\documentclass{article}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{enumerate}
\usepackage{ifthen}

\usepackage{amssymb}

\usepackage{color}

% needs graphicx package. This blows-up math symbols
%\usepackage{graphicx}
%\usepackage{lscape}

\setlength{\unitlength}{0.1in}

\setlength{\oddsidemargin}{-0.50in}
\setlength{\evensidemargin}{-0.50in} 
\setlength{\topmargin}{-0.50in}
\setlength{\headheight}{0.0in}
\setlength{\headsep}{0.0in}
\setlength{\topskip}{0.0in}
\setlength{\textheight}{9.5in}
\setlength{\textwidth}{7.5in}

\newcommand{\comp}{ \,{\scriptstyle \stackrel{\circ}{}}\, } 
\newcommand{\nullset}{\mathrm{O}\!\!\!\!\big/\,}
\newcommand{\divides}{\,\Big|\,}
\newcommand{\myspace}{\mbox{\parbox[c][0.35in]{0.35in}{\hfill}}}
\newcommand{\myspacea}{\mbox{\parbox[c][0.35in]{0.65in}{\hfill}}}
\newcommand{\varspace}[2]{\mbox{\parbox[c][#1]{#2}{\hfill}}}

\newcommand{\force}{{$\mbox{\;}$}}

\newcommand{\bigtimes}[1]{ \mbox{\raisebox{-5pt}{$\stackrel{\mbox{\resizebox{0.2in}{!}{$\times$}}}{\mbox{\resizebox{0.15in}{!}{$#1$}}}$}}}

\begin{document}

%\pagestyle{empty}

\noindent
\parbox{1.5in}{\bf Math 4710/5710} 
\hfill {\Large \bf  In the Midst-term Exam --- In Class} \hfill
\parbox{1.5in}{\bf \hfill November $14^{\mathrm{th}}$, 2014}

\vspace{0.1in}

\noindent
{\bf Standard assumptions:} $X$ is a topological space. Subsets are given the subspace topology. Products are given the product topology.

\vspace{0.1in}

\noindent {\large\bf Name:} \underline{\Large\sc\color{red}\quad Answer Key \quad} \hfill {\bf Be sure to show your work!}

\vspace{0.1in}

\noindent {\bf\large 1. (12 points)} An open question.
\begin{enumerate}[(a)]
\item Using the standard basis for $\mathbb{R}_\ell$ (real numbers with lower limit topology), show $I=(0,1)$ is open.

\vspace{0.1in}

There are two main approaches to this problem. Either we could show that $I$ is the union of intervals of the form $[a,b)$ or we could show that every element of $I$ belongs to some interval $[a,b) \subseteq I$.

\vspace{0.05in}

\noindent {\bf Proof \#1:} $\displaystyle I = (0,1) = \bigcup\limits_{n=1}^\infty \left[ \dfrac{1}{n},1 \right)$. Since $I$ is the union of (basic) open sets, it's open.

\vspace{0.05in}

\noindent {\bf Proof \#2:} Let $x \in I=(0,1)$. Then $0 < x < 1$. Obviously $x \in [x,1)$ and $[x,1) \subseteq (0,1)$ (since $x>0$). So every element of $I$ has a (basic open) neighborhood lying in $I$. Thus $I$ is open.

\vspace{0.1in}

\item The interval $J=(0,1]$ is not open in $\mathbb{R}$. Why? 

\vspace{0.1in}

Suppose $(a,b)$ is a neighborhood of $1$. Then $a<1<b$. So $b/2 >1$ and thus $b/2 \not\in J$. Therefore, $(a,b) \not\subseteq J$. So every open interval containing $1$ spills outside of $J$. This means $J$ is not open. [If $J$ were open, for each of the elements of $J$ we could find a (basic) open set about that element and contained inside of $J$. This cannot be done for $1 \in J$, so $J$ isn't open.]

\vspace{0.1in}

\item The interval $J=(0,1]$ is open in $X=(-\infty,1]$. Why?

