%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%% exercise set 8 - solutions %%% November 30, 2009 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \documentclass{amsart} \usepackage{graphicx} \usepackage{amssymb} \usepackage{epstopdf} \usepackage{nicefrac} \DeclareGraphicsRule{.tif}{png}{.png}{`convert #1 `dirname #1`/`basename #1 .tif`.png} % \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{\cI}{{\mathcal I}} \newcommand{\cP}{{\mathcal P}} \newcommand{\cS}{{\mathcal S}} \newcommand{\cT}{{\mathcal T}} \newcommand{\cU}{{\mathcal U}} \begin{document} \begin{center} Math 445, Fall 2009 \hfill Exercise Set \#8 \hfill Solutions \end{center} \bigskip {\bf 1.} ~ Consider a Hausdorff topological space $T = \{A , \cT\}$. Consider the collection of open subsets $$\cT' = \{U \subset A \mid A- U ~ \text{is ~ compact ~ in} ~ T\} \cup \{\emptyset\}$$ a) Show that $T' = \{A, \cT'\}$ is a topological space. (i.e., that $\cT'$ satisfies the axioms of a topology.) b) Show that $(A, \cT')$ is a compact topological space. {\bf Remarks:} Note that $(A, \cT)$ is Hausdorff implies that $K$ is closed for any compact set $K \subset A$. Thus, $A - K$ is an open set in $\cT$ for any compact $K \subset A$. This is used implicitly in both parts a) and b). Also, $U$ is an open set iff $U = A - K$ for $K \subset A$ compact. This is a variation on the Zariski topology, since every finite point-set is compact. \proof [Proof of a)] Note that $A = A - \emptyset$ is in the topology, since $\emptyset$ is compact. Given two open sets $U = A - K$ and $V = A - L$ then $U \cap V = A - (K \cup L)$. The union of two compact sets is compact, so this is an open set in $\cT'$. Given a collection $\{U_{\alpha} = A - K_{\alpha} \mid \alpha \in \cA\}$ we have $\ds U = \bigcup_{\alpha \in \cA} ~ U_{\alpha} ~ = ~ A - \bigcap_{\alpha \in \cA} ~ K_{\alpha}$ and an arbitrary intersection of compact sets is compact, so $U \in \cT'$. \endproof \proof [Proof of b)] Let $\{U_{\alpha} = A - K_{\alpha} \mid \alpha \in \cA\}$ be an open covering for $A$. Note that each set $U_{\alpha}$ is an open set in the original topology $\cT$ as it is Hausdorff. Pick any $\alpha \in \cA$, and set $U_0 = U_{\alpha}$. The set $K_0 = K_{\alpha}$ is compact in $A$ for $\cT$, and $\{U_{\alpha} = A - K_{\alpha} \mid \alpha \in \cA\}$ restricts to an open covering (in $\cT$) of $K_0$, so there exists a finite subcovering, $\{U_i = A - K_{\alpha_i} \mid i = 1,2, \ldots, n \}$ of $K_0$. But then as each set $U_i$ is also open in $\cT'$ this is a finite covering for $K_0$ in $\cT'$. Then $\{U_0, U_1, \ldots, U_n\}$ is an open subcovering for all of $A$. \endproof {\bf 2.} ~ Let $X = [0,2]$ be the closed interval in $\mR$, but with a topology $\cT$ on $X$ as follows: $$U \in \cT \quad \Longleftrightarrow \quad \text{either} ~ 1 \not\in U, ~ \text{or} ~ (0,2) \subset U$$ Find the closure of the subset $A = \{\frac{1}{2}\}$ of $X$. \proof We have $y \in \overline{A}$ if and only if for every closed set $F \subset X$ with $A \subset F$ we have $y \in F$. The closed sets are the complements of the open sets, so there are two types: $F = \{0\}, \{2\}$ or $\{0,2\}$, corresponding to the case where $(0,2) \subset U$; or $F \subset X$ any subset with $1 \in F$, which is the complement of the condition $1 \not\in U$. Note that $A$ is not a closed subset, but $F = \{\frac{1}{2} , 1\}$ is a closed subset containing $A$. This is clearly a minimal closed subset containing $A$, hence $\overline{A} = \{\frac{1}{2} , 1\}$. \endproof {\bf 3.