Merge branch 'master' of github.com:drowdemon/TopologyMunkres

Conflicts:
	ch2.19/exercises.tex
	ch2.20/exercises.tex
	ch2.21/exercises.tex
	generalNotes.tex
	preamble.sty

.gitignore

@@ -1,4 +1,7 @@
 *~
 *.pdf
 *.aux
-*.log
+*.log
+*.out
+\#*#
+.emacs.desktop*

ch2.19/exercises.tex

@@ -23,8 +23,9 @@
   \indent Suppose the sequence converges to $x$. Then every neighborhood $U$ of $x$ contains every $x_n$ for all $n$ larger than some $m_U$. Hence, every neighborhood $\pi_\alpha(U)$ of $\pi_\alpha(x)$ contains every $\pi_\alpha(x_n)$ for all $n$ larger than some $m_U$, for each $\alpha$. Furthermore, every neighborhood $V$ of $\pi_\alpha(x)$ has a corresponding neighborhood of $x$: $\pi^{-1}_\alpha(x)$. Thus, since neighborhood $\pi_\alpha(U)$ of $\pi_\alpha(x)$ is every neighorhood $\pi_\alpha(x)$. Therefore, every sequence $\pi_\alpha(x_1), \pi_\alpha(x_2),\dots$ converges to $\pi_\alpha(x)$.\n
   \indent Now suppose every sequence $\pi_\alpha(x_1), \pi_\alpha(x_2)\dots$ converges to $\pi_\alpha(x)$. Any open neighborhood $U_\alpha$ of $\pi_\alpha(x)$ in $X_\alpha$ contains all $\pi_\alpha(x_n)$ greater than some $m_{U_\alpha}$. Use finitely many $U_\alpha$s to construct an open neighborhood of $x$; label these by $i$. $U=\prod U_\alpha$, where $U_\alpha=U_i$ for finitely many values of $\alpha$, and $U_\alpha=X_\alpha$ for the others. Then, for all values $n$ over $\max_\alpha(m_{U_\alpha})$, $U$ will contain $x_n$, thus the sequence converges to $x$.\n
   \indent In the box topology, the second result does not hold, since if the value $U_{m_\alpha}$ increases unboundedly, the final sequence does not converge. For example, in the space $\mathbb{R}^\omega$, consider $x=(0,0,\dots)$, its neighborhood $((-1,1),\allowbreak (-1/2,1/2),\allowbreak (-1/4,1/4),\allowbreak (-1/8,1/8),\allowbreak (-1/16,1/16)\dots)$, and the sequence $x_n = (1/n,1/n,1/n,\dots)$.
-  \item Let $\mathbb{R}^\infty$ be the subset of $\mathbb{R}^\omega$ consisting of all sequences that are ``eventually zero,'' that is, all sequences $(x_1, x_2,\dots)$ such that $x_i\neq 0$ for only finitely many values of $i$. What is the closure of $\mathbb{R}^\infty$ in $\mathbb{R}^\omega$ in the box and product topologies?\n
-  \indent GO BACK TO THIS, NOT SURE HOW TO APPROACH IT.
+  \item Let $\mathbb{R}^\infty$ be the subset of $\mathbb{R}^\omega$ consisting of all sequences that are ``eventually zero,'' that is, all sequences $(x_1, x_2,\dots)$ such that $x_i\neq 0$ for only finitely many values of $i$. What is the closure of $\mathbb{R}^\infty$ in $\mathbb{R}^\omega$ in the box and product topologies?\\
+  Suppose $x=x_1,x_2,\dots$ is a point in the product topology, and let $U$ be a neighborhood of $x$. $U=\prod U_i$, where for some $n$, all $U_{i>n}=\R$. Therefore, the point $y = (x_1,x_2,\dots,x_n,0,0,\dots)\in U$ and $y\in R^\infty$, so $U$ intersects $\R^\infty$, so $x$ is a limit point of $\R^\infty$, so $R^\omega$ is the closure of $\R^\infty$\\
+  In the box topology, $\R^\infty$ is its own closure. Suppose a point $x=(x_1,x_2,\dots)\in\R^\omega$ outside of $\R^\infty$ were a limit point. But $U=(\prod_i \min(x_i/2, 2x_i),\max(2x_i,x_i/2))$ is an open set in the box topology that does not intersect $R^\infty$. Correction: if $x_i=0$, replace $U_i$ by $(-1,1)$, it is currently $(0,0)$ which would not work.
   \item Given sequences $(a_1, a_2, \dots)$ and $(b_1, b_2,\dots)$ of real numbers $a_i > 0$ for all $i$, define $h: \mathbb{R}^\omega \rightarrow \mathbb{R}^\omega$ by the equation $h((x_1, x_2, \dots)) = (a_1x_1+b_1, a_2x_2+b_2, \dots)$. Show that if $\mathbb{R}^\omega$ is given the product topology, $h$ is a homeomorphism of $\mathbb{R}^\omega$ with itself. What happens if $\mathbb{R}^\omega$ is given the box topology?\hspace{5in}\n
   \indent First, to prove that $h$ is a homeomorphism, we prove that it is bijective. To do that, we first show that it is injective. Suppose the opposite, there exist distinct $x$ and $y$ such that $h(x)=h(y)$. Then $a_1x_1+b_1 = a_1y_1 + b_1$, so $x=y$, a contradiction. Now we show that $h$ is surjective, again by contradiction. Suppose there is a point $y$ such that there exists no $x$ such that $h(x)=y$. However, the point $x=((y_1-b_1)/a_1, (y_2-b_2)/a_2, \dots)$ is such a point. Thus, $h$ is bijective. \hspace{5in}\n
   \indent Now we show that $h$ is continuous. Each function $h_1(x) = a_1x_1+b_1, h_2(x)=a_2x_2+b_2, \dots$ is clearly continuous, thus, by \hyperref[thm:MapsProducts19.6]{theorem 19.6}, $h$ is continous. Similarly, $h^{-1}$ is continuous, thus $h$ is homeomorphic.\hspace{5in}\n