\vspace{0.1in}

$(0,2)$ is open in $\mathbb{R}$ so $J = (0,1] = (-\infty,1] \cap (0,2) = X \cap (0,2)$ is open in the subspace topology (i.e. open in $X$).

\vspace{0.1in}

\end{enumerate}

\noindent {\bf\large 2. (10 points)} Let $\mathcal{T} = \{ U \subseteq X \;|\; U \mbox{ is finite.} \} \cup \{ X \}$.
\begin{enumerate}[(a)]
\item Explain why $\mathcal{T}$ is not a topology for $X$.

\vspace{0.1in}

Although $\mathcal{T}$ contains the empty set and is closed under finite (in fact arbitrary) intersections, it may not be closed under arbitrary unions.

\vspace{0.1in}

\item Can we restrict $X$ so that $\mathcal{T}$ is a topology for $X$?

\vspace{0.1in}

Yes. If $X$ is finite, we'll have closure under arbitrary unions as well. In this case, all subsets are finite, so $\mathcal{T} = \mathcal{P}(X)$ (it's the discrete topology).

\vspace{0.1in}

\end{enumerate}

\noindent {\bf\large 3. (9 points)} Let $A \subseteq X$. Suppose that $a_n \in A$ for all $n=1,2,3,\dots$ and $a_n \to a$. Show that $a \in \bar{A}$.

\vspace{0.1in}

Let $U$ be an open neighborhood of $a$. Then (by the definition of a convergent sequence) there exists some $N>0$ such that $a_n \in U$ for all $n \geq N$.
In particular, $a_N \in U$, but $a_N \in A$ (this is a sequence in $A$ so all $a_n$'s belong to $A$). Thus $U \cap A \not= \varnothing$. Therefore, every neighborhood of $a$ intersects $A$. This means that $a \in \overline{A}$.

\vspace{0.1in}

\noindent {\bf\large 4. (13 points)} Let $\displaystyle f : A \to \prod\limits_{j \in J} X_j$ ($A$ and $X_j$'s are topological spaces). For each $a \in A$, we have $f(a) = (f_j(a))_{j \in J}$ for some functions $f_j:A \to X_j$. 
\begin{enumerate}[(a)]
\item Suppose  $f$ is continuous. Show that $f_j$ is continuous for all $j \in J$.

\vspace{0.1in}

The projection maps $\pi_k : \prod\limits_{j \in J} X_j \to X_k$ (where $\pi_k\left((a_j)_{j\in J}\right) = a_k$) are continuous. Notice that $f_j(a) = (\pi_j \circ f)(a)$ for all $a$. Thus $f_j$ is continuous since it is the composition of continuous maps ($f$ and $\pi_j$ are continuous).

\vspace{0.1in}

\newpage

\item Suppose that $f_j$ is continuous for all $j \in J$. Show that $f$ is continuous.

\vspace{0.1in}

Let $U$ be open in $\prod\limits_{j \in J} X_j$. So $U = \prod\limits_{j \in J} U_j$ where $U_j$ is open in $X_j$ and $U_k=X_k$ for all but finitely many $k \in J$ (remember we are working with the product topology). 

$$f^{-1}(U) = f^{-1}\left( \prod\limits_{j \in J} U_j \right) = \{ a \in A \;|\; f(a) \in \prod\limits_{j \in J} U_j \} = \{ a \in A \;|\; f_j(a) \in U_j \mbox{ for all } j \in J \} = \bigcap\limits_{j \in J} f^{-1}_j(U_j)$$

Now each $U_j$ is open in $X_j$ so $f^{-1}_j(U_j)$ is open in $A$ (we are assuming that the $f_j$'s are continuous). However, we have a potentially infinite intersection. We're only guaranteed that {\it finite} intersections of open sets yield an open set. This is where the {\it product} topology comes to the rescue. Recall that $U_k = X_k$ for almost all $k$'s. If $U_k=X_k$, then $f^{-1}_k(U_k) = f^{-1}_k(X_k) = A$. So any $f^{-1}_k(U_k)$ where $U_k=X_k$ has no effect on this intersection. Thus the intersection is effectively a finite intersect and thus is an open set. 

