%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%% exercise set 7 - solutions %%% November 20, 2009 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \documentclass{amsart} \usepackage{graphicx} \usepackage{amssymb} \usepackage{epstopdf} \usepackage{nicefrac} \DeclareGraphicsRule{.tif}{png}{.png}{`convert #1 `dirname #1`/`basename #1 .tif`.png} \pagestyle{plain} % \input hw.sty \textwidth = 6 in \textheight = 8.5 in \oddsidemargin = 0.0 in \evensidemargin = 0.0 in \topmargin = 0.0 in \headheight = 0.0 in \headsep = 0.0 in \parskip = 6pt \parindent = 0.0in % \hoffset=-.7in % \voffset=-.7in \newcommand{\ds}{\displaystyle} \newcommand{\mA}{{\mathbb A}} \newcommand{\mB}{{\mathbb B}} \newcommand{\mD}{{\mathbb D}} \newcommand{\mN}{{\mathbb N}} \newcommand{\mQ}{{\mathbb Q}} \newcommand{\mR}{{\mathbb R}} \newcommand{\mS}{{\mathbb S}} \newcommand{\mZ}{{\mathbb Z}} \newcommand{\cA}{{\mathcal A}} \newcommand{\cB}{{\mathcal B}} \newcommand{\cC}{{\mathcal C}} \newcommand{\cE}{{\mathcal E}} \newcommand{\cG}{{\mathcal G}} \newcommand{\cI}{{\mathcal I}} \newcommand{\cO}{{\mathcal O}} \newcommand{\cP}{{\mathcal P}} \newcommand{\cT}{{\mathcal T}} \newcommand{\cU}{{\mathcal U}} \begin{document} \begin{center} Math 445, Fall 2009 \hfill Exercise Set \#7 \hfill Solutions \end{center} \bigskip {\bf 1.} ~ Let $X$ be a topological space, and suppose that $A, B \subset X$ are compact subsets. Show that $A \cup B$ is again compact. \proof We must show that an open cover of $A \cup B$ has a finite subcover. Let $\cU = \{U_{\alpha} \mid \alpha \in \cA\}$ be an open cover of $A \cup B$. Then $\cU$ restricts to an open cover of $A$. As $A$ is compact, there is a finite collection $\{U_{\alpha_1} , U_{\alpha_2} , \ldots , U_{\alpha_n}\} \subset \cU$ which is an open cover for $A$. Similarly, $\cU$ is an open cover for $B$, so there is a finite subcollection $\{U_{\beta_1} , U_{\beta_2} , \ldots , U_{\beta_m}\} \subset \cU$ which is an open cover for $B$. Then the union of these two collections is a finite subcover of $A \cup B$. \endproof \bigskip {\bf 2.} ~ Let $X, Y$ be topological spaces, $f \colon X \to Y$ a continuous map, and assume that $X$ is \emph{compact} and $Y$ is \emph{Hausdorff}. Show that $f$ is a closed map. That is, if $F \subset X$ is closed, then $f(F) \subset Y$ is closed. \proof Let $F \subset X$ be a closed subset. Since $X$ is compact, $F$ is also compact. Since $f$ is a continuous map, the image $f(F)$ is compact. Since $Y$ is Hausdorff, the compact subset $f(F)$ is closed. \endproof \bigskip {\bf 3.} [\#3, page 103] ~ Let $f \colon X \to Y$ be a continuous one-to-one mapping of a compact metric space $X$ onto a metric space $Y$. Prove that $f^{-1} \colon Y \to X$ is continuous (and thus $f$ is a homeomorphism.) \proof We must show that $f^{-1} \colon Y \to X$ is a continuous map. This is the same as showing that $f \colon X \to Y$ is an open map. Let $U \subset X$ be open, then $F = X - U$ is closed. Since $X$ is compact, $F$ is a closed set. Since $Y$ is a metric space, it is Hausdorff. Thus, by the problem 2) above, $f(F)$ is a closed subset. Now, $f$ is a bijection, so $f(U) = f(X - F) = Y - f(F)$ which is open, as $f(F)$ is closed. \endproof \bigskip {\bf 4.