\magnification 1200 \nopagenumbers \parindent 0pt \parskip=10pt \voffset 20pt Topology HW answers (many answers modified from latexed HW of Colin B. McKinney) p. 111: 2.) Suppose that $f: X \mapsto Y$ is continuous. If $x$ is a limit point of the subset $A$ of $X$, is it necessarily true that $f(x)$ is a limit point of $f(A)$? It is NOT necessarily true that $f(x)$ is a limit point of $f(A)$ if $x$ is a limit point of the subset $A$ of $X$. The constant function $f: R \rightarrow R$, $f(x) = 0$ is continuous. 5 is a limit point of $R$, but $f(5) = 0$ is not a limit point of $f(R) = \{0\}$ since $\{0\}$ has no limit points. 9.) Let ${A_\alpha}$ be a finite collection of subsets of $X$; let $X = \cup_\alpha A_\alpha$. Let $f: X \mapsto Y$; suppose that $f|A_\alpha$ is continuous for each $\alpha$. Since each $f|A_\alpha$ is continuous, if we choose any set $V$ closed in $Y$, $(f|A_\alpha)^{-1} (V) = f^{-1} (V) \cap A_\alpha$ is closed in $A_\alpha$. Since $A_\alpha$ in closed in $X$, $f^{-1} (V) \cap A_\alpha$ is closed in $X$. Hence $\bigcup_\alpha (f^{-1} (V) \cap A_\alpha) = f^{-1} (V)\cap (\bigcup_\alpha A_\alpha) = f^{-1} (V)\cap X = f^{-1} (V)$ is closed since finite unions of closed sets are closed. Hence $f$ is continuous. p. 118 3.) Suppose $X_\alpha$ is Hausdorff for all $\alpha$. Show $\Pi X_\alpha$ is Hausdorff Take $(x_\alpha)_{\alpha \in J}$ and $(y_\alpha)_{\alpha \in J} \in Pi X_\alpha $ such that $(x_\alpha)_{\alpha \in J} \not= (y_\alpha)_{\alpha \in J}$. Then there exists $\alpha_0$ such that $ x_{\alpha_0} \not= y_{\alpha_0}$. Since $X_{\alpha_0}$ is Hausdorff there exists $U, V$ open in $X_{\alpha_0}$ such that $U \cap V = \emptyset$ and $ x_{\alpha_0} \in U, ~ y_{\alpha_0} \in V$. Then $(x_\alpha)_{\alpha \in J} \in \Pi U_\alpha$ and $(y_\alpha)_{\alpha \in J} \in \Pi V_\alpha$ are disjoint open sets in both the box and product topology where $U_alpha = \cases{U & $\alpha = \alpha_0$ \cr X & $\alpha \not= \alpha_0$}$. 5: Theorem: Let $f: A \mapsto \Pi_{a \in J} X_\alpha$ be given by the equation $f(a) = (f_\alpha (a))_{\alpha \in J}$, where $f_\alpha : A \mapsto X_\alpha$ for each $\alpha$. Let $\Pi X_\alpha$ have the product topology. Then the function $f$ is continuous if and only if each function $f_\alpha$ is continuous. One of the implications stated in the preceeding theorem holds for the box topology. Which one? The following implication holds: if $f$ is continuous, then each $f_\alpha$ is continuous. This is true since each $f_\alpha$ is a function that restricts the range of $f$. The reverse implication, however, is not necessarily true, as shown in example 2, pg. 117. Prove Theorem 20.4: The uniform topology on ${R}^J$ is finer than the product topology and coarser than the box topology; these three topologies are different if $J$ is infinite. 