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\title{Topology} 
\author{Hamilton College Topology Class, '05}
\begin{document} 

\setcounter{section}{6}
\section{Additional Problems and Material}
This supplemental chapter contains questions, definitions, and theorems related to previous ones, and it also contains new material.  

\subsection{Additional Material for Chapter 1}

\begin{ex}\ \\
\begin{xbox}
Let $X$ denote the set of all lines in $\mathbb{R}^2$ that pass through the origin.  Topologize $X$ in a meaningful way -- i.e. using neither the discrete or trivial topologies.  
\end{xbox}
\end{ex}
\ \\
\begin{ex}\ \\
\begin{xbox}
Let $X$ denote the set of all lines in $\mathbb{R}^3$ that pass through the origin.  Topologize $X$ in a meaningful way -- i.e. using neither the discrete or trivial topologies.  
\end{xbox}
\end{ex}

\ \\
\begin{ex}\ \\
\begin{xbox}
Let $X$ denote the set of all $(n-1)$-dimensional vector subspaces of $\mathbb{R}^n$.  Topologize $X$ in a meaningful way -- i.e. using neither the discrete or trivial topologies.  
\end{xbox}
\end{ex}

\ \\

\begin{defn}
A \emph{metric} or \emph{distance function} on a set $X$ is a function $d:X\times X \rightarrow \mathbb{R}$ satisfying the following properties:
\begin{itemize}
\item $\forall\, x, y \in X,\, d(x, y) \geq 0$ and $d(x,y) = 0 \iff x=y$.
\item $\forall\, x, y \in X,\, d(x, y) = d(y, x)$
\item $\forall\, x, y, z \in X,\, d(x,z) + d(y,z) \leq d(x,y)$
\end{itemize}
A set $X$ equipped with a metric $d$ is referred to as a \emph{metric space}.
\end{defn}

\ \\
\begin{thm} Let $(X, d)$ be a metric space.  Then the sets $$N(x, \epsilon) = \left\{y \in X : d(x,y) < \epsilon\right\}$$ can be used as a basis for a topology on $X$ (here $x \in X$ and $\epsilon \in \mathbb{R}$ is positive). \end{thm}
\begin{proof}\ \\
\begin{xbox} Prove the theorem \end{xbox} \end{proof}
\ \\
\begin{ex}\ \\
\begin{xbox}
Let $X$ be any set.  Find a metric $d$ so that the basis from Theorem 7.4 induces the discrete topology on $X$, or explain why no such metric exists.  Do the same for the trivial topology on $X$.  
\end{xbox}
\end{ex}

\ \\

\begin{ex}\ \\
\begin{xbox}
Recall that the circle (or ``1-sphere") is the space $$S^1 = \left\{(x,y) \in \mathbb{R}^2 : x^2 + y^2 = 1\right\}$$ equipped with the subspace topology of $\mathbb{E}^2$.  Similarly, the ``$n$-sphere" is the space $$S^n = \left\{\vec{x} \in \mathbb{R}^{n+1} : ||\vec{x}||^2 = 1\right\}$$ equipped with the subspace topology in $\mathbb{E}^{n+1}$.  Draw pictures of $S^1$ and $S^2$ and indicate at least one, proper open subset in each space.  Describe how one might visualize $S^3$.
\end{xbox}
\end{ex} \ \\

\begin{thm} (Pasquale's Theorem) A space $X$ has the discrete topology $\iff \, \forall x \in X$ the singleton $\{x\}$ is open. \end{thm}
\begin{proof} \ \\
\begin{xbox} Prove or disprove. \end{xbox} \end{proof}


\begin{thm} Suppose $Y$ is a subspace of $X$.  Then a set $C \subseteq Y$ is closed $\iff C = K\cap Y$ where $K$ is clsoed in $X$. \end{thm}
\begin{proof}\ \\
\begin{xbox} Prove or disprove. \end{xbox} \end{proof}

\begin{ex} Let $X$ be a set and define $$\mathcal{T} = \{S \subseteq X : S^c \text{ is countable }\}\cup\{\emptyset\}.$$  Prove that $\{X, \mathcal{T}\}$ is a topological space. This topology is called the \emph{cocountable} or \emph{countable complement} topology, and when $X = \mathbb{R}$ is equipped with it, we will use the symbol $\C$.  \end{ex}