} ~ Prove that there is no continuous bijective map $f \colon \mS^1 \to [0,1]$. [Hint: think connected.] \proof Suppose that there exists a continuous bijective map $f \colon \mS^1 \to [0,1]$. Then the domain is compact and the range is Hausdorff, so $f$ is a homeomorphism. This means that $\mS^1$ and $[0,1]$ have identical topological properties. We obtain a contradiction by giving a topological property that is not preserved. Let $\frac{1}{2} \in [0,1]$ be the midpoint. Then $Y = [0,1] - \frac{1}{2}$ is a disconnected set. It follows that $X = f^{-1}(Y)$ is also disconnected. Let $\theta = f^{-1}(\frac{1}{2})$, then $f^{-1}(Y) = \mS^1 - \{\theta\}$. Thus, $\theta \in \mS^1$ is a point so that removing it disconnects the circle. It is intuitively obvious that this is impossible; we give a proof. The circle is the quotient space $\mS^1 = [0,1]/ 0 \sim 1$. Let $\theta = 0$ or $\theta = 1$, then $\mS^1 - \{\theta\}$ is a half-open interval, which is connected. On the other hand, if $0 < \theta < 1$ then $\mS^1 - \{\theta\} = [0, \theta) \cup (\theta, 1]$ with $0 \sim 1$. Each interval $[0,\theta)$ and $(\theta,1]$ is connected, and the two intervals have a point in common, $0=1$, in $\mS$, so $\mS^1 - \{\theta\}$ is the union of two connected sets which have a common point, hence is also connected. Thus, in all cases, $\mS^1 - \{\theta\}$ is connected. \endproof \bigskip {\bf 4.} [\#1, page 97] ~ Prove that any subspace of a separable metric space is separable. \proof Recall that $X$ is separable if and only if there exists a dense subset of $X$, and $X$ is second countable if and only if there is a countable basis for the topology of $X$. By Theorems~17.7 and 17.8 of the ``Notes'', a metric space is separable if and only if it is second countable. Let $Y \subset X$. If $X$ is separable, then it is second countable. That is, there is a countable base $\cB$ for the topology $\cT$ of $X$. The metric topology on the subspace $Y$ equals the subspace topology, so the open sets for $Y$ are all of the form $U \cap Y$ where $U \subset X$ is open. Thus, the restriction of the sets in $\cB$ gives a basis for the subspace topology on $Y$. This shows that $Y$ is second countable also. But then $Y$ is separable as it is a metric space. \endproof {\bf 5.} ~ Let $X$ be a locally path-connected space. Show that every \emph{open} connected subset of $X$ is path-connected. \proof $X$ is locally path-connected space implies it has a basis for the topology consisting of path-connected open sets. Suppose that $U \subset X$ is open and connected. Define an equivalence relation $\stackrel{p}{\sim}$ on $U$ where $x \stackrel{p}{\sim} y$ for $x, y \in U$ if and only if there be a continuous path in $U$ from $x$ to $y$. The path components of $U$ are the equivalence classes of $\stackrel{p}{\sim}$. Given $x \in U$, $P_x = \{y \in U \mid y \stackrel{p}{\sim} x\}$ is the path component of $U$ containing $x$. We must show that $P_x = U$. Recall also that each path component $P_x$ of $U$ is a connected set: if $P_x = A \cup B$ is a separation of $P_x$ and there exists $y \in A$ and $z \in B$, then there exists a continuous map $\sigma \colon [0,1] \to U$ with $\sigma(0) = y$ and $\sigma(1) = z$. Then $[0,1] = \sigma^{-1}(A) \cup \sigma^{-1}(B)$ is a separation of the connected set $[0,1]$ which is impossible. So, either $A$ or $B$ must be empty. That is, $P_x$ is connected. For each $x \in U$, we claim that $P_x$ is open. Let $y \in P_x$ and choose a path-connected open set $V \subset U$ with $y \in