We have shown that $f^{-1}(U)$ is open for all open sets $U$. Therefore, $f$ is continuous.

\vspace{0.1in}

\item Let $f: \mathbb{R} \to \mathbb{R}^\omega$ be defined by $f(t)=(t,t,t,\dots)$. Explain why $f$ is continuous if $\mathbb{R}^\omega$ is given the product topology but not continuous if given the box topology.

\vspace{0.1in}

The map $g(t)=t$ is continuous. So since $f_j=g$ is continuous for all $j$, by part (b), $f$ is continuous (when we are using the product topology). 

On the other hand, consider $U = (-1,1) \times (-1/2,1/2) \times (-1/3,1/3) \times \cdots$. $U$ isn't open in the product topology, but it is open in the box topology. 
Notice that if $t \in f^{-1}(U)$ we must have that $f(t) = (t,t,t,\dots) \in (-1,1) \times (-1/2,1/2) \times (-1/3,1/3) \times \cdots$ so $-\dfrac{1}{n} < t < \dfrac{1}{n}$ for all $n=1,2,3,\dots$. Therefore, $t=0$. Thus $f^{-1}(U) = \{ 0 \}$. So we've shown that the inverse image of the open set $U$ is the set $\{0\}$ (which is not open). Therefore, $f$ is not continuous (when dealing with the box topology).

\vspace{0.1in}

\end{enumerate}

\noindent {\bf\large 5. (9 points)} Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces. 
\begin{enumerate}[(a)]
\item Using the language of metric spaces, give a careful definition of what it means for $f:X \to Y$ to be continuous at $x \in X$.

\vspace{0.1in}

For every $\epsilon >0$ there exists some $\delta > 0$ such that for all $y \in X$, $d_X(x,y) < \delta$ implies that $d_Y(x,y) < \epsilon$.

\vspace{0.1in}

\item Using the language of topological spaces, give a careful definition of what it means for $f:X \to Y$ to be continuous at $x \in X$ (this should work if $X$ and $Y$ are just topological spaces and not necessarily metric spaces).

\vspace{0.1in}

For every open set $U$ in $Y$ such that $f(x) \in U$ there exists some open set $V$ in $X$ such that $x \in V$ and $V \subseteq f^{-1}(U)$ (or $f(V) \subseteq U$). In words, the inverse image of an arbitrary neighborhood of $f(x)$ contains a neighborhood of $x$.

\vspace{0.1in}

\item Briefly explain why these concepts are equivalent.

\vspace{0.1in}

The definition in part (b) still works if we use {\it basic} open sets instead of arbitrary open sets. In a metric space a basic open set is an $\epsilon$-ball. In fact, we can demand that our basic open neighborhood of some element $z$ be an $\epsilon$-ball centered at $z$. So our definition in part (b) reduces to: for every $\epsilon$-ball with $B_{\epsilon}(f(x))$ there is a $\delta$-ball with $B_{\delta}(x)$ such that $B_{\delta}(x) \subseteq f^{-1}(B_{\epsilon}(f(x)))$. That is $f(B_{\delta}(x)) \subseteq B_{\epsilon}(f(x))$. Thus $d_X(x,y)<\delta$ implies that $d_Y(f(x),f(y))<\epsilon$.

\vspace{0.1in}

\end{enumerate}

\noindent {\bf\large 6. (7 points)} Basic continuity. $X$, $Y$, and $Z$ are topological spaces. Suppose $f:X \to Y$ and $g:Y \to Z$ are continuous. Show $g \circ f$ is continuous.

\vspace{0.1in}

Let $U$ be an open subset of $Z$. Then $g^{-1}(U)$ is open $Y$ since $g$ is a continuous function (inverse images of open sets are open). Thus $f^{-1}(g^{-1}(U))$ is open in $X$ since $f$ is continuous ($g^{-1}(U)$ is open in $Y$ so its inverse image must be open in $X$). Therefore, $(g \circ f)^{-1}(U) = f^{-1}(g^{-1}(U))$ is open in $X$ whenever $U$ is open in $Z$. Therefore, $g \circ f$ is continuous.

