Show that the connected component of x Y in the subspace topology of Y is a subset of the connected component of x in X. 7. (more exercises) Let A and B be subsets of X. If A is connected, B is open and closed, and U = A B, then A B. Proof. Since B is clopen, U is clopen in A under the subspace topology. Since the only clopen sets of a connected space are the whole set and empty set, it follows that A B. 8. (more exercises) Let A and B be connected subsets of X. If V = A B, then prove that U = A B is connected. Proof. Let P U be clopen and nonempty. If P A and assume B P =, by the previous question, A P. This implies that P B, contradiction; hence P B. It follows that B P ; we then have A B P. Since P = A B, we have P = A B; i.e., the nonempty clopen set is the whole set, A B is connected. 9. (a2015) Let X be a compact space and {C j } j J is a family of closed subsets of X. If there exists open set U such that C = j J C j U, then prove that there exists subfamily {C i } n i=1 such that n j=1 C j U. Proof. Take the complement, we then have U c j J Cc j. U c is closed subset of a compact space; hence U c is compact and clearly {Cj c} j J is an open cover of U c ; hence there exists a finite subcover {C j } n j=1 such that U c n j=1 Cc j. Take the complement, we then have n j=1 C j U. Theorem 0.1. Any closed subset C of a compact topological space X is compact. Proof. Let {C j } j J be a family of subsets with finite intersection property in C. Since C is closed, this is also a family with F.I.P. in X; hence it has non empty intersection. Clearly this intersection is in C; hence C is closed. Remark 0.2. If A and B are two compact subsets of a topological space X, A B may not be compact. See assignment 4. 2

10. (J2015) Let q : X Y be an open quotient map. Show that Y is Hausdorff if and only if A = {(x 1, x 2 ), q(x 1 ) = q(x 2 )} is closed in the product space X X. Proof. Assume Y is Hausdorff. We need to show B = A c is open; Clearly B = {(x 1, x 2 ), q(x 1 ) q(x 2 )}; for any x 1 and x 2 in B, we have q(x 1 ) and q(x 2 ) not equal, there exist open disjoint subsets U and V such that q(x 1 ) U and q(x 2 ) V. By the definition of the quotient map, q 1 (U) and q 1 (V ) are open disjoint in X; and clearly W = q 1 (U) q 1 (V ) is a subset of B; otherwise assume (a, b) A W, then q(a) = q(b); i.e., U V, which is the contradiction. It follows that W is the open set containing a point outside A, then A is closed. Conversely, for any x and y in Y and x y, there exists (a, b) B such that x = q(a) and y = q(b). Since B is open, there exist open disjoint subsets U and V such that a U and b V and U V B. Since q is open, q(u) and q(v ) are open in Y and contain x and y respectively. Clearly q(u) q(v ) = ; otherwise, there exists c U and d V such that q(c) = q(d), this means that U V A, which is the contradiction. Hence Y is Hausdorff. 11. (a2014) Let X be compact Hausdorff space and f : X X be an injective continuous function. Show that there exists a nonempty closed subset A of X such that f(a) = A. Proof. 12. (a2014) Let X be a locally compact space and A be a subset of X such that for every compact subsets K of X, A K is closed in X. Show that A is closed in X. Proof. Let x X A; since X is locally compact, there exists a compact neighborhood x; that is, there exit V open and U compact such that x V U. By the assumption, B = A U is closed in X; it follows that C = V B is open clearly, we have x C and C A =. since x is arbitrary in X A, this means that X A is open. 13. (J2013) Let X and Y be topological spaces. Let f : X Y be surjective and satisfy int(f(a)) f(int(a)) for any A X. Show that f is continuous. Remark 0.3. I cannot solve this problem; I guess we need more assumption to prove it. Proof. Let U Y be open. we need to show f 1 (U) is open in X. Since f 1 (U) is a subset of X, by the assumptions, we have int(f(f 1 (U))) = int(u) f(int(f 1 (U))). Meanwhile we have int(f 1 (U)) f 1 (U), it follows that f(int(f 1 (U))) f(f 1 (U)) = U = int(u); hence we have U = f(int(f 1 (U))). 14. (a2008) Show that if a map f : X Y is closed and f 1 (y) is compact for all y Y, then f is proper. Proof. Let A Y be compact and take A as Y in the question below and use the same method. 15. (similar question) Let f : X Y be closed map. If f 1 (y) is compact for any y Y. Prove that X is compact if Y is compact. Proof. Let y Y. Let Ûy = {Uα} y α J be an arbitrary open cover of f 1 (y); hence U = Û y is an arbitrary cover of X. Since f 1 (y) is compact, there exists a finite subcover which I denote as U y = n i=1 U y i, such that f 1 (y) U y. Let V y = X U y ; clearly V y is closed and since f is a closed map, we have f(v y ) is closed in Y. It follows that C y = Y f(v y ) is open in Y. Clearly, y C y ; hence there exists an open set W y such that y W y and C y W y =. Clearly y Y W y is an open cover of Y ; since Y is compact, there is a finite subcover y W i=1 y i covering Y. 3