8.1A Show the uniform topology is finer than the product topology. Take a basis element in the product topology: $\Pi U_\alpha$, where each $U_\alpha$ is open in ${R}$. Show $\Pi U_\alpha$ is open in the uniform topology. Take a point $x$ in ${R}^J$; represent $x$ as $x = (x_\alpha)$ for $\alpha \in J$. By definition of the product topology, there only finitely many of the $U_\alpha \neq {R}$, so denote let $\alpha_1, \alpha_2, ..., \alpha_n$ be the values of $\alpha$ for which $U_\alpha \neq {R}$. Next, for each $i$, choose $\epsilon_i > 0$ so that $B_{\overline{\rho}}(x_{\alpha_i}, \epsilon_i) \subset U_{\alpha_i}$, which is possible since by definition each $U_{\alpha_i}$ is open in ${R}$. Claim: if we let $\epsilon = \min \{ \epsilon_1, ..., \epsilon_n \}$, $x \in B_{\overline{\rho}}(x, \epsilon) \subset \Pi U_\alpha$. If a point $z \in B_{\overline{\rho}}(x, \epsilon)$ then $\overline{d}(x_\alpha,z_\alpha) < \overline{\rho}(x,z) < \epsilon \leq \epsilon_i$ for all $i$ and $\alpha$. Hence $z_{\alpha_i} \in B_{\overline{\rho}}(x_{\alpha_i}, \epsilon_i) \subset U_{\alpha_i}$ for all $\alpha_i$ and $z_\alpha \in R = U_\alpha$ for all $\alpha \not= \alpha_1, ..., \alpha_n$, so $z \in \Pi U_\alpha$. Hence the uniform topology is finer than the product topology. 8.1B For $J$ infinite, show that the uniform topology is different than the product topology. $B_{\overline{\rho}}({\bf 0}, 1) = \Pi_{\alpha \in J} (-1, 1)$ which is not open in the product topology when $J$ is infinite. Thus, the uniform topology is different than the product topology when $J$ is infinite. 8.1C Show that the box topology is finer than the uniform topology. Note the book's proof is incorrect. Begin by letting $B$ be the $\epsilon_1$-ball around $y$ in the uniform topology. Show $B$ is open in the box topology. Take $x \in B$. Then there exists a $B_{\overline{\rho}}(x, \epsilon) \subset B_{\overline{\rho}}(y, \epsilon_1)$ Then we could define a neighbourhood of $x$ in the box topology as $U = \prod (x_\alpha - {1 \over 2}\epsilon, x_\alpha + {1 \over 2}\epsilon)$. Claim $U$ is contained in $B$. For any element $y$ of $U$, $\overline{d}(x_\alpha,y_\alpha) < {1 \over 2}\epsilon$ for all $\alpha$. This implies that $\overline{\rho}(x,y) \leq {1 \over 2}\epsilon$, implying that the box topology is indeed finer than the uniform topology. 8.1D Show that for $J$ infinite, the uniform topology and the box topology are different. Take $I \subset J$ where $I$ is a countable subset of $J$. Let $x_{n_\alpha} = \cases { {1 \over n} & $\alpha \in I$ \cr 0 & $\alpha \not\in I$}$. If ${\bf x}_n = \Pi{\alpha \in J} x_\alpha$ and ${\bf x} = \Pi_{\alpha \in J} 0$, then $\overline{\rho}({\bf x_n}, {\bf 0}) = {1 \over n}$. Hence ${\bf x_n} \rightarrow {\bf 0}$ in the uniform topology, but ${\bf x_n}$ does not converge to ${\bf 0}$ in the box topology. Thus, the uniform topology and the box topology are different when $J$ is infinite. 