James R. Munkres

Second Edition (2000)

\documentclass[12pt]{article} \begin{document} \title{Host} \author{Sándor Szabó (Dept of Analysis)}

Chapter 3

\maketitle \textbf{Section 23 Connected Spaces} \textbf{1}. \textcolor{green}{Solution}. If $X$ is connected in $\tau'$ then so is in $\tau$. \textbf{3}. \textcolor{green}{Solution}. Suppose $A\cup\left(\cup A_{\alpha}\right)=B\cup C$ is a separation. Since $A$ is connected we may assume $A\subseteq B$. Since $A\cap A_{\alpha}\neq\emptyset$ we have $A_{\alpha}\subseteq B$ which is a contradiction. \textbf{4}. \textcolor{green}{Solution}. Assume $X=A\cup B$ is a separation. If $A$ is cofinite then $B$ is finite, similarly we obtain $A$ is finite which is a contradiction. \textbf{5}. \textcolor{green}{Solution}. $\mathbf{R}_{\ell}$ is totally disconnected (see Exe. 7), although it is not the discrete topology. \textbf{6}. \textcolor{green}{Solution}. $X=Int A\cup^{\star}Bd A\cup^{\star}Int (X-A)$. If $C$ intersects $A$ but not $Bd A$ then $C$ intersects $Int A$, similarly intersects $Int (X-A)$. Then we obtain a separation of $C$\[ C=(C\cap Int A)\cup(C\cap Int (X-A))\] which is a contradiction. \textbf{7}. \textcolor{green}{Solution}. $\mathbf{R}=(-\infty,r)\cup[r,\infty)$ is a separation for any real number $r$. It follows that any subspace of $\mathbf{R}_{\ell}$ containing more than one point is disconnected, hence $\mathbf{R}_{\ell}$ is totally disconnected. \textbf{8}. \textcolor{green}{Solution}. We use the idea of Example 6. Let $A\subset\mathbf{R}^{\omega}$ consist of all bounded sequences of real numbers, and the set $B$ of all unbounded sequences. These sets are disjoint, and each is open in the uniform topology. For, if $\mathbf{a}$ is a point of $\mathbf{R}^{\omega}$, the open set $B_{\overline{\varrho}}(\mathbf{a},1)$ consists entirely of bounded sequences if $\mathbf{a}$ is bounded, and of unbounded sequences if $\mathbf{a}$ is unbounded. Hence $\mathbf{R}^{\omega}$ is not connected in the uniform topology. \textbf{9}. \textcolor{green}{Solution}. We can write $(X\times Y)-(A\times B)=[X\times(Y-B)]\cup[(X-A)\times Y]$. Assume $v\times w\in(X\times Y)-(A\times B)$. Then $X\times w$ is connected (it is homeomorphic to $X$), and $x\times Y$ is connected. Denote $T_{x}:=(X\times w)\cup(x\times Y)$, where $x\in X-A$. By Theorem 23.3 it is connected because $x\times w$ is a common point. Now form the union $U:=\cup_{x\in X-A}T_{x}$ which is connected because $v\times w\in T_{x}$. Here $U_{w}:=U=(X\times w)\cup[(X-A)\times Y]$. Thus $[X\times(Y-B)]\cup[(X-A)\times Y]=\cup_{w\in Y-B}U_{w}$ is connected because $(X-A)\times Y$ is a common set. \textbf{10}. \textcolor{green}{Solution}. (a) $X_{K}$ is homeomorphic to $\prod_{\alpha\in K}X_{\alpha}$, hence is connected. (b) Since $\mathbf{a}$ is common point, $\cup_{K\subset J}X_{K}$, $card\,K\lt\infty$ is connected. (c) Obviously $\cup_{K\subset J}X_{K}\subset X$ ($card\,K\lt\infty$). Let $U:=\prod U_{i}$ be a basis element for the product topology that contains $\mathbf{a}$. We show that $u$ intersects $\cup_{K\subset J}X_{K}$. There is an integer $N$ such that $U_{i}=X_{i}$ for $i\gt N$. Then the point $\mathbf{x}:=(a_{1},\ldots,a_{N},0,0,\ldots)$ of $\cup_{K\subset J}X_{K}$ belongs to $U$, since $a_{i}\in U_{i}$ for all $i$, and $0\in U_{i}$ for $i\gt N$. \textbf{11}. \textcolor{green}{Solution}. Assume $X=A\cup B$ is a separation. $Y_{1}:=\{y\,|\, p^{-1}(y)\subseteq A\}$, $Y_{2}:=\{y\,|\, p^{-1}(y)\subseteq B\}$. (Since $p^{-1}(y)$ is connected it is entirely in either $A$ or $B$.) Obviously $Y_{1}\cap Y_{2}=\emptyset$. From these $p^{-1}(Y_{1})=A$ and $p^{-1}(Y_{2})=B$. Since $A,B$ are open and $p$ open map, we obtain $Y_{1},Y_{2}$ are open, that is $Y_{1},Y_{2}$ is a separation which is impossible. \textbf{12}. \textcolor{green}{Solution}. We proof $Y\cup A$ is connected, the proof for $Y\cup B$ is the same with meaningful modifications. Assume $Y\cup A=C\cup^{\star}D$ is a separation. Then \[ X=(B\cup C)\cup D\] is a partition of $X$. Since $Y$ is connected, by Lemma 23.2 we may assume $Y\subset C$. By Lemma 23.1 we have $\overline{A}\cap B=\emptyset$, $\overline{B}\cap A=\emptyset$, $\overline{C}\cap D=\emptyset$, $\overline{D}\cap C=\emptyset$. Now we can write\[ \overline{B\cup C}\subseteq\overline{B}\cup\overline{C}\subseteq(X-A)\cup(X-D)=(Y\cup B)\cup(B\cup C)=B\cup C.\] Since $\overline{D}\subseteq\overline{A}$ we can write\[ \overline{D}\subseteq(X-C)\cup(X-B)=D.