Let us claim f 1 (W y ) U y ; otherwise assume there exists x f 1 (W y ) V y, it follows that f(x) f(v y ) W y ; however W y f(v y ) = ; this gives the contradiction; hence f 1 (W y ) V y =. It follows that f 1 (W y ) U y, that is, it can be covered by U y, which is a finite subcover. Now let us consider all y i Y such that {U yi } n i=1 form a finite cover of Y. Since f is surjective, f 1 (W yi ) for all i cover X. By the claim, f 1 (W yi ) can be covered by by a finite subcover; this implies that X is compact. (Since the number of W yi is finite due to the compactness of Y.) Remark 0.4. This question is quite interesting and not so easy; but it is not hard to do. To get the closed set, I consider take the compliment of U y ; to use f is closed, I consider the complement of the image is open. 16. (a2008) Assume all spaces below are at least T 1. (a) Fix a subbasis S of X. Show that X is completely regular if and only if for any x X and V S with x V, there exists a map f : X [0, 1] such that f(x) = 0 and f(y) = 1 for y X V. Definition 0.1. Let X be a topological space with topology τ. A collection B of open sets is called a subbasis if any open set in τ is the union of some collection of finite intersections of members in B. Proof. Suppose X is regular. V is a subbasis set and thus V is open; hence V c is close. Since X is completely regular, there exists a continuous function f such that f(x) = 0 and f(v c ) = 1; i.e., f(y) = 1 for any y X V. Conversely, let x X and A X closed with x X A. it suffice to show there exists V S with x V such that V A =. Assume such V does NOT exist, i.e., for any V S with x V, we have V A. Since X A is open, it must the union of some collection J of open sets U α which is the finite intersection of members in S; hence there exists U such that x U. Let U = n i=1 V n where V n S; by the assumption, V n A for all n; this implies that U A ; it follows that {U α } α J A, which is the contradiction. (b) Show that the arbitrary product of completely regular spaces is completely regular. Proof. I do not know. (c) Show that a subspace of a completely regular space is completely regular. Proof. Suppose X is complete regular and A X. Let x A and V A be closed and x V ; There exists U closed subset in X such that V = U A. Clearly x U; otherwise we have x V. Since X is complete regular, there exists f such that f(u) = 1 = f(v ) and f(x) = 0; hence A is also completely regular. 17. (easy exercise) Let f : X Y be a continuous surjective closed map. Show that Y is normal if X is normal. Proof. Let A Y be closed and U Y open such that A U; it follows that C = f 1 (A) W = f 1 (U) and C is closed and W is open. Since X is normal, there exists V X open such that C V V W. Since f is surjective, it follows that f f 1 (A) = A f(v ) f(v ) U. Now let us claim f(v ) is open. Since f is surjective, Y is the disjoint union of f(v ) and f(x V ). Clearly f(x V ) is closed; hence f(v ) is open. Moreover since f is closed, f(a) is closed; it follows that f(ā) = f(a). We finally have This implies that Y is normal. A f(v ) f(v ) f( V ) = f(v ) U. 18. (a2012) Suppose X is normal and U and V are open such that X = U V. Show that one can find open subsets U 1 and V 1 with U 1 U and V 1 V such that X = U 1 V 1. Proof. I do not know at the moment. 4

19. (before the following question) Suppose X is locally compact and Hausdorff, then for any point, there exists an open neighborhood whose closure is compact. Proof. Let x X. Since X is locally compact, there exists a compact neighborhood C of x; i.e., U open such that x U C. Clearly compact in Hausdorff is closed; it follows that x U U C; hence U is the desired set (closed in compact is compact). 