8.2: Using the metric $D$ defined in Theorem 20.5, describe $B_D (0, {1 \over3})$. First off we note that $D(0,x) < {1 \over3}$ for $x$ to be an element of $B_D$. But $D(0,x) = sup \{ { min \{ d(0,x_i),1 \} \over i} \} = \sup \{ { min \{|x_i|,1 \} \over i} \} < c{1 \over 3} $ for each $i$. For $i = 1$, any point $x$ where $|x_1| < {1 \over 3}$ will satisfy the above equation. For $i=2$, any point $x$ where $|x_2| < {2 \over 3}$ will work. For $i=3$, any point $x$ where $|x_3| < 1$ will work. For $4 \leq i$, any point $x$ where $x_4 \in {R}$ will work. Hence $B_D(0,{1 \over3}) = (-{1 \over 3}, {1 \over 3}) \times (-{2 \over 3}, {2 \over 3}) \times (-1, 1) \times {R} \times {R} \times ... $ p. 145: Repeat 4A for the equivalence relation $x_0 \times y_0 ~ x_1 \times y_1$ if $x_0^2 + y_0^2 = x_1^2 + y_1^2$. $f: X* \rightarrow [0, \infty), ~f((x, y) ) = \sqrt(x^2 + y^2)$ is a homeomorphism where $[0, \infty)$ is a subspace of $R$. Thus $X*$ is equivalent to the subspace $[0, \infty)$ of $R$. pg 152 1.) Let ${\cal T}'$ and ${\cal T}$ be two topologies on $X$, where ${\cal T}'$ is finer than ${\cal T}$. What does the connectedness of $X$ in one imply about its connectedness in the other? $R_l$ is finer than $R$ which is finer than $R$ with the indiscrete topology, $R$ and $R$ with the indiscrete topology are connected but $R_l$ is not connected by problem 7. Thus connectedness of $X$ in ${\cal T}$ does not imply anything about the connectedness of $X$ in ${\cal T}'$ Suppose ${\cal T}$ is not connected, then $X = U \cup V$ where $U, V \in {\cal T} \subset {\cal T}'$. Thus, ${\cal T}'$ is not connected. Hence connectedness of $X$ in ${\cal T}'$ implies connectedness of $X$ in ${\cal T}$. 7.) $R_l$ is not connected since $R_l = (\infty, 0) \cup [0, \infty)$ and $(\infty, 0)$ and $[0, \infty)$ are both open in $R_l$. p. 157 8A: If two spaces $X$ and $Y$ are path connected, is the space $X \times Y$ path connected? Let $a \times b$ and $c \times d$ be distinct points in $X \times Y$. Since $X$ is path connected, there exists a path $\gamma_1: [0, 1] \rightarrow X$ from $a$ to $c$. Similarly, since $Y$ is path connected, there exists a path $\gamma_2 [0, 1] \rightarrow Y$ from $b$ to $d$, Hence $\gamma = (\gamma_1, \gamma_2): [0, 1] \rightarrow X \times Y$ is a path from $a \times b$ to $c \times d$, so $X \times Y$ is indeed path connected. 8B, If $A \subset X$ and $A$ is path connected, is $\overline{A}$ also path connected? Not necessarily. Take $X = {R}^2$ and $A = S$, where $S$ is the topologist's sine curve given in example 7, pg 157. $S$ is path connected, but $\overline{S}$ is not. 8CIf $f: X \rightarrow Y$ is continuous and $X$ is path connected, is $f(X)$ necessarily path connected? Fix $x,y$ in $f(X)$. Take $c \in f^{-1}(x)$ and $d \in f^{-1}(y)$, the inverse images of $x$ and $y$. Then since $X$ is path connected, there exists a continuous function $\gamma_X: [a,b] \mapsto X$, where $[a,b]$ is a subset of ${R}$ such that $\gamma_X(a) = c$ and $\gamma_X(b) = d$. Since both $f$ and $\gamma_X$ are continuous, the composition of $f$ with $\gamma_X$ will also be a continuous function. Let this function be $\gamma$. Thus $\gamma = f \circ \gamma_X: [a,b] \mapsto f(X)$, where $\gamma(a) = f \circ \gamma_X(a) = f(c) = x$ and $\gamma (b) = f \circ \gamma_X(b) = f(d) = y$, so $f(X)$ is path connected. 