\] Thus $X=(B\cup C)\cup D$ is separation of $X$ which is a contradiction. \\ \textbf{Section 24 Connected Subspaces of the Real Line} \textbf{1}. \textcolor{green}{Solution}. (a) $(0,1)$: $\varphi^{-1}(1)\in(0,1)$, so $(0,1)-\varphi^{-1}(1)$ is not connected, it is impossible $\varphi((0,1)-\varphi^{-1}(1))$ would be $(0,1]$ or $[0,1]$. $(0,1]$ and $[0,1]$: if $\varphi^{-1}(1)\in(0,1)$ the same argument works. If $\varphi^{-1}(1)=1$ then we can consider $\varphi^{-1}(0)$. (b) $f:(0,1)\to(0,1]$, $f(x):=x$, $g:(0,1]\to(0,1)$, $g(x):=\frac{x}{2}$. (c) $\mathbf{R}-\varphi^{-1}(\mathbf{0})$ is not connected, however $\mathbf{R}^{n}-\{\mathbf{0}\}$ is connected. \textbf{2}. \textcolor{green}{Solution}. $g(x):=f(x)-f(-x)$. Then $f$ is a continuous odd function. \textbf{3}. \textcolor{green}{Solution}. $g(x):=f(x)-x$. $g(0)\geq0$ and $g(1)\leq1$. If $X=[0,1)$ or $X=(0,1)$ then the function $f:=1$ has no fixed point. \textbf{4}. \textcolor{green}{Solution}. (2) If $x\lt y$, there exists $z$ such that $x\lt z\lt y$. Indirectly, assume there exist $x\lt y$ such that there is no $z$with this property. Then $X=(-\infty,y)\cup^{\star}(x,\infty)$ is a separation which is a contradiction. (1) $X$ has the least upper bound property. Indirectly, assume the set $A$ has no least upper bound. Define $L:=\{x\,|\,\exists a\in A:x\lt a\}$, $R:=\{x\,|\,\forall a\in A:x\geq a\}$. Obviously $L\cup^{\star}R=X$. We show they are open. Let $x\in L$. By (2) there exists $x\lt z\lt a$, so $(-\infty,z)\subset L$ is a neighborhood of $x$. Let $x\in R$. Since $A$ has no least upper bound we can find $x\gt z\geq a$. Then $(z,\infty)\subset R$ is a neighborhood of $x$. \textbf{5}. \textcolor{green}{Solution}. (a) linear continuum. (b), (c), (d) they have no least upper bound. \textbf{6}. \textcolor{green}{Solution}. (2) Assume $(x_{1},r_{1})\prec(x_{2},r_{2})$. If $x_{1}\lt x_{2}$, then since there exists $x_{1}\lt z\lt x_{2}$ we obtain $(x_{1},r_{1})\prec(z,0)\prec(x_{2},r_{2})$. If $x_{1}=x_{2}$, then since there exists $r_{1}\lt r\lt r_{2}$ we obtain $(x_{1},r_{1})\prec(x_{1},r)\prec(x_{1},r_{2})$. (1) Let $A\subset X\times[0,1)$. Denote $B:=\{x\,|\, x\geq a,\forall a\in\pi_{1}(A)\}$. Then denote $b$ the smallest element of $B$. If $A\cap(b\times[0,1))=\emptyset$ then $(b,0)$ is the least upper bound of $A$. If $C:=A\cap(b\times[0,1))\neq\emptyset$ then $C$ has the order type of $[0,1)$ and $C=\{(b,r)\,|\,(b,r)\in A\}$. Denote $s:=\sup\pi_{2}(C)$. Then $(b,s)$ is the least upper bound of $A$. \textbf{7}. \textcolor{green}{Solution}. (a) $f$ is injective because of order preserving, so it is bijection. It is enough to prove $f$ is continuous, because $f^{-1}$ have the same properties as $f$. Now $f^{-1}((a,b))=(f^{-1}(a),f^{-1}(b))$ is open so $f$ is continuous. (b) If $0\lt a\lt b$ then $a^{n}\lt b^{n}$, because $\left(\frac{a}{b}\right)^{n}\lt 1$. Define $g_{a}(t):=ta^{n}$. Obviously it is continuous in $t$. We know $0=0^{n}\lt 1^{n}\lt 2^{n}\lt \cdots$. If $k^{n}\lt r\lt (k+1)^{n}$ then for the function $g_{k}:\left[1,\left(\frac{k+1}{k}\right)^{n}\right]\to\mathbf{R}$ we can apply the intermediate value theorem (Theorem 24.3) which gives the surjective property. By (a) the $nth$ root function is continuous. (c) $f$ is not homeomorphism, because $\mathbf{R}$ is connected, but $(-\infty,-1)\cup[0,\infty)$ is not. It does not contradict to (a) because if $Y$ is a subset of an ordered set $X$, then restricted the order to $Y$, the resulting order topology on $Y$ is not the same as the topology that $Y$ inherits as a subspace of $X$. Indeed, $[0,\infty)$ is open as a subspace, but not in order topology. \textbf{8}. \textcolor{green}{Solution}. (a) Let $\mathbf{x},\mathbf{y}\in\prod X_{i}$, $\mathbf{x}=(x_{i})$, $\mathbf{y}=(y_{i})$. In the definition of path, we may assume that $[a,b]=[0,1]$. Hence $f_{i}:[0,1]\to X_{i}$ continuous and $f_{i}(0)=x_{i}$, $f_{i}(1)=y_{i}$. By Theorem 19.6 $\mathbf{f}=(f_{i})$ is continuous, hence the product is path connected. (b) Not, see Example 7. Since $S$ is connected so it is path connected. (c) Let $a,b\in f(X)$, $a=f(x)$, $b=f(y)$. Since $X$ is path connected there exists $g:[0,1]\to X$ continuous such that $g(0)=x$, $g(1)=y$. Then $f\circ g$ is a path between $a$ and $b$. (d) Let $p\in\cap A_{\alpha}$ and $a,b\in\cup A_{\alpha}$. Then $a,p\in A_{\alpha_{1}}$ and $b,p\in A_{\alpha_{2}}$. Since $A_{\alpha_{1}},A_{\alpha_{2}}$ are path connected, there exists path $f_{1}$ between $a$ and $p$, and there exists path between $p$ and $b$. By pasting lemma there exists a path ($f_{1}\cup f_{2}$) between $a$ and $b$. \textbf{9}. \textcolor{green}{Solution}. Let $x,y\in\mathbf{R}^{2}-A$. Obviously we can choose