\subsection{Additional Material for Chapter 2}

\begin{thm} Suppose $f, g$ are continuous functions from $\mathbb{E}^n$ to $\mathbb{E}^1$.  Prove that $p(\vec{x}) = f(\vec{x})\cdot g(\vec{x})$ is also continuous.  Prove that $f(\vec{x})\pm g(\vec{x})$ is also continuous.  \end{thm}
\begin{proof}\ \\
\begin{xbox} Prove the theorem.  Is this result true if one uses \emph{other} topologies on $\mathbb{R}$? \end{xbox} \end{proof} \ \\
%\ \\ Hint (i.e. most of an answer): We want to show that $p^{-1}\left(N(\epsilon, y)\right)$ is open, i.e. that for every point $\vec{x}_0$ in this pre-image, there exists a ball $N(\delta, \vec{x}_0) \subseteq p^{-1}\left(N(\epsilon, y)\right).$   We know that a statement like this is true individually for the functions $f$ and $g$, and somehow we want to use that information to conclude it holds for the new function $p$.  So, suppose $\vec{x}_0 \in p^{-1}\left(N(\epsilon, y)\right)$ so that $y = p(\vec{x}_0) = f(\vec{x}_0)\cdot g(\vec{x}_0)$.  We must find a $\delta$ so that $N(\delta, \vec{x}_0) \subseteq p^{-1}\left(N(\epsilon, y)\right).$  By our assumptions, we know that given any positive numbers $\epsilon_1$ and $\epsilon_2$ there exist $\delta_1$ and $\delta_2$ exist satisfying 
%
%\begin{align*} N\left(\delta_1, \vec{x}_0\right) \subseteq f^{-1}\left(N(\epsilon_1, f(\vec{x}_0))\right) &\iff \Big(||\vec{x}-\vec{x}_0|| < \delta_1 \Rightarrow |f(\vec{x}) - f(\vec{x}_0)| < \epsilon_1 \Big) \\ \\ 
%N\left(\delta_2, \vec{x}_0\right) \subseteq g^{-1}\left(N(\epsilon_2, g(\vec{x}_0))\right) &\iff \Big(||\vec{x}-\vec{x}_0|| < \delta_2 \Rightarrow |g(\vec{x}) - g(\vec{x}_0)| < \epsilon_2 \Big)\end{align*} \ \\
%If we therefore choose $\delta = \text{min}\{\delta_1, \delta_2\}$ then we find that 
%
%$$||\vec{x}-\vec{x}_0|| < \delta \Rightarrow \left\{ \begin{array}{c} |f(\vec{x})-f(\vec{x}_0)| < \epsilon_1 \\ \\ \\ |g(\vec{x})-g(\vec{x}_0)| < \epsilon_2\end{array} \right.$$ \ \\
%
%The question is now something like this: Given an arbitrary $\epsilon$, is this the $\delta$ that we want to ensure that $|p(\vec{x}) - p(\vec{x}_0)| < \epsilon$? The $\delta$-ball was chosen in terms of the functions $f$ and $g$, and so we should probably relate everything.  Of course, we assume $\vec{x} \in N(\delta, \vec{x}_0)$, i.e. that $||\vec{x}-\vec{x}_0|| < \delta$.  And here.  we.  go:
%
%\begin{align*} |p(\vec{x}) - p(\vec{x}_0)| &= |f(\vec{x})g(\vec{x}) - f(\vec{x}_0)g(\vec{x}_0)| = \left|\left(f(\vec{x})-f(\vec{x}_0)\right)g(\vec{x}_0) + \left(g(\vec{x}_0) - g(\vec{x})\right)f(\vec{x})\right| \\ \\ &\leq |f(\vec{x})-f(\vec{x}_0)|\cdot|g(\vec{x}_0)| + |g(\vec{x}) - g(\vec{x}_0)|\cdot|f(\vec{x})| \\ \\ &<\epsilon_1\,|g(\vec{x}_0)| + \epsilon_2\,|f(\vec{x})| = \epsilon_1\,|g(\vec{x}_0)| + \epsilon_2|f(\vec{x})-f(\vec{x}_0)+f(\vec{x}_0)| \\ \\ &\leq \epsilon_1|g(\vec{x}_0)| + \epsilon_2\cdot\epsilon_1 + \epsilon_2|f(\vec{x}_0)|\end{align*}
%
%Given that $g(\vec{x}_0)$ and $f(\vec{x}_0)$ are fixed numbers, one simply needs to argue that $\epsilon_1$ and $\epsilon_2$ can be chosen so that the expression above is less than (or equal to) $\epsilon$.  Assuming $g(\vec{x}_0) \neq 0 \neq f(\vec{x}_0)$, we can set 
%