} ~ Show that a topological space $X$ is Hausdorff if and only if the diagonal $$\Delta(X) = \{(x,x) \mid x \in X\} \subset X \times X$$ is closed for the product topology. \proof Suppose that $X$ is Hausdorff. We must show that $\Delta(X)$ is a closed set, or that $X \times X - \Delta(X)$ is an open set. Let $(x,y) \in X \times X$ with $x \ne y$. Since $X$ is Hausdorff, there exists open sets $U, V \subset X$ with $x \in U$, $y \in V$ and $U \cap V = \emptyset$. Then $U \times V$ is an open set for the product topology on $X \times X$. Also, $U \times V \cap \Delta(X) = \emptyset$, as $(x,x) \in U \times V$ implies that $x \in U \cap V$. Thus, $(x,y) \in U \times V \subset X - \Delta(X)$ is an open neighborhood. Conversely, suppose that $\Delta(X)$ is a closed subset, then $X \times X - \Delta(X)$ is an open set in the product topology. Thus, for each $x \ne y$, there exists an element $U \times V$ of the basis for the product topology such that $(x,y) \in U \times V \subset X \times X - \Delta(X)$. Each of $U, V \subset X$ is open by definition. The inclusion $(x,y) \in U \times V$ implies that $x \in U$ and $y \in V$. Finally, $U \times V \cap \Delta(X) = \emptyset$ implies that $U \cap V = \emptyset$. Thus, we have shown there exists disjoint open neighborhoods of the distinct points $x \ne y$. Hence, $X$ is a Hausdorff space. \endproof \bigskip The graph of a function $f \colon X \to Y$ is the set $\ds \cG_f ~ = ~ \{ (x,f(x)) \mid x \in X\}$. {\bf 5.} ~ Let $X, Y$ be topological spaces, and suppose that $Y$ is compact Hausdorff. a) Show that if $f \colon X \to Y$ is continuous, then the graph $\cG_f $ of $f$ is closed in $X \times Y$. \proof We show that $(X \times Y) - \cG_f$ is an open set in the product topology. Let $(x,y) \in (X \times Y) - \cG_f$. Then $u = f(x) \ne y$ as $(x,y) \not\in \cG_f$. Since $Y$ is Hausdorff, there exists disjoint open sets $U,V \subset Y$ with $u \in U$ and $y \in V$. As $f$ is continuous, $W = f^{-1}(U) \subset X$ is an open set, and $x \in W$ as $u \in U$. We claim that $(W \times V) \cap \cG_f = \emptyset$: let $\xi \in W$ and $\zeta \in V$ then $f(\xi) \in U$ which is disjoint from $V$, hence $f(\xi) \ne \zeta$. It follows that $W \times V$ is an open neighborhood of $(x,y) \in (X \times Y) - \cG_f$. Thus $(X \times Y) - \cG_f$ is open, so $\cG_f$ is closed. \endproof \medskip b) Show that if the graph $\cG_f $ of $f$ is closed, then $f \colon X \to Y$ is continuous. \proof For $x \in X$ with $y = f(x) \in V$, we show there exists open $U \subset X$ with $x \in U$ and $f(U) \subset V$. If $X$ and $Y$ are metric spaces, the basic idea of the proof is clear - we use the graph is closed to show that $f$ maps converging sequences to converging sequences, which is equivalent to continuity for the metric topology. We give this proof first, to show the basic idea. Suppose that there exists $x \in X$ with $f(x) = y \in V$ but for every open neighborhood $x \in U$ we have $f(U) \cap (Y - V) \not= \emptyset$. In particular, this holds for each $U = B(x, 1/n)$, so there exists a sequence $\{x_n\} \subset X$ with $x_n \to x$ and $f(x_n) \in (Y-V)$. The set $(Y-V)$ is closed, hence is compact metric, so is sequentially compact. Thus, the sequence $\{y_n = f(x_n)\}$ has a convergent subsequence $y_{n_i} \to y_* \in Y- V$. It follows that the sequence $\{(x_{n_i}, y_{n_i}) \mid n = 1,2, \ldots \}$ converges to $(x, y_*) \in X \times (Y - V)$. As $\cG_f$ is closed, we must have that $(x,y_*) \in \cG_f$, or that $y_* = f(x) = y$, which is a contradiction, as $y_* \in Y -V$ but $y \in V$ was assumed. The proof for the general case uses the same idea, but we cannot use sequences, so must argue that $f^{-1}(Y-V)$ misses an open neighborhood of $x \in X$. Let $V \subset Y$ be an open set and $x \in f^{-1}(V)$, then set $y = f(x) \in V$. The complement $K = Y - V$ is closed, so $X \times K$ is also closed in the product topology, as $X \times V$ is open. The graph $\cG_f$ is a closed subset of $X \times Y$, so the intersection $\cG_f \cap K$ is closed. So far, the proof is exactly the same as in the metric case. It is the next step where it differs, as we cannot use the metric to choose a sequence in $\cG_f \cap K$ as above and consider a cluster point for the sequence to get a contradiction. Instead, we must use the compactness of $Y$ directly. Since $f(x) \not\in K$, we have $(\{x\} \times Y) \cap (\cG_f \cap K) = \emptyset$. Thus, for each $\zeta \in Y$ we have $(x,\zeta) \not\in \cG_f \cap K$. As $\cG_f \cap K$ is closed, there exists an open set (in the product topology on $X \times Y$) containing $(x,\zeta)$ and disjoint from $\cG_f \cap K$. That is, for each $(x,\zeta)$ there exists open sets $x \in U_{\zeta} \subset X$ and $\zeta \in V_{\zeta} \subset Y$ with $(U_{\zeta} \times V_{\zeta}) \cap (\cG_f \cap K) = \emptyset$. The collection $\{V_{\zeta} \mid \zeta \in Y\}$ is an open cover for $Y$, so by compactness of $Y$ we can choose a finite subcover, $\{V_{\zeta_1} , \ldots , V_{\zeta_m}\}$ where $\{\zeta_1, \ldots , \zeta_m\} \subset Y$. Then the collection $\{U_{\zeta_1} \times V_{\zeta_1} , \ldots , U_{\zeta_m} \times V_{\zeta_m}\}$ is a finite open cover of $\{x\} \times Y$. Set $U = U_{\zeta_1} \cap \cdots \cap U_{\zeta_m}$, then $U \times Y$ is an open set in $X \times Y$ with $(U \times Y) \cap (\cG_f \cap K) = \emptyset$, as each open set $U_{\zeta_i} \times V_{\zeta_i}$ is disjoint from $(\cG_f \cap K)$. But this implies that for each $\xi \in U$ we have $f(\xi) \not\in K$, or $U \subset f^{-1}(V)$ as was to be shown. \endproof \vfill \eject {\bf 6.