a line passing through $a$ which does not intersect $A$. Similarly, we can choose a line passing through $b$ which does not intersect $A$ and not parallel to the other line, consequently they intersect each other in a common point. \textbf{10}. \textcolor{green}{Solution}. Let $x_{0}\in U$ be fixed. Define $G:=\{x\in U\,|\, x_{0}\to x\,path\}$. We show that $G$ is open in $U$. Let $x\in G$. Since $U$ is open there is a $V:=\prod(a_{i},b_{i})\subset U$ neighborhood of $x$. Obviously $V$ is path connected, so $V\subseteq G$. Now we show that $G$ is closed. Let $x\in G'$. Since $U$ is open there is a $V:=\prod(a_{i},b_{i})\subset U$ neighborhood of $x$. Since $x\in G'$ there is $y\neq x\in G\cap V$. Obviously $V$ is path connected, so $x\in G$, hence $G=G'$, that is, $G$ is closed. \textbf{11}. \textcolor{green}{Solution}. $A$ connected but $ Int A$ and $ Bd A$ not connected. $ Int A$ and $ Bd A$ are connected but $A$ is not connected. Let $A:=B(0,1)\cup^{\star}((\mathbf{R}^{2}-D(0,1))\cap(\mathbf{Q}\times\mathbf{Q}))$. Then $ Int A=B(0,1)$ and $ Bd A=\mathbf{R}^{2}-B(0,1)$ are connected but $A$ not. \textbf{12}. \textcolor{green}{Solution}. (a) Assume $[a,c)$ has the order type of $[0,1)$. That is, $\varphi:[a,c)\to[0,1)$ order isomorphism. Then $\varphi:[a,b)\to[0,\varphi^{-1}(b))$ and $\varphi:[b,c)\to[\varphi^{-1}(b),1)$ are order isomorphisms. Assume $\pi:[a,b)\to\left[0,\frac{1}{2}\right)$ and $\psi:[b,c)\to\left[\frac{1}{2},1\right)$ are order isomorphisms. Then $\varphi:=\pi\cup\psi$ is an $\varphi:[a,c)\to[0,1)$ order isomorphism. (b) Assume $\varphi:[x_{0},b)\to[0,1)$ is order isomorphism. Then by (a) $[x_{0},x_{i+1})\cong[0,1)$, and again by (a) $[x_{i},x_{i+1})\cong[0,1)$. Assume $\varphi_{i}:[x_{i},x_{i+1})\to\left[\sum_{k=1}^{i}\frac{1}{2^{k}},\sum_{k=1}^{i+1}\frac{1}{2^{k}}\right)$ are order isomorphisms. Then $\varphi:=\cup_{i=0}^{\infty}\varphi_{i}$ is an $\varphi:[x_{0},b)\to[0,1)$ order isomorphism. (c) First we show that if $a$ has no an immediate predecessor in $S_{\Omega}$ then there is an increasing sequence $a_{i}$ in $S_{\Omega}$ such that $\sup\{a_{i}\}=a$. The set of predecessors $S_{a}=\{b\in S_{\Omega}\,|\, b\lt a\}=\{b_{1},b_{2},\ldots\}$ is countable and the sets $(b_{n},a]=(b_{n},a^{+})$ is a neighborhood basis of $a$. Since $b_{1}$ is not an immediate predecessor, there is an element $a_{1}\in(b_{1},a]$. Since $\sup\{a_{1},b_{2}\}$ is not an immediate predecessor, there is an element $a_{2}\in(\sup\{a_{1},b_{2}\},a]$. Continuing this process we obtain a sequnce of elements $a_{n}\lt a$ such that $a_{n-1}\lt a_{n}$ and $a_{n}\gt b_{n},b_{n-1},\ldots,b_{1}$ for all $n$. It implies, the sequence $a_{i}$ converges to $a$. We show that the interval $[a_{0}\times 0,a\times 0)$ has the order type of $[0,1)$. If $a=a_{0}^{+}$ then $[a_{0}\times 0,a_{0}^{+}\times 0)$ has the order type of $[0,1)$. Suppose the statement holds for all $b\lt a$. We show it holds for $a$. If $a$ has an immediate predecessor $a^{-}$ then $[a_{0}\times 0,a\times 0)=[a_{0}\times 0,a^{-}\times 0)\cup^{\star}[a^{-}\times 0,a\times 0)$. By induction hypothesis $[a_{0}\times 0,a^{-}\times 0)$ has the order type of $[0,1)$. Since $[a^{-}\times 0,a\times 0)$ has the order type of $[0,1)$, applying (a) we obtain the statement. If $a$ has not it, then, as we have seen above, there exists an increasing sequence $a_{i}$ in $S_{\Omega}$ such that $\sup\{a_{i}\}=a$. Applying induction hypothesis for $[a_{0}\times 0,a_{i+1}\times 0)$ and (a) ($[a_{0}\times 0,a_{i+1}\times 0)=[a_{0}\times 0,a_{i}\times 0)\cup^{\star}[a_{i} \times 0,a_{i+1}\times 0)$) we obtain $[a_{i}\times 0,a_{i+1}\times 0)$ has the order type of $[0,1)$. Then by (b) we obtain $[a_{0}\times 0,a_{0}^{+}\times 0)$ has the order type of $[0,1)$. (d,e) Let $x\in L$, $x\neq a_{0}\times 0$. Choose an element $a\in S_{\Omega}$ such that $x\prec a\times 0$. Then $x$ lies in the open interval $(a_{0}\times 0,a\times 0)$ of $L$, which has the order type of the open interval $(0,1)$ of $\mathbf{R}$. By Theorem 16.4 the order topology on $(a_{0}\times 0,a\times 0)$ is the same as the topology $(a_{0}\times 0,a\times 0)$ inherits as a subspace of $L$. By Exe. 7 (a) we obtain $(a_{0}\times 0,a\times 0)$ is homeomorph to $(0,1)$. Since $(0,1)$ is path connected, $(a_{0}\times 0,a\times 0)$ is so. However $L=\cup_{a\gt a_{0}}(a_{0}\times 0,a\times 0)$, each of which is path connected and have a common point $\left(a_{0}\times\frac{1}{2}\right)$, so $L$ is path connected. (f) Obviously $\left(a\times\frac{1}{4},a\times\frac{1}{2}\right)$, $a\in S_{\Omega}$ are open disjoint sets. If there would be imbedding then we obtained uncountable many disjoint open sets in $\mathbf{R}^{n}$. \textbf{Section 25 Components and Local Connectedness} \textbf{1}. \textcolor{green}{Solution}. $\mathbf{R}_{\ell}$ is totally disconnected. Its components are points, and by Theorem 25.5 its path components are also points. Since $\mathbf{R}$ is connected, therefore its continuous map is also connected, so only the constant maps are continuous. \textbf{2}. \textcolor{green}{Solution}. (a) Since the product of connected and path-connected spaces are connected and path-connected respectively, so the components and path components are the space itself. (b) Let $X$ denote $\mathbf{R}^{\omega}$ in the uniform topology. Then $X$ is not connected, because $X=B\cup^{\star}U$, where $B$ is the set of bounded sequences, $U$ is the set of unbounded sequences. They are open, since any sequence within distance $1$ of a (un)bounded sequence is (un)bounded. We shall now determine the path components of $X$. $\mathbf{0}=(0,\ldots,0)$ and $\mathbf{y}=(y_{n})$ are in the same path component $\Longleftrightarrow$ $\mathbf{y}=(y_{n})$ is a bounded sequence. $\Longrightarrow:$ Let $u:[0,1]\to X$ be a path from $\mathbf{0}$ to $\mathbf{y}$. Since $u(0)=\mathbf{0}\in B$ and $u([0,1])$ is connected it must be $u([0,1])\subset B$, hence $u(1)=\mathbf{y}\in B$. $\Longleftarrow:$ We show that $u(t):=t\mathbf{y}$, $t\in[0,1]$ is a path from $\mathbf{0}$ to $\mathbf{y}$. Obviously $u$ is bijection. To see that $u$ is continuous we note that $\overline{\varrho}(u(t_{0}),u(t_{1}))=\sup\{n\in\mathbf{Z}^{+}\,|\,\min(|(t_{0}-t_{1})y_{n}|,1)\}=|t_{0}-t_{1}|M$ when $|t_{0}-t_{1}|\lt \frac{1}{M}$, where $M=\sup\{|y_{n}|\,|\, n\in\mathbf{Z}^{+}\}$. Since $\mathbf{z}\to\mathbf{z}-\mathbf{x}$ is an isometry it is a homeomorphism (see Exe. 21.2) of $X$ onto itself. It implies $\mathbf{x}$ and $\mathbf{y}$ are in the same path component $\Longleftrightarrow$ $\mathbf{y}-\mathbf{x}=(y_{n}-x_{n})$ is bounded. This describes the path components of $X$. It also shows that balls with radius less than $1$ are path connected. Thus $X$ is locally path connected and by Theorem 25.5 the components and the path components of $X$ are the same. (c) Let $X$ denote $\mathbf{R}^{\omega}$ in the box topology. Then $X$ is not connected since the box topology is finer than the uniform topology (see Theorem 20.4 and Exe. 23.1). In fact, $X=B\cup^{\star}U$, where $B$ is the set of bounded sequences, $U$ is the set of unbounded sequences. They are open since they are open in the uniform norm topology. The (path) components of $X$ can be described in the following way. $\mathbf{x}$ and $\mathbf{y}$ are in the same (path) component $\Longleftrightarrow$ $x_{n}=y_{n}$ for all but finitely many $n$. $\Longrightarrow:$ We show that if $\mathbf{x}-\mathbf{y}$ is not eventually zero, then $\mathbf{x}$ and $\mathbf{y}$ are not in the same (path) component. For each $n$ choose a homeomorphism $h_{n}:\mathbf{R}\to\mathbf{R}$ such that $h_{n}(x_{n}):=0$, and $h_{n}(y_{n}):=n$ if $x_{n}\neq y_{n}$. Then $\mathbf{h}=\prod h_{n}:X\to X$ is a homeomorphism (see Exe. 19.8) with $\mathbf{h}(\mathbf{x})=\mathbf{0}$ and $(\mathbf{h}(\mathbf{y}))_{n}=n$ for infinitely many $n$. Since a homeomorphism takes (path) components to (path) components and $\mathbf{h}(\mathbf{x})=\mathbf{0}\in B$ and $\mathbf{h}(\mathbf{y})\in U$ are not in the same (path) component, $\mathbf{x}$ and $\mathbf{y}$ are not in the same (path) component. $\Longleftarrow:$ The map $u(t):=(1-t)\mathbf{x}+t\mathbf{y}$, $t\in[0,1]$ is constant in all but finitely many coordinates. From this it follows $u:[0,1]\to X$ is a continuous path from $\mathbf{x}$ to $\mathbf{y}$, therefore $\mathbf{x}$ and $\mathbf{y}$ are in the same (path) component. By Theorem 25.3 we note that $X$ is not locally connected because there is a component of $X$ which is not open in $X$. Indeed, the component of $\mathbf{0}$ is $\mathbf{R}^{\infty}$, which is closed in box topology (see Exe. 19.7). Thus $\mathbf{R}^{\omega}$ in the box topology is an example of a space where the components and the path components are the same however the itself is not locally path connected, compare with Theorem 25.5. \textbf{3}. \textcolor{green}{Solution}. We know that the ordered square is a linear continuum (Exa. 24.1). We know that any linear continuum is locally connected (the basis of topology consists of intervals which are connected in a linear continuum, Theorem 24.1). We also know that the ordered square is connected