ch2.20/exercises.tex

@@ -53,7 +53,9 @@
       This converges in each topology to $(0,0,0,\dots)$. Every open set $U$ in the box topology must contain an interval around $0$ for each coordinate, and there will be a $z$ within that interval.
     \end{itemize}
   \end{enumerate}
-  \item Let $\R^\infty$ be the subset of $\R^\omega$ consisting of all sequences that are eventually zero. What is the closure of $\R^\infty$ in $\R^\omega$ in the uniform topology?
+  \item Let $\R^\infty$ be the subset of $\R^\omega$ consisting of all sequences that are eventually zero. What is the closure of $\R^\infty$ in $\R^\omega$ in the uniform topology? \\
+  We claim that $\bar{\R^\infty}$ is the set of all sequences which converge to $0$.  Suppose $x=(x_1,x_2,\dots)$ is such a point, and let $B=B_{\bar{p}}(x,\epsilon)$ be a basis element of the uniform topology containing $x$. There exists some $n$ such that $\bar{d}(x_{i>n},0)<\epsilon$, and so $x_0,\dots,x_n,0,0,\dots$ is a point in $B$ and in $\R^{\infty}$, which means that $x$ is a limit point of $\R^\infty$.\\
+  If $x$ is not a sequence which converges to $0$, then there is an $\epsilon$ such that $\bar{d}(x_i,0)>\epsilon$ for infinitely many $x_i$, and so $x$ is not a limit point of $\R^\infty$. Thus the set of all sequences converging to 0 is the closure of $\R^\infty$ in the uniform topology.
   \item Let $\bar{p}$ be the uniform metric on $\R^\omega$. Given $x = (x_1, x_2, \dots) \in \R^\omega$, and $0<\epsilon < 1$, let $U(x,\epsilon) = (x_1 - \epsilon, x_1+\epsilon) \times \dots \times (x_n-\epsilon, x_n+\epsilon)\times \cdots$
   \begin{enumerate}
     \item Show that $U(x,\epsilon)$ is not equal to the $\epsilon$-ball $B_{\bar{p}}(x,\epsilon)$. \\