} [\#5, page 104] ~ Let $A, B \subset X$ be disjoint subsets of a metric space $X$. Suppose that $A$ is closed, and $B$ is compact. Prove that the distance between $A$ and $B$ is positive. That is, show that $$\inf ~ \{D(a,b) \mid a \in A , ~ b \in B\} > 0$$ \proof Suppose not, then there exists a sequence of points $\{(x_n , y_n) \in A \times B\}$ with $D(x_n , y_n) \to 0$. As $B$ is compact, there exists a convergent subsequence $\{y_{n_i}\}$ with $y_{n_i} \to y_* \in B$ and $n_1 < n_2 < \cdots$. Note that $n_i < n_{i+1}$ implies, by induction, that $n_i \geq i$. Claim: $\ds \lim_{i \to \infty} ~ D(x_{n_i}, y_*) = 0$. Given $\epsilon > 0$ there exists $N$ such that $n \geq N$ implies $D(x_n , y_n) < \epsilon/2$. Also, there exists $I$ such that $i \geq I$ implies $D(y_{n_i}, y_*) < \epsilon/2$. Let $J = \max \{N, I\}$. Then $i \geq J$ implies $n_i \geq N$ and $i \geq I$, hence $$D(x_{n_i} , y_*) \leq D(x_{n_i} , y_{n_i}) + D(y_{n_i}, y_*) < \epsilon/2 + \epsilon/2 = \epsilon$$ It follows that $x_{n_i} \to y_*$ and thus $y_*$ is in the closure of $X$. But $X$ is closed, hence $y_* \in X$. But $A \cap B = \emptyset$ is given, so this is a contradiction. It follows that there is no sequence of points $\{(x_n , y_n) \in A \times B\}$ with $D(x_n , y_n) \to 0$. This implies that $\ds \inf \{D(a,b) \mid a \in A , ~ b \in B\} > 0$, as was to be shown. \endproof \medskip {\bf 7.} ~ Let $\{X,d\}$ be a compact metric space. Suppose that $f \colon X \to X$ is a function which satisfies $$d(f(x), f(y)) ~ < ~ d(x, y) ~~ {\rm for\;\; all} ~~ x \ne y$$ Prove that $f$ has a {unique fixed point}. \proof Define a sequence of subsets inductively: $K_0 = X$, $K_1 = f(X) = f(K_0)$, and $K_n = f(K_{n-1})$. Note that $K_1 = f(X) \subset X = K_0$, so by induction we have $K_{n+1} \subset K_n$. As $X$ is compact and $f$ is continuous, each $K_n \subset X$ is a compact subset. Since $X$ is a metric space, it is Hausdorff, hence $K_n$ is also closed. Thus, we have a descending chain of closed subsets of $X$, $$K_0 \supset K_1 \supset K_2 \supset \cdots$$ As $X$ is compact, the intersection $\ds K_* = \bigcap_{n=1}^{\infty} ~ K_n \ne \emptyset$. Since $f(K_n) = K_{n+1}$ we have $$f(K_*) ~ = ~ f(\bigcap_{n=1}^{\infty} ~ K_n) ~ = ~ \bigcap_{n=1}^{\infty} ~ f(K_n) ~ = ~ \bigcap_{n=1}^{\infty} ~ K_{n+1} ~ = ~ K_*$$ We claim that $K_*$ consists of a single point. Suppose not, that there exists $x, y \in K_*$ with $x \ne y$. Then $\ds \text{diam}(K_*) = \sup \{D(x,y) \mid x,y \in K_*\} > 0$. The distance function $D \colon K_* \times K_* \to \mR$ is continuous. As $X$ is compact, the closed subset $K_*$ is also compact, hence $K_* \times K_*$ is compact in the product topology. Thus, the distance function realizes its maximum on $K_* \times K_*$. That is, there exists $u,v \in K_*$ such that $D(u,v) = \text{diam}(K_*) > 0$. Now, $f \colon K_* \to K_*$ is onto, as $f(K_*) = K_*$. Choose $x , y \in K_*$ with $f(x) = u$ and $f(y) = v$. Now, we use the contraction property of $f$, $$ \text{diam}(K_*) = D(u,v) = D(f(x), f(y)) < D(x,y) \leq \text{diam}(K_*)$$ which is a contradiction. Thus, $K_* = \{x_*\}$ consists of a single point. And $f(K_*) = K_*$ implies $f(x_*) = x_*$. Uniqueness follows as usual. Suppose there exists distinct points $x_* \ne y_*$ with $f(x_*) = x_*$ and $f(y_*) = y_*$. Then $0 < D(x_*, y_*) = D(f(x_*) , f(y_*) ) < D(x_* , y_*)$ which is a contradiction. \endproof \vfill \end{document}