20. (before the following question)(hard question) Let x be any point in a locally compact Hausdorff space X. Show that for any neighborhood U of x, there exists an open set V such that x V V U; moreover, V is compact. Proof. It is suffice to consider the open set only. By the last question, for any x X, there exists an open neighborhood W whose closure is compact. Let F = W U. Clearly W is compact and Hausdorff, hence W is regular. We have x F ; it follows that there exists disjoint sets H and G open in subspace topology such that x H and F G. Now we have, x H W G W F U. Moreover, since G is open, W G is closed; it follows that H W G U. Here the closure is taken under subspace topology. By the definition, there exists open set O in X such that H = W O; let V = W O; clearly x V and V H W ; we then have V W H H W. Moreover, since W is compact, V is compact. Remark 0.5. Something you need to know about this important question. (a) The question tells us that all closed compact neighborhoods of x forms a neighborhood basis for x in a locally compact space Hausdorff space. Conversely, if every point of a Hausdorff space has a compact closed neighborhood basis, then X is locally compact. (b) We need to find an open set. The open set here is generated due to the regularity of a subspace. The open here is in subspace topology; hence we need to find its open set in X. (c) The Hausdorff property is usually used indirectly (prove some other results like the regularity in this question) in the questions concerning locally compact. Theorem 0.2. Let X be Hausdorff; then the following statements are equivalent, (a) X is locally compact. (b) For any point x X, there exists an open neighborhood whose closure is compact. (c) For any neighborhood U of x X, there exists an open set V such that x V V U; moreover, V is compact. (d) X has a basis consisting of open sets whose closure are compact. 21. (J2012) Let x be a second countable locally compact Hausdorff space. (a) Show that the one point compactification X + is compact and second countable. Definition. Let X be Hausdorff with topology τ. Define X + = X { } where X and τ + = {τ} {X + Kwhere K is compact in τ}. The compact space X + with τ + is called the one point compactification of X. Proof. Since X is second countable, there exists a countable basis; but X is locally compact, that means we can find a countable open basis {U n } of X such that U n is compact for all n = 1, 2,... Let V n = X + n i=1 U n. Now let me claim that {V n } is a countable local basis for. For each n, clearly n i=1 U n is compact in X, by the definition of the τ +, we have V n is open and contains. Now let O be a neighborhood of ; then there exists a compact set V X such that O = X V. Since {U n } is an open cover of V, there exists a finit subcover denoted as n i=1 U k i contains V. It follows that, V kn X k n i=1 U i X n U ki O. This implies {V n } is a local basis for ; i.e., {U n } {V n } is a countable basis for X +. i=1 5

(b) Every subspace is is paracompact. Proof. 22. (more exercises) Show that every open subspace A of a locally compact Hausdorff space X is locally compact. Proof. For any x A, x X; let U be a neighborhood of x; i.e., there exists open set V such that x V U. it follows that x V A U A; i.e., U A is a neighborhood of x. Since X is locally compact Hausdorff, there exists an open neighborhood W of x whose closure is compact, that is, x W W U A A. This means for any x A, there exists open neighborhood whose closure is compact in A; since A is Hausdorff, A is locally compact. Remark 0.6. What if A is closed in X? What if X is not Hausdorff? 23. (simple exercise about metric spaces) Show that if a metric space X has a countable dense subset D, then X is second countable. Proof. Let B = {B r (x), for all x D with r Q + }. Let us claim that B is a countable basis for X; that is for any U open neighborhood of x X, there exists a ball Q B such that x Q U. Clearly, there exists ɛ > 0 and small enough such that x B 3ɛ (x) U. Since D is dense, for the given ɛ, there exists q D such that d(x, q) < ɛ. Now choose δ Q such that ɛ < δ < 2ɛ. By definition B δ (q) B and x is in it. For any y B δ (q), d(x, y) < d(y, q) + d(q, x) < 3ɛ; this implies that B δ (q) U; hence we showed that B is a countable basis. Remark 0.7. X is metric and D is dense in X. For any x X and any ɛ > 0, there exists q D such that d(x, q) < ɛ. Remark 0.8. The converse is also true; that is a metric space is second countable then it is separable. To prove this, let {B n } be a countable basis for the space X; Claim D = {x n, x n B n } is countable and dense. The countable part is trivial since {B n } is countable. For any B open ball in X, there is B n such that B n B; this means x n B; i.e. B D. 24. (simple exercise about metric spaces) Every metric space X is regular. Proof. Let x X and A X be closed and x A. A c is open, there exists ɛ > 0 such that B ɛ (x) A c. This implies that for any q A, we have d(q, x) ɛ; therefore B ɛ/2 (x) B ɛ/2 (q) =. Let U = B ɛ/2 (x) and V = q A B ɛ/2(q); clearly U V =. 