8D If $\{ A_\alpha \}$ is a collection of path-connected subspaces of $X$, and if $\cap A_\alpha \neq \emptyset$, is $\cup A_\alpha$ necessarily path connected? Since $\cap A_\alpha \neq \emptyset$, by the definition of intersection there exists a point $y$ such that $y \in A_\alpha$ for all $\alpha$. Let $x, z$ be two points in $\cup A_\alpha$. $x$ therefore must be an element of $A_\alpha$ for some $\alpha$ by the definition of union; similarly for $z$. Let $x \in A_1$ and $z \in A_2$. Since $A_1$ is path connected, and $x,y \in A_1$, there exists a path $\gamma_1: [0, 1] \rightarrow A_1$ between $x$ and $y$. Similarly, there exists a path $\gamma_2: [1, 2] \rightarrow A_2$ between $y$ and $z$ in $A_2$. Hence $\gamma[0, 2] \rightarrow \cup A_\alpha, \gamma(x) = \cases{\gamma_1(x) & $x \in [0, 1]$ \cr \gamma_2(x) & $x \in [1, 2] $}$ is a path connecting $x$ and $z$ via $y$, a point in common. p. 170 1a.) (R, indiscrete topology) is compact, but (R, discrete topology) is not compact. Thus ${\cal T} \subset {\cal T}'$, $X$ compact w.r.t ${\cal T}$ does not imply $X$ is compact w.r.t ${\cal T}$. Suppose ${\cal T} \subset {\cal T}'$ and $X$ is compact w.r.t ${\cal T}'$. Let $X \subset \cup_{a \in A} U_a$, where $U_a \in {\cal T} \subset {\cal T}'$. Since $X$ is compact w.r.t ${\cal T}'$ and $U_a \in {\cal T}'$, there exists $a_1, ..., a_n \in A$ such that $X \subset \cup_{i = 1}^n U_{a_i}$. Thus, $X$ is compact w.r.t ${\cal T}$ 1b.) Suppose that $X$ is compact Hausdorff under both ${\cal T}$ and ${\cal T}'$ and that ${\cal T}$ and ${\cal T}'$ are comparable. Since ${\cal T}$ and ${\cal T}'$ are comparable, then either ${\cal T} \subset {\cal T}'$ or ${\cal T}' \subset {\cal T}$. Without loss of generality, assume ${\cal T} \subset {\cal T}'$. Suppose $A$ is a closed set in ${\cal T}'$. Then $A$ is compact w.r.t ${\cal T}'$. Hence by 1a, $A$ is compact w.r.t ${\cal T}$. Hence $A$ is closed in ${\cal T}$. Thus, ${\cal T} = {\cal T}'$. 2a.) Let $A \subset R$ where $R$ has finite complement topology. Let $A \subset \cup_{d \in D} U_d$ be an open cover of $A$. Take $d_0 \in D$ such that $U_{d_0} \not= \emptyset$. Since $R - U_{d_0}$ is finite, $A- U_{d_0}$ is also finite and hence $A- U_{d_0}= \{a_1, ..., a_n\}$. For each $i = 1, ..., n$, take $d_i \in D$ such that $x_i \in U_{d_i}$. Then $A \subset \cup_{i = 0}^n U_{d_i}$. Hence $A$ is compact. 2b.) Let $U_n = R - \{ n, n + 1, n + 2, ...\}$. Then $R \subset \cup_{i = 1}^\infty U_{n}$, but there does not exist a finite subcover. Hence $R$ with the co-countable topology is not compact. 4.) Suppose $A \subset X$ is not bounded where $X$ is metrizable. Take $a \in A$. Then $A \subset \cup_{n = 1}^\infty B(a, n)$, but there does not exist a finite subcover of this cover (for otherwise $A$ would be bounded). Hence $A$ is not compact. Suppose $A$ is not closed. Then there exists $x_0 \in A' - A$. Since $x_0$ is a limit point of $A$. There exists $a_n \in B(x_0, {1 \over n}) \cap A - x_0$. Note that $U = \{x \in X ~|~ d(x, x_0) > {1 \over 4}\}$ is open in $X$. Thus $A \subset \cup_{n = 1}^\infty B(a_n, {1 \over 2n}) \cup U$, but there does not exist a finite subcover of this cover. Hence $A$ is not compact. $R$ with the discrete metric is closed and bounded, but not compact. 5.) Since $A$ and $B$ are closed, you can use thm 32.4 \end