ch2.22/exercises.tex

@@ -0,0 +1,34 @@
+\documentclass[12pt,letterpaper]{article}
+\input{../preamble.sty}
+
+\begin{document}
+\RaggedRight
+\begin{enumerate}
+  \item Check the details of Example 3, repeated here. Let $p$ be the map of the real line $\R$ onto the three point set $A=\{a,b,c\}$ defined by $p(x)=\begin{cases} a &\quad \text{if } x>0 \\ b &\quad \text{if } x<0 \\ c &\quad \text{if } x=0 \end{cases}$. Check that the quotient topology on $A$ induced by $p$ is the following set of subsets of $A$: $\{\{a\},\{b\},\{a,b\},\{a,b,c\}\}$. \\
+  We wish to create a topology such that a set $U$ in $A$ is open if and only if $p{-1}(U)$ is open in $\R$. $p^{-1}(\{a\}) = (0,\infty)$, which is open in $\R$, so $\{a\}$ must be open. Similarly, $\{b\}$ and $\{a,b\}$ must be open. $p^{-1}(\{c\}) = \{0\}$, which is not open, so $\{c\}$ is not open. Similarly for other sets which include $c$, except the entire set $A$, since $p^{-1}(A)=\R$, $A$ is open. 
+  \item \begin{enumerate}
+    \item Let $p: X\rightarrow Y$ be a continuous map. Show that if there is a continuous map $f: Y\rightarrow X$ such that $p\circ f$ equals the identity map of $Y$, then $p$ is a quotient map. \\
+    Suppose there is such a map $f$. Let $V\subset Y$ be open; since $p$ is continuous, $p^{-1}(V)$ is open in $X$. In the other direction, we must show that if $p^{-1}(V)$ is open, then $V$ is open. $f$ is continuous, therefore assuming $p^{-1}(V)$ is open, $f^{-1}(p^{-1}(V))$ will also be open. Since $p\circ f = i$, $f^{-1}\circ p^{-1} = (p\circ f)^{-1} = i^{-1} = i$, $f^{-1}(p^{-1}(V) = V$, so $V$ is open. Hence $p$ is a quotient map.
+    \item If $A \subset X$, a retraction of $x$ onto $A$ is a continuous map $r: X\rightarrow A$ such that $r(a)=a$ for each $a\in A$. Show that a retraction is a quotient map. \\
+    A retraction is clearly surjective. %$r$ is continuous by definition, so for each open subset $V$ of $A$, $U=r^{-1}(V)$ is open.
+    We must show that if $r^{-1}(V)$ is open, then $V$ is open, where $V$ is a subset of $A$. Because of the way $r$ is defined, clearly $r^{-1}(V)$ is some set $U$ such that $U\cap A = V$. Therefore, if $U$ is open, in the subspace topology $U\cap A$ will be open, so $V$ is open, so $r$ is a quotient map.
+  \end{enumerate}
+  \item Let $\pi_1:\R\times\R \rightarrow \R$ be the projection on the first coordinate. Let $A$ be the subspace of $\R\times\R$ consisting of all points $x\times y$ for which either $x\geq 0$ or $y=0$ (or both); let $q: A\rightarrow \R$ be obtained by restricting $\pi_1$. Show that $q$ is a quotient map that is neither open not closed. \\
+  $q$ is clearly surjective. We show that $q$ is continuous. Let $V$ be an open set of $\R$, $V=(a,b)$, with $a,b\in\R$, then $U=q^{-1}(V)=(a,b)\cup (\max(0,a),b)\times\R$ is open, since $U=(a,b)\times\R \cap A$. Now we show the converse, that if $p^{-1}(V)=U$ is open, then $V$ must be open. $U$ is open, therefore it is of the form $(a,b)\times(b,c)\cap A$, and $V$ will be of the form $(a,b)$, an open set. Therefore $q$ is a quotient map.
+  \item \begin{enumerate}
+    \item Define an equivalence relation on the plane $X=\R^2$ as follows: $x_0\times y_0 \sim x_1\times y_1 \text{ if } x_0+y_0^2 = x_1 + y_1^2$. Let $X^*$ be the corresponding quotient space. It is homeomorphic to a familiar space, what is it? [Hint: Set $g(x\times y) = x+y^2$]