%$$\epsilon_i = \min\left\{\frac{\epsilon}{3|g(\vec{x}_0)|}, \sqrt{\frac{\epsilon}{3}}, \frac{\epsilon}{3|f(\vec{x}_0)|}\right\}$$ ensuring that both $\epsilon_1$ and $\epsilon_2$ are bounded above by each expression in the list.  Judicious use of these bounds allows us to find that
%
%$$||\vec{x}-\vec{x}_0|| < \delta \Rightarrow |p(\vec{x}) - p(\vec{x}_0)| < \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon.$$
%
%\noindent That is, when its all said and done, if we want to show that given an $\epsilon > 0$ we can find a $\delta > 0$ so that $$N(\delta, \vec{x}_0) \subseteq N(\epsilon, p(\vec{x}_0))$$ then we choose $\delta = \min\{\delta_1, \delta_2\}$ where 
%
%\begin{align*} ||\vec{x}-\vec{x}_0|| < \delta_1 &\Rightarrow |f(\vec{x}) - f(\vec{x}_0)| < \epsilon_1 = \min\left\{\frac{\epsilon}{3|g(\vec{x}_0)|}, \sqrt{\frac{\epsilon}{3}}, \frac{\epsilon}{3|f(\vec{x}_0)|}\right\} \\ \\ ||\vec{x}-\vec{x}_0|| < \delta_2 &\Rightarrow |g(\vec{x}) - g(\vec{x}_0)| < \epsilon_2 = \min\left\{\frac{\epsilon}{3|g(\vec{x}_0)|}, \sqrt{\frac{\epsilon}{3}}, \frac{\epsilon}{3|f(\vec{x}_0)|}\right\} \end{align*}
%These choices will ensure that $N(\delta, \vec{x}_0) \subseteq N(\epsilon, p(\vec{x})_0)$, and the case when either function $f$ or $g$ vanishes at $\vec{x}_0$ can be adjusted.  \ \\
%
%%%%%%%%% SUPERIOR PROOF %%%%%%%%%%%%%%%%%
%
%And, actually, all of that stuff above goes to show you just how awful $\epsilon-\delta$ proofs can be.  Let's try the open set version!
%
%So, we get to assume that for every open set $O \subseteq \mathbb{R}$ the pre-images $f^{-1}(O)$ and $g^{-1}(O)$ are open in $\mathbb{R}^n$.  Given such an open set $O$ we now want to analyze the pre-image
%
%$$p^{-1}(O) = \left\{\vec{x} \in \mathbb{R}^n : f(\vec{x})g(\vec{x}) \in O\right\}.$$  \ \\
%In fact, it suffices to understand this pre-image when $O$ is an open interval $O = (a, b)$.  We first consider the case wherein $(a, b) \cap (-1, 1) \neq \emptyset$.  Let $U = (a, b) \cap (-1, 1)$, and note that $U$ is itself an interval $(c, d)$ which is contained in $(-1, 1)$.  In fact, any such interval is closed under multiplication: for every $s, t \in U$ it follows that $s\cdot t \in U$.  By the assumed continuity of both functions, if $\vec{x}$ is in the open set $f^{-1}(U)\cap g^{-1}(U)$, then both $f(\vec{x}), g(\vec{x}) \in U \subseteq (a, b)$ and, by the above sentence, their product remains in $U$.  This shows that every $\vec{x} \in p^{-1}(O)$ is an interior point since $$\vec{x} \in f^{-1}(U) \cap g^{-1}(U) \subseteq p^{-1}(O).$$    
%
%Now, the only remaining case occurs when $(a,b) \cap (-1, 1) = \emptyset$.  This leaves us with two situations, one in which both $a$ and $b$ are elements of $(-\infty, -1]$ or of $[1, \infty)$.  In the first situation, we consider open interval $U_1 = (-\sqrt{|a|}, -\sqrt{|b|})$, while in the second one we set $U_2 = (\sqrt{a}, \sqrt{b})$.  Laws of arithmetic confirm that for every $s, t \in U_i$ the product $s\cdot t \in O = (a, b)$.  That is, multiplying any two elements from the open set $U_i \subseteq O$ keeps their product in the open set $O$, which allows us to conclude that any point $\vec{x} \in p^{-1}(O)$ is an interior point since
%
%$$\vec{x} \in f^{-1}(U_i) \cap g^{-1}(U_i) \subseteq p^{-1}(O).$$  Nice, huh? \ \\  

