Math 550A, Homework 4 Solutions April 2, 2014 Problem 1. Prove that if {Xα}α∈J is a collection of Hausdorff spaces, then α∈J Xα is Hausdorff in both the product and box topologies. Q Proof. Let {Xα}α∈J be a collection of Hausdorff spaces. Consider (xα)α∈J = (yα)α∈J in α∈J Xα. Then by definition, there exists α0 ∈ J such that xα0 = yα0 in Xα0 . But sinceQ Xα0 is Hausdorff by hypothesis, there exists Uα0 ∋ xα0 and Vα0 ∋ yα0 open in Xα0 such that Uα0 ∩ Vα0 = ∅. But then α∈J (Uα ∩ Vα) such that Uα = Xα and Vα = Xα for all α ∈ J except when α = α0,Q in which case Uα = Uα0 and Vα = Vα0 as defined above. Since Uα0 ∩ Vα0 = ∅, it follows that U ∩ V = (U ∩ V )= ∅ α α α α αY∈J αY∈J αY∈J Observe that U = α∈J Uα and V = α∈J Vα are open in both the box and product topologies. Hence,Q we have shown thatQ given x, y ∈ α∈J Xα such that x = y there exists U ∋ x and V ∋ y open such that U ∩ V = ∅. Q Problem 2. Let R∞ be the subset of Rω consisting of all sequences that are eventually zero, that is, all sequences (x1, x2,...) such that xi = 0 for only finitely many values of i. What is the closure of R∞ in Rω in the box and product topologies? Justify your answer. Proof. In the product topology, R∞ = Rω. Recall that x ∈ A if and only if for all U ∋ x ∞ Rω Rω ∞ open, then U∩A = ∅. Let xn n=1 ∈ and let U be open in such that xn n=1 ∈ U. ∞ R Since U is open in the product topology, U = n=1 Un where Un = for all but finitely many n ∈ N. For the n ∈ N such that Un = QR, Un = (an, bn) where an < bn. Define ∞ R sn n=1 ∈ U as follows: if Un = , sn = 0, otherwise, sn ∈ Un = (an, bn). Observe ∞ R∞ ∞ Rω ∞ that sn =1 ∈ . Hence, for any xn =1 ∈ and any U ∋ xn =1, it follows that n ∞ n n R∞ R∞ R∞ Rω U ∩ = ∅ since sn n=1 ∈ U ∩ . Conclude by definition that = . In the box topology, R∞ = R∞. This means that R∞ is closed, so to show this, Rω R∞ ∞ Rω R∞ it would suffice to show that \ is open. Let xn n=1 ∈ \ . We want to ∞ Rω R∞ show that there exists U ∋ xn n=1 open such that U⊆ \ . Define U as follows: 1 R R ∞ if xn = 0, then Un = , if xn = 0, then Un = \{0} and U = n=1 Un. Observe that all Un are open as we have defined them and thus U is open inQ the box topology. ∞ ∞ Rω R∞ If yn n=1 ∈ U, then yn = 0 if xn = 0 and thus xn n=1 ∈ \ . Conclude that ∞ Rω R∞ Rω R∞ R∞ xn n=1 ∈ U ⊆ \ and thus \ is open and thus is closed and thus R∞= R∞. Problem 3. Given sequences (a1, a2,...) and (b1, b2,...) of real numbers ai > 0 for ω ω all i, define h : R → R by the equation h (x1, x2,...) = (a1x1 + b1, a2x2 + b2,...). Show that if Rω is given the product topology, h is a homeomorphism of Rω with itself. What happens if Rω is given the product topology? Proof. We first show h is a bijection (Note: this is topology invariant). Suppose ∞ ∞ ∞ ∞ N h (cn)n=1 = h (dn)n=1 , then ancn + bn n=1 = andn + bn n=1 and so for any n ∈ , ancn + bn = andn + bn which implies thatcn = dn (since we subtract bn for both sides ∞ ∞ and divide both sides by an since an > 0). Conclude by (cn)n=1 =(dn)n=1 and thus h ∞ Rω ∞ Rω yn−bn is injective. Let (yn)n=1 ∈ and define (xn)n=1 ∈ by xn = an (which is defined ∞ ∞ since an > 0. Then h (xn)n=1 =(yn)n=1 and h is surjective. ∞ Let’s consider the product topology with basis BProd = U : U = R for all n=1 n n Q but finitely many n ∈ N . Let’s consider what h does to basis elements. We show h−1 ∞ B R is continuous by showing that h is an open map. Let U = n=1 Un ∈ Prod. If Un = , then Un =(cn,dn) for some cn <dn and h maps that componentQ to (ancn+bn, andn+bn) which is also open, other wise Un = R and h maps that component to R. Hence the image of a basis element via h is open in the product topology. Conclude that h is open and thus h−1 is continuous. By linearity, similar reasoning can show that h is continuous. Conclude that h is a homeomorphism in the product topology. Similar reasoning shows that h is a homeomorphism in the box topology. 2.