Very briefly: $(g \circ f)^{-1}(\mbox{open}) = f^{-1}(g^{-1}(\mbox{open})) = f^{-1}(\mbox{open})=\mbox{open}$.

\vspace{0.1in}

\newpage

\noindent {\bf\large 7. (10 points)} Let $f:X \to Y$ be a continuous map between topological spaces $X$ and $Y$ where $X$ is connected. Show $f(X)$ is connected.

\vspace{0.1in}

Suppose that $U$ and $V$ are open in $Y$. In addition suppose that $U \cap V= \varnothing$ and $f(X) \subseteq U \cup V$. Then $X = f^{-1}(f(X)) \subseteq f^{-1}(U \cup V) = f^{-1}(U) \cup f^{-1}(V)$. Also, $f^{-1}(U) \cap f^{-1}(V) = f^{-1}(U \cap V) = f^{-1}(\varnothing) = \varnothing$. Recall, $f$ is continuous, so $f^{-1}(U)$ and $f^{-1}(V)$ are open in $X$ (they are inverse images of open subsets of $Y$). 

This means that $f^{-1}(U)$ and $f^{-1}(V)$ are disjoint open sets which cover $X$. $X$ is connected, so it cannot have a separation. Therefore either $f^{-1}(U)$ or $f^{-1}(V)$ is empty. Without loss of generality, assume $f^{-1}(U)$ is empty. Then nothing maps to $U$. So $U \cap f(X) = \varnothing$ (so $f(X) \subseteq V$). Thus $U$ and $V$ do not separate $f(X)$. Since $f(X)$ has no separation, it is connected.

\vspace{0.1in}

\noindent {\bf\large 8. (10 points)} Not connected. A statement or a hint?
\begin{enumerate}[(a)]
\item Show $X = [0,1] \cup [2,3]$ is not connected.

\vspace{0.1in}

Consider $U = (-\infty,1.5)$ and $V=(1.5,\infty)$. Then $U$ and $V$ are non-empty, disjoint, open sets such that $X = [0,1] \cup [2,3] \subseteq (-\infty,1.5) \cup (1.5,\infty) = U \cup V$. Therefore, $U$ and $V$ form a separation of $X$ (it is disconnected).

\vspace{0.1in}

\item Explain why $\mathbb{R} \not\cong S^1$ (the unit circle).

\vspace{0.1in}

Suppose that $\mathbb{R} \cong S^1$. Then there is some homeomorphism $f:\mathbb{R} \to S^1$. Thus $f$ restricted to $\mathbb{R}-\{0\}$ is a homeomorphism from $\mathbb{R}-\{0\}$ to $S^1-\{f(0)\}$. But this is impossible since $\mathbb{R}-\{0\} = (-\infty,0) \cup (0,\infty)$ is disconnected while $S^1 - \{ \mbox{any point}\}$ is still connected. Therefore, no such map $f$ can exist. Therefore, $\mathbb{R} \not\cong S^1$.

\vspace{0.05in}

Alternatively, notice that $\mathbb{R}$ is not compact while $S^1$ is compact ($S^1$ is a closed and bounded subset of $\mathbb{R}^2$). Therefore, they cannot be homeomorphic (continuous  maps, such as homeomorphisms, send compact sets to compact sets). 

\vspace{0.1in}

\end{enumerate}

\noindent {\bf\large 9. (10 points)} Compact and not compact.
\begin{enumerate}[(a)]
\item Let $X$ have the finite complement topology. Show $X$ is compact.

\vspace{0.1in}

Let $\mathcal{U}$ be an open cover of $X$. Pick some $U_0 \in \mathcal{U}$. Then $U_0$ is open in $X$ so $X-U_0$ is finite, say $X-U_0 = \{x_1,\dots,x_n\}$. 
Since $\mathcal{U}$ is a cover of $X$ there exists sets $U_1,\dots,U_n \in \mathcal{U}$ such that $x_j \in U_j$ for each $j=1,\dots,n$. Therefore, everything is $X$ except for $x_1,\dots,x_n$ is covered by $U_0$ and these points are covered by $U_1,\dots,U_n$. Thus $\{U_0,U_1,\dots,U_n \}$ is a covering for $X$. This is our desired finite subcover of $X$. Therefore, every open cover has a finite subcover. This means $X$ is compact.