but not path connected (Exa. 24.6). From Theorem 25.5 it follows that a connected and no path connected space can not be locally path connected. Now we determine the path components. The subsets $\{x\}\times[0,1]=[x\times0,x\times1]$, $x\in[0,1]$, are convex in the ordered square. By Theorem 16.4 the order topology on $[x\times0,x\times1]$ is the same as the topology inherits as a subspace of the ordered square. Hence $[x\times0,x\times1]$ is homeomorphic to $[0,1]$, so it is path connected. There is no continuous path starting in $[x\times0,x\times1]$ and ending in $[y\times0,y\times1]$ when $x\neq y$. It can be proved the same way as there is no path from $0\times0$ to $1\times1$, see Exa. 24.6. Therefore the sets $\{x\}\times[0,1]=[x\times0,x\times1]$, $x\in[0,1]$ are the path components of the ordered square. (Since the path components are not open we see again by Theorem 25.4 that $I_{o}^{2}$ is not locally path connected.) The space $I_{o}^{2}$ is an example of a space with one component and uncountable many path components. \textbf{4}. \textcolor{green}{Solution}. In a locally path connected space any open set, as a subspace, is locally path connected. By Theorem 25.5 in a locally path connected space the components and the path components are the same. Hence the connected open set is path connected. \textbf{5}. \textcolor{green}{Solution}. (a) The line segments $[r\times0,0\times1]$, $r\in\mathbf{Q}$ are path connected and since they have a common point $p$, $T$ is path connected. Since path connectedness implies connectedness we obtain $T$ is locally connected at $0\times1$. At a different point the ``small enough'' neighborhoods consist of disjoint (open or half open) line segments, so $T$ is not locally connected at other points. (b) \textbf{6}. \textcolor{green}{Solution}. By Theorem 25.3 it is enough to show that components of open sets are open. Let $U\subseteq X$ be open and $C\subseteq U$ be a component. Let $x\in C$. Since $X$ is weakly locally connected at $x$, there is a connected subspace $K\subset U$ that contains a $x\in V_{x}$ neighborhood. Since $x\in C\cap K$ by Theorem 25.1 $K\subset C$. So we have $x\in V_{x}\subset C$. From this $C=\cup_{x\in C}\{x\}\subseteq\cup_{x\in C}V_{x}\subseteq C$, that is $C=\cup_{x\in C}V_{x}$, hence $C$ is open. \textbf{8}. \textcolor{green}{Solution}. Let $p:X\to Y$ be a quotient map where $X$ is locally (path) connected. We prove that $Y$ is locally (path) connected. . By Theorem 25.3 (25.4) we have to show if $U$ is an open set of $Y$ and $C$ a (path) component of $U$, then $C$ is open in $Y$. Since $p$ is a quotient map, it is equivalent to $p^{-1}(C)$ is open in $X$. Since $p$ is continuous, $p^{-1}(U)$ is open. By Theorem 25.3 (25.4) the (path) components of $p^{-1}(U)$ are disjoint open (in $X$) subsets with union $p^{-1}(U)$. If $G$ is a (path) component of $p^{-1}(U)$ such that $G\cap p^{-1}(C)\neq\emptyset$ then $G\subset p^{-1}(C)$. Indeed, since $p$ is continuous, $p(G)$ is (path) connected, furthermore $p(G)\cap C\neq\emptyset$, thus by Theorem 25.1 (25.2) $p(G)\subset C$. Thus we obtain that $p^{-1}(C)=\cup_{j}^{\star}G_{j}$ where $G_{j}$ are open (in $X$), consequently $p^{-1}(C)$ is open in $X$. \textbf{9}. \textcolor{green}{Solution}. Since multiplication is homeomorphism therefore $xC$ and $Cx$ are the components of $G$. Since $x\in xC\cap Cx$ thus $xC=Cx$ for every $x$, that is, $C$ is a normal subgroup of $G$. \textbf{10}. \textcolor{green}{Solution}. (a) Obviously $x\sim x$, and $x\sim y$ implies $y\sim x$. Now consider the transitivity. Assume that $x\sim y$ and $y\sim z$, and prove that $x\sim z$. Indirectly, assume that there exists an $x\in A$, $z\in B$ separation. If $y\in A$\textbf{ }then $y$ and $z$ are separated, if $y\in B$ then $x$ and $y$ are separated which is a contradiction. (b) Let $C$ be a component of $X$, and assume that $C\cap Q_{i}=q_{i}$, $i=1,2$, where $Q_{i}$ are quasicomponents of $X$. It means there exists a $q_{1}\in A$, $q_{2}\in B$ separation. By Lemma 23.2 it is a contradiction, so $C\subset Q_{1}$ or $C\subset Q_{2}$. Now assume that $X$ is locally connected. We have seen above that each component of $X$ lies in a quasicomponent of $X$. Now we show that each quasicomponent lies in a component of $X$. Let $Q$ be a quasicomponent and assume that $c_{i}\in Q\cap C_{i}$, $i=1,2$, where $C_{i}$ are components of $X$. Now we have $c_{2}\in X-C_{1}$, $c_{1}\in C_{1}$. Since in a locally connected space the components are clopen sets (see Theorem 25.3), therefore $X-C_{1}$ and $C_{1}$ are open