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FUNDAMENTOS DE TOPOLOGIA E ANÁLISE REAL 1o Semestre 2010/2011 1o Teste (Recuperação) 3-2-2011 – Take-home 1. Let X be a a topological space, ∼ an equivalence relation on X and p : X → X/ ∼ the quotient map. Let R := {(x, y) | x ∼ y} ⊂ X × X. a) p × p : X × X → (X/ ∼) × (X/ ∼) is continuous. Answer: p is continuous, by the definition of quotient map. The sets U × V, with U, V open in X/ ∼ form a basis for the product topology on (X/ ∼) × (X/ ∼) and (p × p)−1 (U × V) = p−1 (U) × p−1 (V) is open, being the product of open sets. Hence p × p is continuous. b) If X/ ∼ is Hausdorff then R is closed in X × X. (Hint: determine A such that R = (p × p)−1 (A).) Answer: Recall that X/ ∼ is Hausdorff iff the diagonal ΛX/∼ = {([x], [x]) x ∈ X} is closed in the product (X/ ∼) × (X/ ∼). Since p × p : X × X → (X/ ∼) × (X/ ∼) is continuous, and (p × p)−1 (ΛX/∼ ) = {(x, y) ∈ X × X p(x) = p(y)} = R, we have that R is closed if ΛX/∼ is closed. Alternatively: Let (x, y) < R ⇔ [x] , [y], where [x] denotes the equivalence class of x ∈ X. We check that there is a neighborhood of (x, y) which does not intersect R, hence (x, y) < R, so R ⊂ R and R is closed1 . Since X/ ∼ is Hausdorff, there exist disjoint neighborhoods U of [x] and V of [y] in X/ ∼. Since p is continuous, p−1 (U) and p−1 (V) are open, hence neighborhoods of x, y in X, respectively, so W := U × V is a neighborhood of (x, y) in X × X. If (x0 , y0 ) ∈ W then [x0 ] ∈ U and [y0 ] ∈ V. Since U ∩ V = ∅, [x0 ] , [y0 ], that is, (x0 , y0 ) < R. Hence W ∩ R = ∅. 2. a) Show that there is a topology τ over R such that A ⊂ R is closed if it satisfies at least one of the following conditions: 1) 0 ∈ A, 2) A is finite. Answer: It follows from the definition of topology and from De Morgan’s laws that τ is a topology iff 1 or equivalently, (X × X) \ R is open. 1 i) R, ∅ are closed; ii) ∩i∈J Ai is closed, if Ai closed, arbitrary J; p iii) ∪i=1 is closed, if Ai closed, p ∈ Z+ . As for i), since 0 ∈ R and ∅ is finite, both R, ∅ are closed. Now let Ai be closed, for any i ∈ J. If Ai is finite for some i ∈ J then ∩i∈J Ai ⊂ Ai is finite, hence closed. If not, then 0 ∈ Ai , for all i ∈ J, hence 0 ∈ ∩i∈J Ai , which gives that ∩i∈J Ai is closed and ii) is proved. As for finite unions, if 0 ∈ AI for some i = 1, ..., p, then p p 0 ∈ ∪i=1 Ai , hence ∪i=1 Ai is closed. If not, then Ai is finite, for all i = 1, .., p, hence p ∪i=1 Ai is also finite, hence closed. NOTE: U ⊂ R open iff 0 < U or U = R \ {x1 , ..., xp }. b) In (R, τ) any {x}, x , 0, is open but τ is not the discrete topology. Answer: If x , 0, then 0 ∈ (R \ {x}), hence (R \ {x}) is closed, that is, {x} is open. On the other hand, {0} is not open, since R \ {0} is not closed: 0 < R \ {0} and R \ {0} is not finite, so it doesn’t satisfy any of the conditions 1), 2). Hence, τ is not the discrete topology. c) Is (R, τ) connected? Answer: No, since {x}, x , 0 is both open (by b)) and closed (since it is finite), and not R, ∅. Equivalently: R = {x} ∪ (R \ {x}), x , 0 is a separation. (Can take any finite A with 0 < A as open and closed.) d) For infinite A ⊂ R, A = A ∪ {0} under τ. Is (R, τ) separable? Answer: By definition, A is the intersection of all the closed sets containing A. Since A is infinite, any X closed with A ⊂ X is infinite, so 0 ∈ X and we have that 0 ∈ A, so that A ⊃ A ∪ {0}. Since A ∪ {0} is closed and contains A, we have also A ⊂ A ∪ {0}, and we conclude that A = A ∪ {0}, for infinite A. (R, τ) separable iff there is a countable dense subset. But if A is finite, it is closed and A = A , R, and if A is infinite with A = A ∪ {0} = R then A ⊃ R \ 0 is uncountable, so R is not separable. e) (R, τ) is regular and not first countable. Is it metrizable? Answer: (R, τ) is regular iff it is T1 and for any x ∈ R and any closed set A such that x < A, there are open, disjoint sets U, V such that x ∈ U and A ⊂ V. It is T1, since any one point set is finite, hence closed. Let now first x , 0. Since {x} is open, we can take U = {x} and, since {x} is finite, V = R \ {x} is open and A ⊂ V. 2 If x = 0 and A closed with 0 < A, then since 0 ∈ R \ A, have that V = A is open, and 0 ∈ U = R \ A is also open. Hence (R, τ) is regular. We check that there is no countable basis of neighborhoods at 0. First note that U is a neighborhood of 0 iff 0 ∈ U and R \ U is finite. Let B = {Bi } be a basis of neighborhoods at 0, so that R \ Bi is finite, for any i, and for any neighborhood U of 0, we have U ⊃ Bi for some i, that is, R \ U ⊂ R \ Bi . If B is countable, then ∪R \ Bi is also countable (countable union of finite sets). Take x ∈ R \ ∪(R \ Bi ), with x , 