+    \item Repeat $(a)$ for the equivalence relation $x_0\times y_0 \sim x_1\times y_1 \quad \text{if } x_0^2+y_0^2 = x_1^2 + y_1^2$.
+  \end{enumerate}
+  \item Let $p: X\rightarrow Y$ be an open map. Show that if $A$ is open in $X$, then the map $q: A\rightarrow p(A)$ obtained by restricting $p$ is an open map.\\
+  Suppose $U$ is open in $X$, then if $A$ is open in $X$, $U\cap A$ is open. Furthermore, $p(U)$ is open, $p(A)$ is open, so $p(U\cap A)$ is open, or equivalently $q(U\cap A)$ is open. $U\cap A$ is a basis for $A$, so $q$ is an open map.
+  \item Recall that $\R_K$ denotes the real line in the \hyperref[dfn:KTopology]{$K$-topology}. Let $Y$ be the quotient space obtained from $\R_K$ by collapsing the set $K$ to a point; let $p: \R_K\rightarrow Y$ be the quotient map.
+  \begin{enumerate}
+    \item Show that $Y$ satisfies the $T_1$ axiom but is not Hausdorff. \\
+    The $K$-Topology is Hausdorff, since it is finer than the standard topology on $\R$, which is Hausdorff. To show that $Y$ satisfies the $T_1$ axiom, we show that each element of the partition $Y$ is closed. Each element besides the point $p(K)$ is clearly closed, it is a one point set in $\R_K$. In the $K$-topology, $K$ is closed, it contains all of it's limit points ($0$ is not a limit point, $(-\epsilon, \epsilon)-K$ is a neighborhood of $0$ that does not intersect $K$). Thus, $Y$ satisfies the $T_1$ axiom.\\
+    Suppose $Y$ were a Hausdorff space, then every pair of distinct points $y_1,y_2\in Y$ have disjoint neighborhoods $V_1, V_2$. Let $y_1=p(K)$ and $y_2=p(0)$. $V_1$ and $V_2$ are disjoint open sets, and $p$ is a quotient map, so $U_1=p^{-1}(V_1)$ and $U_2=p^{-1}(V_2)$ are disjoint open sets of $R_K$. $U_2$ is a neighborhood of $0$, it has an open subset of the form $U_2'=(-\epsilon,\epsilon)-K$. $U_1$ is a neighborhood of $K$, and there exists some point $1/n$ in $K$ such that $1/n<\epsilon$. Any neighborhood of $K$ must have as a subset a neighborhood of this $1/n$, but this neighborhood intersects $U_2'$, and so intersects $U_2$, a contradiction, since these sets should be disjoint. Therefore, $Y$ is not Hausdorff.
+    \item Show that $p\times p : \R_K\times \R_K \rightarrow Y\times Y$ is not a quotient map. [Hint: The diagonal is not closed in $Y\times Y$, but its inverse image is closed in $\R_k\times\R_K$.] \\
+    Consider the set $D=\{y\times y\;|\;y\in Y\}$.  By the previous part, every neighborhood of $p(0)$ intersects some neighborhood of $p(K)$. Therefore, the point $p(K)\times p(0)$ will have neighborhoods $U\times V$, where $U$ is a neighborhood of $f(K)$, and $V$ is a neighborhood of $f(0)$ and so necessarily intersects the $U$ at some point $y$. Therefore, $U\times V$ contains a point $y\times y\in D$, so $f(K)\times f(0)$ is a limit point of $D$ not contained in $D$, so $D$ is not closed.\\
+    Now consider $p^{-1}(D)$. It is Hausdorff, therefore any $x,y$ where $x\neq y$ have disjoint neighborhoods $U,V$, so $x\times Y$ has a neighborhood $U\times V$ which does not intersect $p^{-1}(D)$, so there are no limit points of $p^{-1}(D)$ which are not in $p^{-1}(D)$, so it is closed. Since $D$ is not closed in $Y\times Y$ but its inverse image is closed in $\R_k\times \R_K$, $p\times p$ is not a quotient map.
+  \end{enumerate}
+\end{enumerate}
+\end{document}