sets, they form a separation. However it is impossible that two different points ($c_{1}$ and $c_{2}$) of a quasicomponent can be separated. Hence $Q\subset C_{1}$ or $Q\subset C_{2}$. If $Q$ had no two different points, then $Q$ is a component of $X$ at the same time. (c) $A$: first we investigate the quasicomponents of $A$. Let $x_{1}\times y_{1}$ and $x_{2}\times y_{2}$ be in $A$ with $x_{1}\lt x_{2}$. Then there exists $x\notin K$, $x_{1}\lt x\lt x_{2}$. Then $A\cap\left((-\infty,x)\times\mathbf{R}\right)$ and $A\cap\left((x,\infty)\times\mathbf{R}\right)$ is a separation of $A$, hence $x_{1}\times y_{1}$ is not $\sim x_{2}\times y_{2}$. Thus each quasicomponent (and by (b) each component) of $A$ lies in $\frac{1}{n}\times[0,1]$ for some $n\in\mathbf{Z}_{+}$ or in $\{0\times0,0\times1\}$. Since $\frac{1}{n}\times[0,1]$ is connected (in fact, path connected), it is both a component and a quasicomponent of $A$. Since $\{0\times0,0\times1\}$ is disconnected, the remaining components of $A$ are the singletons $\{0\times0\}$ and $\{0\times1\}$. Suppose that $U,V$ is a separation of $A$ with $0\times0\in U$ and $0\times1\in V$. Then there is $n\in\mathbf{Z}_{+}$ such that $\frac{1}{n}\times0\in U$ and $\frac{1}{n}\times1\in V$. Then the connected set $\frac{1}{n}\times[0,1]$ meets both $U$ and $V,$ which is a contradiction. Hence, the remaining quasicomponent of $A$ is $\{0\times0,0\times1\}$. $B$: The set $B-\{0\times1\}$ is path connected, hence connected. Since $B=\overline{B-\{0\times1\}}$ is connected by Theorem 23.4, and is therefore its own unique component and quasicomponent. $C$: Let $C_{1}:=K\times[0,1]$, $C_{2}:=-K\times[-1,0]$, $C_{3}:=[0,1]\times-K$ and $C_{4}:=[-1,0]\times K$. We investigate the quasicomponents of $C$. Let $m\neq n\in\mathbf{Z}_{+}$, and suppose that $U,V$ is a separation of $C$ with $\frac{1}{m}\times0\in U$ and $\frac{1}{n}\times0\in V$. Then there is some $k\in\mathbf{Z}_{+}$ such that $\frac{1}{m}\times\frac{-1}{k}\in U$ and $\frac{1}{n}\times\frac{-1}{k}\in V$. However $[0,1]\times\frac{-1}{k}$ is connected so it is impossible that it meets both $U$ and $V$. Hence $\frac{1}{m}\times0\sim\frac{1}{n}\times0$. Since $\frac{1}{n}\times[0,1]$ is connected (in fact, path connected), we have $\frac{1}{n}\times0\sim\frac{1}{n}\times r$, $0\leq r\leq1$ . Consequently, $C_{1}$ is contained in a single quasicomponent of $C$. Similarly, each of $C_{2},C_{3},C_{4}$ is contained in a quasicomponent. Suppose, $U,V$ is a separation of $C$, and $C_{1}\subseteq U$, we have $cl_{C}(C_{1})\subseteq U$. By Theorem 17.4 $cl_{C}(C_{1})=cl_{\mathbf{R}^{2}}(C_{1})\cap C$. Thus we obtain $0\times K\subseteq cl_{C}(C_{1})\cap C_{4}$, so $C_{1},C_{4}$ belongs to the same quasicomponent. Following a similar argument, we obtain that $C$ is its own unique quasicomponent. By (b) there is no separation of $C$, so it is also its own unique component. The path components are the line segments of $C_{1},C_{2},C_{3},C_{4}$. \textbf{Section 26 Compactness} \textbf{1}. \textcolor{green}{Solution}. (a) If $X$ is compact under $\tau'$, then it is compact under $\tau$, but not conversely. For example $[0,1]$ in the usual topology is compact, but if $\tau'$ is the discrete topology then not ($[0,1]=\cup_{0\leq x\leq1}\{x\}$). (b) Assume that $\tau$ and $\tau'$ are comparable, say $\tau'\supset\tau$. Then the identity map $(X,\tau')\to(X,\tau)$ is a bijective continuous map, so by Theorem 26.6 it is a homeomorphism, so $\tau'=\tau$. \textbf{2}. \textcolor{green}{Solution}. (a) Assume $X$ is an arbitrary (non-empty) set. If $card\,X\lt \infty$ then every covering is finite and every set can be covered. If $card\,X=\infty$ then let $Y\subseteq X$ be arbitrary set and $Y\subseteq\cup_{i\in J}A_{i}$ be an open covering. We may assume that $Y\cap A_{i}\neq\emptyset$. Then take $A_{1}$. If $Y\subseteq A_{1}$ then we are done. Otherwise $A_{1}$ contains all but finitely many points of $Y$, say, $a_{1},\ldots,a_{n}\in Y$ is not covered. Since $\{A_{i}\}_{i\in J}$ covers $Y$, there must exist, say, $A_{k_{1}},\ldots,A_{k_{n}}$ such that $a_{i}\in A_{k_{i}}$, $i=1,\ldots,n$. Then $A_{1}\cup_{i=1}^{n}A_{k_{i}}$ is a finite covering of $Y$. (b) We prove a more general result. Let $X$ be an uncountable set, and let\[ \tau_{c}:=\{A\subset X\,|\, X-A\, countable\, or\, equal\, X\}.\] This topology is the countable complement topology. The compact subspaces of $X$ are exactly the finite subspaces. Since they are compact sets, assume $A$ is infinite. Let $B:=\{b_{1},b_{2},\ldots\}$ be a countable subset of $A$. Denote\[ A_{n}:=(X-B)\cup\{b_{1},\ldots,b_{n}\}.