25. (simple exercise about metric spaces) Compact subspace K in metric space X is closed and bounded. Proof. X is Hausdorff, hence X is closed. For r > 0, K x K B r(x). Since K is compact, K can be covered by finitely many balls B r (x i ) for i = 1, 2,..., n. Let M = max{d(x i, x j )}; then clearly K B M+ɛ (x 1 ). 26. ( simple exercise about metric spaces) Let X be totally bounded metric space; then X is separable. Proof. For any n, let S n be the finite subset such that X = s S n B 1/n (s). Let S = n S n. Claim S is dense and countable. Let B r (x) be any open ball in X. Let m N + be such that 1/m < r; then there exists s S m such that x B 1/m (s). Since d(x, s) < 1/m < r, we have x B r (x); i.e., B r (x) S ; hence it is dense. Clearly, S is countable. 27. (a2015) Let X be the interval [0, 1] with the following topology. A subset U of X is open if and only if it contains the interval (0, 1) or it does not contain 1/2. (a) Can you compare this topology τ with the standard topology on unit interval. 6

Proof. (1/4, 3/4) is open in the standard topology in unit interval; but 1/2 (1/3, 3/4) and it does not contains (0, 1); hence it is not open in X under τ. [1/2, 1] is open under τ but closed in the standard topology. Conclusion, we cannot compare two topologies. (b) Determine the closure of 1/4. Proof. It is not {1/4}. (c) Show that X is T 0 but not T 1. Proof. It is easy to see it is T 0. Let us consider 1/2 and 3/4. Clearly the open sets containing 1/2 contains (0, 1); that means it must contains 3/4; hence it is not T 1. 28. (J2015) Let R l denote the real line with the right open left closed topology τ. (a) Find the closure of (a, b). Proof. 29. (a2009) Let X and Y are locally compact Hausdorff. Show that any proper map is closed. Proof. Have no idea; but the one point compactification is useful. 30. (a2017) Let X be a topological space and Y be Hausdorff compact generated space (any A Y is closed if and only if A K is closed in K for any K compact subset of Y ). Show that a continuous proper function f from X to Y is closed. Proof. Let C X be closed. We need to show A = f(c) is closed in Y. Let K Y be compact and arbitrary. If K A = for all K, it is clear that A is compact since is closed in K. Now let B = A K and B. We need to show B is closed in K. Since Y is Hausdorff, K is closed; it follows that f 1 (K) is closed in X and compact; denote D = f 1 (K) C. Clearly D is closed and f(d) = B; moreover since D is compact, we know B is compact and hence closed in Hausdorff space Y. This implies that B is closed in K. Since K is arbitrary, by the definition of compact generated space, we know f is closed. 31. Let f : X Y be proper and continuous. Show that if Y is locally compact Hausdorff, then f is closed. Proof. Let A X be closed; we need to show Y f(a) is open. Let x Y f(a); since Y is locally compact, there exists an open neighborhood V whose closure is compact. Since f is proper, f 1 (V ) is compact; it follows that C = f 1 (V ) A is closed both in X and f 1 (V ) and hence compact. Since f is continuous, f(c) is compact and closed in Y which is Hausdorff. Now let U = V f(c); then U is open and has no intersection with f(c). 32. (J2009) If X is countably compact (every open countable cover has a finite subcover) and Y is second countable Hausdorff. Show that a continuous bijection is a homeomorphism. Lemma 0.1. Every second countable space is Lindelof (every open cover has a countable subcover). Proof. Let α J U α be an open cover of X and {B n, n N} be a countable basis. Clearly for any U α non empty in the collection, there exists x U α ; i.e., U α is an open neighborhood of x; hence there exists B α such that x B α U α. The collection of non empty members of U α is countable and covers X. Lemma 0.2. The continuous image of a countably compact subset is countably compact. Proof. Let {U} be a family of open sets in Y such that f(x) U. Then U = f 1 (U) is an open cover of X and hence has finite subcover U 1, U 2,..., U n. It follows that f(x) = f(u 1 U 2... U n) = ff 1 (U 1 ) ff 1 (U 2 )... ff 1 (U n ) U 1... U n. Lemma 0.3. Closed subset of a countably compact set is countably compact. 7