ch3.23/exercises.tex

@@ -0,0 +1,33 @@
+\documentclass[12pt,letterpaper]{article}
+\input{../preamble.sty}
+
+\begin{document}
+\RaggedRight
+\begin{enumerate}
+  \item Let $\T$ and $\T'$ be two topologies on $X$. If $\T'\supset\T$, what does connectedness of $X$ in one topology imply about connectedness in the other?\\
+  If $\T$ is connected, then the courser $\T'$ is clearly connected. The reverse relationship does not imply anything about connectivity, because a finer set can always be made by adjoining a separation of $X$ to $\T$.
+  \item Let $\{A_n\}$ be a sequence of connected subspaces of $X$, such that $A_n\cap A_n+1 \neq \emptyset$ for all $n$. Show that $\bigcup A_n$ is connected. \\
+  $A_1$ and $A_2$ are connected by \hyperref[thm:unionConnected]{Theorem 23.3}, since their intersection is not null. Likewise, the union of the sets $(A_!\cup A_2)$ and $A_3$ is connected, since $A_2\cap A_3$ is connected. Proceeding by induction, $\bigcup A_n$ is connected.
+  \item Let $\{A_\alpha\}$ be a collection of connected subspaces of $X$; let $A$ be a connected subspace of $X$. Show that if $A\cap A_\alpha\neq\emptyset$ for all $\alpha$, then $A\cup(\bigcup(A_\alpha))$ is connected.\\
+  For each $\alpha$, $A$ intersects $A_\alpha$, so $A\cup A_\alpha$ is connected. Similarly, for $\beta\neq\alpha$, $A\cup A_\alpha \cup A_\beta$ is connected, and so by induction we get that $A\cup(\bigcup A_\alpha)$ is connected.
+  \item Show that if $X$ is an infinite set, it is connected in the \hyperref[dfn:finiteComplementTopology]{finite complement topology}. \\
+  Suppose it is not connected, there exists a subset $U$ of $X$ which is both open and closed. However, for $X-U$ to be finite, $U$ must be an infinite set. Therefore, $X-U$ is not open, since $X-U$ is finite, so $U$ is not closed, a contradiction. $X$ is connected.
+  \item A space is totally disconnected if its only connected subspaces are one-point sets. Show that if $X$ has the discrete topology, then $X$ is totally disconnected. Does the converse hold? \\
+  Let $A$ be a connected subspace of $X$, suppose $A$ contains more than one distinct point, let $x$ be an element $A$. $\{x\}$ and $X-\{x\}$ are open sets of $X$, and when intersected with $A$ remain open sets. They are nonempty, and so form a separation of $X$. A one point subset is obviously connected, and it is the only type of connected subspace of $X$. FIX!%The converse does hold, if there is no subset with more than one point in it that is connected, then every one point set must be open: Suppose there is a set $A\subset X$ with more than one point. By hypothesis, it is disconnected, it has a separation $C$ and $D$, so $C$ and $D$ are open. If $C$ is not a one point set, repeat. Eventually, this procedure yields $C$ and $D$ that are one-point sets, and they are open. Hence, all one point sets are open in a totally disconnected space, which means that the topology is the discrete topology. Messy.
+  \item Let $A\subset X$. Show that if $C$ is a connected subspace of $X$ that intersects both $A$ and $X-A$, then $C$ intersects $\text{Bd } A$. [\hyperref[dfn:boundary]{Boundary}].\\
+  Consider the subsets of $C$, $U=A\cap C$ and $V=(X-A)\cap C$. $U$ and $V$ are cleary a pair of disjoint nonempty sets whose union is $Y$. If $x$ is a limit point of both $A$ and $X-A$, then $x$ will be in the closure of each of these sets, and thus in the intersection of their closure, and so in the boundary of $A$. $C$ must contain $x$; it must be in one of $U$ or $V$, otherwise neither will contain a limit point of the other which would imply that $C$ is not connected.
+  \item Is the space \hyperref[dfn:lowerLimitTopology]{$\R_\ell$} connected? \\
+  It is not. The set $(-\infty,0)$ is open in $\R_\ell$, it is the union of basis elements $[-n,-n+1)$ for $n\in\mathbb{Z}_+$. $[0,\infty)$ is also open, constructed similarly. Together these form a separation of $\R_\ell$.
+  \item Determine whether or not $\R^\omega$ is connected in the uniform topology. \\
+  %If $\R^\omega$ is not closed, then it contains a set that is open and closed. If a typical open set, $B=B_{\bar{p}}(x,\epsilon)$ is also closed, then $A = \R^\omega - B$ is open. $A$ contains some point $y$ such that $\bar{p}(x,y)=\epsilon$. $y$ is a limit point of $B$, every neighborhood of $y$ intersects $B$, therefore $A$ cannot be open and disjoint from $B$, so $B$ is not closed, so $\R^\omega$ is connected in the uniform topology. WRONG
+  \item Let $A$ be a proper subset of $X$, and let $B$ be a proper subset of $Y$. If $X$ and $Y$ are connected, show that $(X\times Y) - (A\times B)$ is connected.
+  \item Let $\{X_\alpha\}_{\alpha\in J}$ be an indexed family of connected spaces; let $X$ be the product space $X=\prod_{\alpha\in J}X_\alpha$. Let $a=(a_\alpha)$ be a fixed point of $X$.
+  \begin{enumerate}
+    \item Given any finite subset $K$ of $J$, let $X_K$ denote the subspace of $X$ consisting of all points $x=(x_\alpha)$ such that $x_\alpha=a_\alpha$ for $a\not\in K$. Show that $X_K$ is connected.
+    \item Show that the union $Y$ of spaces $X_K$ is connected.
+    \item Show that $X$ equals the closure of $Y$; conclude that $X$ is connected.
+  \end{enumerate}
+  \item Let $p:X\rightarrow Y$ be a quotient map. Show that if each set $p^{-1}(\{y\})$ is connected, and $Y$ is connected, then $X$ is connected.
+  \item Let $Y\subset X$; let $X$ and $Y$ be connected. Show that if $A$ and $B$ form a separation of $X-Y$, then $Y\cup A$ and $Y\cup B$ are connected.
+\end{enumerate}
+\end{document}