\begin{ex}\ \\
\begin{xbox}
Consider the function $D:\mathbb{R}^4 \rightarrow \mathbb{R}$ given by $D(a, b, c, d) = ad - bc.$  With respect to the Euclidean topologies for both $\mathbb{R}^4$ and $\mathbb{R}$, prove that $D$ is continuous.
\end{xbox}
\end{ex}
\ \\

\begin{ex}\ \\
\begin{xbox}
Let $\text{GL}_2(\mathbb{R})$ denote the set of all invertible $2 \times 2$ matrices.  Topologize $\text{GL}_2(\mathbb{R})$  in a meaningful way -- i.e. using neither the discrete or trivial topologies.  
\end{xbox}
\end{ex} \ \\

\begin{defn}
An \emph{embedding} is a map $f:X\rightarrow Y$ that is a homeomorphism onto its image; i.e. a map for which $X \approx f(X)$.  Here we regard the subset $f(X) \subseteq Y$ as being equipped with the subspace topology.  
\end{defn}\ \\

\begin{ex}\ \\
\begin{xbox}
Let $S^n$ denote the $n$-sphere (as defined in the previous subsection), and let $k < m$ be natural numbers.  Find an embedding $f:S^k \rightarrow S^m$, and prove your function is an embedding.   
\end{xbox}
\end{ex} \ \\

\begin{ex}\ \\
\begin{xbox}
Find an embedding $f:\mathbb{R}^n \rightarrow S^n.$
\end{xbox}
\end{ex} \ \\

\begin{ex}\ \\
\begin{xbox}
Let $SO(2)$ denote the set of matrices $A \in \text{GL}_2(\mathbb{R})$ satisfying $\det A = 1$ and $A\,A^T = I$.  Explain how $SO(2) \approx S^1$, where $SO(2)$ is equipped with the subspace topology from the topology defined in 7.9.  Can you prove that these two spaces are homeomorphic?  Is it true that $SO(3) \approx S^2$?
\end{xbox}
\end{ex} \ \\