Proof. Let X be countably compact and A U be closed. Let U = n N U n be a countable open cover of A; clearly A c U is a countable cover of X. Since X is countably compact, there exists a finite subcover U of U such that X = A c U. Clearly U covers A; hence A is also countably compact. Proof. The proof of the problem. We need to show f 1 is continuous; that is for A X closed, f(a) is closed. Since X is countably compact, A is also countably compact; hence f(a) is countably compact. Since Y is Hausdorff, for any x f(a) and y Y f(a) fixed, there exist U x and V x open disjoint subsets such that x U x and y V x. Clearly x f(a) U x is a cover of f(a) and hence has a countable subcover. By countably compactness, there exists a finite subcover, which coresponds to x 1, x 2,..., x n. Let V = n i=1 V x i ; clearly V f(a) = and V is open. This implies f(a) is closed; hence f is a homeomorphism. 33. (similar to the following problem) Suppose X is compact. Show that every locally finite family is finite. Proof. For any x X, there is an open neighborhood U x such that U x intersects only finitely many members of the the locally finite family F nontrivially. Clearly U x is an open cover of X and hence has a finite subcover. It follows that the subcover intersects finitely many members in F. Since this is an cover of X, this implies F is finite. 34. (a2011) Show that X is countably compact, then every locally finite family of noneempty subsets is finite. Definition 0.2. A family of subsets is locally finite if for every point in the space, there is an open neighborhood that intersects nontrivially only finitely many of members in the family. Proof. Since countably compact is compact, the above proof also works for this problem. 35. (a2011) Define paracompact and show that if X is countably compact and paracompact then it is compact. Definition 0.3. A Hausdorff space X is paracompact if every open cover has a locally finite refinement. Definition 0.4. Let X be a space and U and V be two open covers. U is called a refinement of V if each member of U is contained in some member of V. Proof. Let U be an open cover of X. Since X is paracompact and countably compact, there exists a finite locally finite refinement F = {V i, i = 1,..., n} which is also a cover of X and satisfies V i U i for some U i U. Clearly {U i, i = 1,..., n} is the finite subcover of U; hence X is compact. 36. (similar question) A locally finite family F of a second countable space X is always countable. Proof. For any x X, there is an open neighborhood U x that intersects only finitely many members of F. Clearly U x is an open cover of X and hence has a countable subcover U = n i=1 U x i (X is second countable; then Lindelof). Since F X U and each U intersects only finitely many members of F, F is countable. 37. (similar question) A locally finite family F of a Lindelof space X is always countable. Proof. Same proof as above. 38. (J2009)Let X be a paracompact Hausdorff space. Show that if X contains a dense Lindelof subspace S, then X is Lindelof. Proof. Let U = U α be an open cover of X; since X is paracompact(it implies normal), it has a locally finite refinement V = V α ; i.e. X = V α and V α U β for some U β U. Moreover, it is known that V = V α = W α where V α = W α is also locally finite. Since S is a Lindelof subspace of X, V also covers S and hence there exists a countable subcover which I denote as W = i=1 W i. It is not hard to show W is closed(see J2017). Since S is dense, we have W = X; that is W is the countable. Since W i U i, it follows that {U i, i = 1,..., n} will be the countable subcover of U;i.e., X is Lindelof. 39. Every paracompact space is regular. 8

Proof. see notes. 40. If a collection F of subsets of a space X is locally finite and A is compact for any A F, then any A F intersects only finitely many members of F. Proof. 41. Dense and locally compact subset of a Hausdorff space is open. 42. Let A be a connected subset of a connected space X and B X A be an open and closed set in the subspace topology of X A. Prove that A B is connected. 43. Union of connected subsets X and Y is connected if X Y. Proof. Let f be arbitrary discrete map on X Y. Since X is connected, it is known that X is connected. It follows that the restriction of f on X and Y are constant assume f(x) = a and f(y ) = b. For x X Y, f(x) = f(x) = a; meanwhile f(x) = f(y ) = b this implies that f is constant on X Y ; i.e., the union is connected. 44. The closure of a connected subset is connected. Lemma 0.4. Let Y be a topological space and every singleton is closed in Y if and only if for every topological space X and map f : X Y, f is constant on A implies f is constant on A. Proof. Assume all singletons are closed and f(a) = c. 9