\] Then $\cup_{n=1}^{\infty}A_{n}$ is an open covering of $A$, but there is no finite subcovering. This shows, that $[0,1]$ is not compact in the countable compliment topology. We note, that $(X,\tau_{c})$ is not Hausdorff, because there are no two nonempty open disjoint subsets $A,B$ of $X$. Indeed, if $A\cap B=\emptyset$, then $X-(A\cap B)=(X-A)\cup(X-B)$, hence $X$ is countable which is a contradiction. \textbf{3}. \textcolor{green}{Solution}. Let $K=\cup_{i=1}^{n}K_{i}$. If $\cup_{i\in J}A_{i}$ is an open covering, then we can choose a finite subcovering of $K_{1}$ and so on, and the union of these finite subcovering is a finite subcovering of $K$. \textbf{4}. \textcolor{green}{Solution}. Let $x\in X$ be arbitrary but fixed point in the metric space $(X,d)$. Then for the compact set $K$ we have $K\subseteq X=\cup_{n=1}^{\infty}B(x,n)$. Since $K$ is compact there exists a finite subcovering which implies $K$ is bounded. Since every metric space is Hausdorff, by Theorem 26.3 $K$ is closed. In infinite dimensional Banach spaces the closed unit ball is not compact. \textbf{5}. \textcolor{green}{Solution}. Choose $x\in B$. Then by Lemma 26.4 there exist disjoint open sets $A\subset U_{x}$ and $x\in V_{x}$. Since $B\subset\cup_{x\in B}V_{x}$ is an open covering we can choose a finite $B\subset\cup_{i=1}^{n}V_{x_{i}}:=V$ subcovering. Then $A\subset\cap_{i=1}^{n}U_{x_{i}}=:U$ is an open set disjoint from the open set $V$. \textbf{6}. \textcolor{green}{Solution}. We have to show if $A$ is closed in $X$, then $f(A)$ is closed in $Y$. By Theorem 26.2 $A$ is compact, and by Theorem 26.5 $f(A)$ is compact. Now Theorem 26.3 gives the statement. \textbf{7}. \textcolor{green}{Solution}. Let $C$ be a closed subset of $X\times Y$. If $\pi_{1}(C)=X$ then the statement is true. If not, then we show that $X-\pi_{1}(C)$ is open. Let $x\in X-\pi_{1}(C)$. Then $\{x\}\times Y$ is a slice contained in the open set $X\times Y-C$, so by the tube lemma there is a neighborhood $U$ of $x$ such that $U\times Y\subseteq X\times Y-\pi_{1}(C)$. Here $U\subseteq X-\pi_{1}(C)$, so $X-\pi_{1}(C)$ is open. \textbf{8}. \textcolor{green}{Solution}. Assume that $f$ is continuous. In this case it is enough to assume that $Y$ is Hausdorff. Let $(x,y)\in X\times Y$ be a point such that $y\neq f(x)$. Then by the Hausdorff property there exist $y\in V$ and $f(x)\in W$ disjoint neighborhoods. Since $f$ is continuous, there exists $x\in U$ neighborhoods such that $f(U)\subset W$. Then $(x,y)\in U\times V$ and $(x,f(x))\in U\times W$. Since $V,W$ are disjoint, therefore, $(U\times V)\cap(U\times W)=\emptyset$. Here $U\times V$ is an open neighborhood of $(x,y)$ and disjoint from $G_{f}$. Assume that $G_{f}$ is closed. Let $V$ be a neighborhood of $f(x)$. Then $G\cap(X\times(Y-V))$ is closed in $X\times Y$. By Exe.7 $\pi_{1}\left(G\cap(X\times(Y-V))\right)$ is closed and does not contain $x$. It means, there exists a $x\in U$ neighborhoods such that $U\cap\pi_{1}\left(G\cap(X\times(Y-V))\right)=\emptyset$. From this we obtain $(U\times Y)\cap\left(G\cap(X\times(Y-V))\right)=\emptyset$. It implies $f(U)\cap(Y-V)=\emptyset$, that is, $f(U)\subset V$. Thus $f$ is continuous at $x$, an arbitrary point of $X$. \textbf{9}. \textcolor{green}{Solution}. Let $a\in A$ be arbitrary. In the first step we find an $a\in U$ neighborhood and $B\subset V$ open set such that $a\times B\subset U\times V\subset N$. In the product space $X\times Y$ the sets $G\times H$ ($G\subset X$, $H\times Y$ are open) form a basis. Now consider an open covering of $a\times B=\{a\times b_{i}\,|\, b_{i}\in B\}$ with the sets $a\times b_{i}\in G_{i}\times H_{i}\subset N$. Since $a\times B$ compact there is a finite subcovering, say $G_{i}\times H_{i}$, $i=1,\ldots,n$. Define $U:=\cap_{i=1}^{n}G_{i}$, $V:=\cup_{i=1}^{n}H_{i}$. Then $U\times V$ is the desired covering of $a\times B$. Since $U,V$ depend on $a$, we write $U_{a},V_{a}$. Now the sets $U_{a}\times V_{a}$, $a\in A$ form a covering of $A\times B$. Since it is compact there exists a finite subcovering, say $U_{i}\times V_{i}$, $i=1,\ldots,n$. Then $U:=\cup_{i=1}^{n}U_{i}$, $V:=\cap_{i=1}^{n}V_{i}$, $U,V$ are open and $A\times B\subset U\times V\subset N$. \textbf{10}. \textcolor{green}{Solution}. (a) Let $x$ be fixed. Since $f_{n}$ is monotone increasing for every $\varepsilon\gt 0$ there exists $n_{0}(\varepsilon,x)$ such that $|f_{n}(x)-f(x)|\lt\varepsilon$, $n\gt n_{0}(\varepsilon,x)$. In fact, we can write $0\leq f(x)-f_{n}(x)\lt\varepsilon$, $n\gt n_{0}(\varepsilon,x)$. Since $f-f_{n}$ is continuous, there exists $x\in U_{x,n}$ open set such that $0\leq f(z)-f_{n}(z)\lt\varepsilon$ for every $z\in U_{x,n}$, $n\gt n_{0}(\varepsilon,x)$. Since $f_{n}$ is monotone increasing we can write $z\in U_{x,n_{0}(\varepsilon,x)+1}$. Since $X$ is compact there exists a finite subcovering $X\subseteq\cup_{k=1}^{n}U_{x_{k},n_{0}(\varepsilon,x_{k})+1}$. If we choose $n_{0}(\varepsilon):=\max_{k}\, n_{0}(\varepsilon,x_{k})$ then $|f_{n}(x)-f(x)|\lt\varepsilon$, $n\gt n_{0}(\varepsilon)$, independently of $x$. (b) $X$ is not compact: $X:=[0,\infty)$, $f_{n}(x):=-\frac{x}{n}$. Then $f(x)=0$, but the convergence is not uniform, because taking $x_{n}:=n$ we have $f_{n}(x_{n})=-1$ not tends to $0$. $f_{n}$ is not monotone: see Exe. 21.9. \textbf{11}. \textcolor{green}{Solution}. By Theorem 26.9 and the note after the proof, the set $Y$ is not empty. Since $X$ is compact, so is $Y$. Assume $Y=C\cup^{\star}D$ is a separation. Then $C,D$ are disjoint compact sets in $X$. By Exe. 5 there exist disjoint open sets $U$ and $V$ containing $C$ and $D$, respectively. Then $Y-(U\cup^{\star}V)=\emptyset$. On the other hand $Y-(U\cup^{\star}V)=\cap_{A\in\mathcal{A}}(A-(U\cup^{\star}V))$. We show that the intersection is not empty. Since $A-(U\cup^{\star}V)$ are closed nested sets it is enough to prove that $A-(U\cup^{\star}V)\neq\emptyset$. If it would be empty, then since $A$ is connected we have $A\subset U$ or $A\subset V$, say $A\subset U$. It implies $Y\subset U$, and $Y\cap V=\emptyset$, that is, $Y\cap D=\emptyset$ which is impossible. Hence the intersection is not empty that is a contradiction. \textbf{12}. \textcolor{green}{Solution}. We prove a little bit more. Let $p:X\to Y$ be closed continuous surjective map such that $p^{-1}(y)$ is compact for each $y\in Y$. Then $p^{-1}(C)$ is compact for any compact set $C\subseteq Y$. \begin{lemma} Let $p:X\to Y$ be a closed map. (1) If $p^{-1}(y)\subset U$ where $U$ is an open subspace of $X$, then $p^{-1}(W)\subset U$ for some neighborhood $W\subset Y$ of $y$. (2) If $p^{-1}(B)\subset U$ for some subspace $B$ of $Y$ and some open subspace $U$ of $X$, then $p^{-1}(W)\subset U$ for some neighborhood $W\subset Y$ of $B$. \end{lemma} \begin{proof} We note that $p^{-1}(W)\subset U\Leftrightarrow p(X-U)\cap W=\emptyset$. (1) Taking $W:=y$ we obtain $y\notin p(X-U)$. Since $p(X-U)$ is closed there exists a $y\in W$ neighborhood such that $p(X-U)\cap W=\emptyset$. (2) Each point $y\in B$ has a neighborhood $W_{y}$ such that $p^{-1}(W_{y})\subset U$. Then $W:=\cup_{y\in B}W_{y}$ is a neighborhood of $B$ such that $p^{-1}(W)\subset U$. \end{proof} Now we prove the statement. For this we use only $(1)$. Let $C\subset Y$ be compact. Consider an open covering $\cup_{\alpha\in J}U_{\alpha}$ of $p^{-1}(C)$. For each $y\in C$ the compact set $p^{-1}(y)$ is covered by a finite subcovering $\cup_{\alpha\in J(y)}U_{\alpha}$. Applying $(1)$ there is an $y\in W_{y}\subset Y$ neighborhood such that $p^{-1}(W_{y})$ is contained in this finite union. By the compactness of $C$, finitely many $W_{y_{1}},\ldots,W_{y_{n}}$ cover $C$. Thus the finite subcovering $\cup_{i=1}^{n}\left(\cup_{\alpha\in J(y_{i})}U_{\alpha}\right)$ covers $p^{-1}(C)$ which means it is compact. \textbf{13}. \textcolor{green}{Solution}. (a) Assume $c\notin AB=\cup_{b\in B}Ab$. By Exe. 22.7 there are open disjoint sets $c\in W_{b}$ and $Ab\subset U_{b}$. Then $b\in A^{-1}U_{b}$ is an open neighborhood. Since $B$ is compact there exists a finite subcovering, say $B\subset A^{-1}U_{b_{1}}\cup\cdots\cup A^{-1}U_{b_{k}}=A^{-1}U$, where $U=\cup_{i=1}^{k}U_{b_{i}}$. Taking $W:=\cap_{i=1}^{k}W_{b_{k}}$ we have $c\in W$ is an open neighborhood and by the construction $W\cap(AB)\subset W\cap U=\cup_{i=1}^{k}(W\cap U_{b_{i}})\subset\cup_{i=1}^{k}(W_{b_{i}}\cap U_{b_{i}})=\emptyset$. (b) Suppose that $A\subset G$ is closed. We show that $G/H-p(A)$ is open. Consider $x\in G$ such that $xH$ does not belong to $p(A)$, then $x$ does not belong to $AH$. Since by (a) $AH$ is closed in $G$, there is an $x\in U$ open neighborhood such that $U\cap(AH)=\emptyset$. By Exe. 22.5 (c) $\{uH\,|\, u\in U\}$ is an open set in $G/H$ which contains $xH$, and it is disjoint from $p(A)$. Hence $G/H-p(A)$ is open. (c) We show that $p$ satisfies the conditions of Exe. 12. By (b) the quotient map $p$ is closed. Let $y\in G/H$. Then $y=gH$ for some $g\in G$. Hence $p^{-1}(y)=gH$ is compact. So by Exe. 12 we obtain $G$ is compact. \textbf{Section 27 Compact Subspaces of the Real Line}
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