\subsection{Additional Material for Chapter 3}



\begin{thm} $\C$ is connected. \end{thm}
\begin{proof}\ \\
\begin{xbox} Prove or disprove. \end{xbox} \end{proof}

\begin{thm} $\C$ is Hausdorff. \end{thm}
\begin{proof} \ \\
\begin{xbox} Prove or disprove. \end{xbox} \end{proof}

\begin{defn} A sequence of points $\{x_n \in X : n \in \mathbb{N}\}$ in a topological space $X$ is said \emph{to converge} to the point $x \in X$ if for every open set $U \subseteq X$ which contains $x$, there exists $N \in \mathbb{N}$ such that $\{x_n : n \geq N\} \subseteq U.$ \end{defn} %in more intuitive terms, a sequence is said to converge to a point $x$ if every open set containing $x$ contains the ``tail" of the sequence.

\begin{ex} \ \\
\begin{xbox}
Let $X$ be a trivial topological space.  Explain why / prove every sequence converges to \emph{every} point $x \in X$. \end{xbox} \end{ex}
\
\begin{ex} \ \\
\begin{xbox} Let $X = \F$ and consider the sequence $\{ n : n \in \mathbb{N} \}$.  To which points, if any, does this sequence converge? Determine all (other?) divergent sequences.  \end{xbox} \end{ex}
\

\begin{thm} Let $X$ be a Hausdorff space.  If the sequence $\{x_n\}$ converges to $x$ and $y$, then ... \end{thm}
\begin{proof} \ \\
\begin{xbox} Complete and prove the theorem. \end{xbox} \end{proof}
\

\noindent \begin{xbox} \begin{thm} State the converse to the above theorem. \end{thm} \end{xbox} 

\begin{proof} \ \\
\begin{xbox} Prove or disprove. %hint: consider $\C$. 
\end{xbox} \end{proof}

\begin{defn} A space $X$ is said to be \emph{locally compact} if for every point $x \in X$ there exists a compact, open set $U$ that contains $x$. \end{defn}

\begin{thm} Every compact space is a locally compact space. \end{thm}
\begin{proof} \ \\
\begin{xbox} Prove or disprove, but oh my god, if you disprove it then go away. \end{xbox} \end{proof}

\begin{ex} Explain why $\mathbb{Q}$, endowed with the subspace topology from $\E$, is \emph{not} locally compact. \end{ex}

\begin{thm} All open and closed subsets of a locally compact, Hausdorff space $X$ are locally compact. \end{thm}
\begin{proof} \ \\
\begin{xbox} Prove or disprove. \end{xbox} \end{proof}

\begin{defn} A \emph{compactification} of a topological space $X$ is a topological space $Y$ that contains $X$ as a subspace \end{defn}

\begin{defn} Suppose $X$ is a topological space with topology $\mathcal{T}$.  Let $\infty$ denote some abstract point that is not in $X$, and let $\hat{X} = X\cup\{\infty\}.$  Define a topology $\hat{\mathcal{T}}$ on $\hat{X}$ by declaring $U \subseteq \hat{X}$ to be open if either 

\begin{enumerate} \item $\infty \notin U$ and $U \in \mathcal{T}_X$ or \ \item $\infty \in U$ and $X-U$ is a closed, compact subset of $X$. \end{enumerate}  The new space $\{\hat{X}, \hat{\mathcal{T}}\}$ is called \emph{the one point compactification of $X$}. \end{defn}

\begin{thm} $\{\hat{X}, \hat{\mathcal{T}}\}$ is a topological space. \end{thm}
\begin{proof} \ \\
\begin{xbox} Prove. \end{xbox} \end{proof}
\

\begin{thm} Endow $X$ with the subspace topology it inherits from $\mathcal{X}$.  This topology coincides with the original topology $\mathcal{T}$ on $X$. \end{thm}
\begin{proof} \ \\
\begin{xbox} Do the opposite of disprove. \end{xbox} \end{proof}

\begin{thm} Given any space $X$, the one point compactification $\hat{X}$ is compact. \end{thm}
\begin{proof} \ \\
\begin{xbox} Prove. \end{xbox} \end{proof}




 \end{document}