V$. For any point $z \in V$ there exists a path from $y$ to $z$. We are given there exists a path from $x$ to $y$, so combining these paths we get a path from $x$ to $z$, hence $z \in P_x$. Thus, $y \in V \subset P_x$. Thus, the path components of $U$ are connected open sets. We claim they are disjoint. Suppose there exists $z \in P_x \cap P_y$. Then there exists a path in $U$ from $x$ to $z$, and a path in $U$ from $z$ to $y$, and combining them we get a path in $U$ from $x$ to $y$. That is, $y \stackrel{p}{\sim} x$ and so $P_x = P_y$. Now suppose that $U$ is not path-connected. Then there exists $x, y \in U$ with $P_x \cap P_y = \emptyset$. Define a separation of $U$ as follows: $$A = P_x ~ \text{ and } B = \bigcup_{y \in U - P_x} ~ P_y$$ As each $P_y$ is open, the set $B$ is open. And $A \cap B = \emptyset$, as $P_x \cap P_y \ne \emptyset$ implies that $y \in P_x$. So if there exists $y \in U - P_x$ then $U$ has a non-trivial separation, hence is not connected. \endproof \vfill \eject {\bf 6.} [\#3a, page 97] ~ Let $X$ be a separable metric space, and let $h \colon X \to Y$ be a continuous onto map. Prove that $Y$ is separable. \proof $X$ is separable if and only if there is a countable subset $\cS \subset X$ such that $\cS$ is dense. We claim that $h(\cS) = \{f(x) \mid x \in \cS\} \subset Y$ is a countable dense subset. It is countable, as $\cS$ is countable. Let $y \in Y$, the $h$ is onto so there exists $x \in X$ with $h(x) = y$. Choose a sequence $\{x_n \} \subset \cS$ such that $x_n \to x$. Then $y_n = f(x_n)$ is a sequence in $h(\cS)$ such that, by the continuity of $h$, $$ \lim_{n\to \infty} ~ y_n ~ = ~ \lim_{n\to \infty} ~ h(x_n) ~ = ~ h(\lim_{n\to \infty} ~ x_n) ~ = ~ h(x) = y$$ so $h(\cS)$ is also dense in $Y$. \endproof {\bf 7.} [\#3b, page 97] ~ Let $X$ be a complete metric space, and let $h \colon X \to Y$ be a continuous onto map. Either prove that $Y$ is complete, or give a counter-example. \proof Let $X = \mR$, with the Euclidean metric. This is complete. Let $Y = (0,1) \subset \mR$, with the Euclidean metric. This is not complete, since $x_n = 1/n \to 0 \not\in Y$. Define $f \colon \mR \to (0,\infty)$ by $h(x) = e^x$. Define $g \colon (0, \infty) \to (0,1)$ by $g(y) = \frac{y}{y+1}$. Then $h = g \circ f$ is a homeomorphism, from the complete space $\mR$ to the incomplete space $(0,1)$. \endproof {\bf Remark:} This exercise shows that completeness is not a topological property. It is not even a diffeomorphism property. Rather, it is a metric property, as one see the problem with the map $h$ is that it distorts distances by a greater and greater amount as $x \to \pm \infty$. \bigskip {\bf 8.} [\#4*, page 97] ~ Let $M$ be a metric space. Prove that $M$ is separable if and only if every collection of disjoint open sets of $M$ is countable. \proof Assume that $M$ is separable, with countable dense subset $\cS = \{x_n \mid n = 1,2,3, \ldots\} \subset M$. Let $\cU = \{U_{\alpha} \mid \alpha \in \cA\}$ be a collection of disjoint open sets in $M$. We define an injection $s \colon \cA \to \mN$. For each $\alpha \in \cA$ select $n \in \mN$ such that $x_n \in U_{\alpha}$. Note that if $s(\alpha) = n = s(\beta)$ then $x_n \in U_{\alpha} \cap U_{\beta}$. As $U_{\alpha}$ and $U_{\beta}$ are disjoint if $\alpha \ne \beta$, this shows that $s(\alpha) = s(\beta)$ implies $\alpha = \beta$. Thus, $s$ is injective, hence $\cA$ is countable. \medskip Conversely, suppose that every collection of disjoint open sets of $M$ is countable. \endproof \vfill \end{document}