0. Then U = R \ {x} is a neighborhood of 0 but R \ U = {x} is not contained in R \ Bi , for any i, which is a contradiction. Hence B is uncountable and (R, τ) is not first countable. It is not metrizable, since any metric space is first countable (taking for instance B(x, 1/n), n ∈ N as a basis of neighborhoods at x). f) A ⊂ R is compact under τ if and only if it is closed. Answer: Since (R, τ) is regular, it is in particular Hausdorff, so any compact set under τ is closed. Conversely, let A be closed. If A is finite, then it is compact, since any open covering of a finite set has a finite subcovering. If not, then 0 ∈ A. Let {Uα } be an open covering of A, and pick α0 such that 0 ∈ Uα0 . Then Uα0 is a neighborhood of 0, so that R \ Uα0 is finite, and A \ Uα0 is also finite, hence compact. Since {Uα }, α , α0 is an open covering for A \ Uα0 , there is a finite subcover Uα1 , ..., Uαp , and then A ⊂ Uα0 ∪ Uα1 ∪ ... ∪ Uαp . 3. Let X be a T1 topological space. Assume that every continuous, bounded, realvalued function on a closed set has a continuous extension to all of X. Show that X is normal. Is the converse also true? Answer: X is normal iff it is T1 and for any given closed, disjoint sets A, B, there are open, disjoint sets U, V such that A ⊂ U, B ⊂ V. Given that X is T1, we only need to check the second condition. Let then A, B be closed, disjoint sets. Since A ∩ B = ∅, the function f : A ∪ B → R such that f (A) = 0 and f (B) = 1 is continuous (pasting lemma). It is also bounded and defined on a closed set, so it has a continuous extension g : X → R, with g(A) = 0 and g(B) = 1. Define U := g−1 (] − 1/2, 1/2[) and V := g−1 (]1/2, 3/2[). Then U, V are inverse images of open sets under a continuous function, hence open. Moreover, they are disjoint, and A ⊂ g−1 (0) ⊂ U, B ⊂ g−1 (1) ⊂ V. The converse is true, by Tietze’s theorem. 3 4. Let Y be a Hausdorff topological space, X a set, and consider the function space YX with the product topology τp . a) Show that τp coincides with the topology of pointwise convergence, that is, fn → f in YX if, and only if, fn (x) → f (x) in Y, for all x ∈ X. Answer: Can use a general result on the product topology: an → a in Πα∈J Xα iff πβ (an ) → πβ (a) in Xβ , for all β ∈ J, where πβ : Πα∈J Xβ is the projection. In our case: J = X and πx ( f ) = f (x). Alternatively: if fn → f then πx ( fn ) = fn (x) → πx ( f ) = f (x), since πx is continuous. Conversely, let fn , f ∈ YX be such that πx ( fn ) → πx ( f ), for all x ∈ X. Let U = ΠUx be a neighborhood of f in the product topology, with Ux = Y, for x , x1 , ..., xp and Uxi is a neighborhood of f (xi ) in Y. Since fn (xi ) → f (xi ), for each i = 1, ..., p, have that there is Ni such that if n > Ni then fn (xi ) ∈ Uxi . Let N = max Ni . Then n > N gives that fn (xi ) ∈ Uxi , for any i = 1, ..., p, that is, fn (x) ∈ Ux , for any x ∈ X, that is, fn ∈ U. Hence, fn → f . b) F ⊂ YX is compact if and only if F is closed and, for any x ∈ X, the set f (x) | f ∈ F has compact closure in Y. (Hint: write F = Πx∈X πx (F ), with πx : YX → Y the projection map.) Answer: Note that πx (F ) = f (x) | f ∈ F . Let F ⊂ YX be compact. Since YX is Hausdorff, being the product of Hausdorff spaces, F is closed. Moreover, since the projection πx is continuous, for any x ∈ X, have that πx (F ) is compact in Y. Since Y is Hausdorff, it is also closed, so πx (F ) = πx (F ) is compact. Conversely: if F is closed, then F = F = Πx∈X πx (F ) = Πx∈X πx (F ) is compact, by Tychonoff’s theorem, since πx (F ) is compact. c) Let τ be a topology on YX finer than τp . If YX is compact under τ, then τ = τp . Answer: The identity map (YX , τ) → (YX , τp ) is continuous, since τ ⊃ τp , and any continuous bijection from a compact space to a Hausdorff space is a closed map, therefore a homeomorphism. Hence τ = τp . d) Let (Y, d) be a compact metric space, X be a topological space. What can you say about compactness of YX in the pointwise convergence, uniform, and compact convergence topologies? Answer: In the pointwise convergence topology: it follows from b) 2 that YX is compact iff Y is compact. In the uniform and compact convergence topologies, it may not be compact. 2 or Tychonoff’s theorem together with πx continuous 4 Let τc , τu denote the compact convergence and uniform topologies, respectively. Then we have τp ⊂ τc ⊂ τu . Hence, it follows from c) that YX is compact in τc , respectively, in τu , iff τc = τp , respectively, τu = τp . It follows from the definition of τc that τc = τp iff the only compact sets in X are finite (this holds if X is discrete). On the other hand, τu = τp iff X is finite ( τu = τc iff X is compact). 5