\vspace{0.1in}

\item Show the interval $I=(0,1) \subset \mathbb{R}$ is not compact.

\vspace{0.1in}

Consider $\mathcal{O} = \left\{ \left(\dfrac{1}{n},1-\dfrac{1}{n} \right) \;\Big|\; n=1,2,\dots \right\}$. So $\mathcal{O}$ is a collection of open sets such that $\displaystyle \bigcup\limits_{O \in \mathcal{O}} O = \bigcup\limits_{n=1}^\infty \left(\dfrac{1}{n},1-\dfrac{1}{n}\right) = (0,1)$. Thus $\mathcal{O}$ is an open cover of $I=(0,1)$. 

Consider any finite subcover of $\mathcal{O}$. We have $(1/n_1,1-1/n_1),\dots,(1/n_k,1-1/n_k)$. But the union of these interval is $(1/m,1-1/m)$ where $m$ is the maximum of $n_1,\dots,n_k$. Since $1-1/m <1$, we have $(1/m,-1/m) \not= (0,1)=I$. Thus $\mathcal{O}$ has is no finite subcover. 

Since we found an open cover with no finite subcover, $I$ is not compact.

\vspace{0.1in}

\end{enumerate}

\noindent {\bf\large 10. (10 points)} Let $A \subset X$ where $X$ is Hausdorff and $A$ is compact. Suppose that $x_0 \not\in A$. Show that there exists open sets $U$ and $V$ such that $U$ and $V$ are disjoint, $x_0 \in U$ and $A \subseteq V$.

\vspace{0.1in}

Select $a \in A$. Then $x_0 \not= a$ (because $x_0 \not\in A$). $X$ is Hausdorff so we can find disjoint open sets $U_a$ and $V_a$ such that $a \in U_a$ and $x_0 \in V_a$. Now consider $\mathcal{U} = \{ U_a \;|\; a \in A \}$. This is an open cover of $A$. Now $A$ is compact, so there exists a finite subcover $\{U_{a_1},\dots,U_{a_n}\}$. 

Let $U = U_{a_1} \cup \cdots \cup U_{a_n}$ and $V = V_{a_1} \cap \cdots \cap V_{a_n}$. $U$ and $V$  are open since unions and finite intersections of open sets are open. The $U_{a_i}$'s cover $A$ so $A \subseteq U$. Each of the $V_{a_i}$'s is a neighborhood of $x_0$ so $x_0 \in V$. 

Suppose $z \in U \cap V$. Then $z \in U_{a_j}$ for some $j = 1,\dots,n$. But $z \in V$ so $z \in V_{a_j}$ as well. However, $U_{a_j} \cap V_{a_j} = \varnothing$ (contradiction). Thus $U$ and $V$ are disjoint. So we've separated $A$ and $x_0$ with open sets.


\vspace{0.1in}


\newpage

\noindent
\parbox{1.5in}{\bf Math 4710/5710} 
\hfill {\Large \bf  In the Midst-term Exam --- Take Home} \hfill
\parbox{1.5in}{\bf \hfill November $14^{\mathrm{th}}$, 2014}

\vspace{0.1in}

The following are True/Possible/False questions. If the statement is always true, state ``{\sc True}'' and then prove the statement. If the statement is never true, state ``{\sc False}'' and then prove the statement cannot ever hold. If the state is sometimes true and sometimes false, state ``{\sc Possible}'' and then give an example of it holding and an example of it failing to hold.

\vspace{0.05in}

Each problem is worth {\bf 10 points}. Undergrads should complete {\bf 5} out of {\bf 7} (you may do them all for potential extra credit). Grad students must complete all of these.

\vspace{0.1in}

\begin{enumerate}
\item Let $\mathcal{T}$ be a topology for the set $\mathbb{Z} = \{ \dots, -2,-1,0,1,2,\dots \}$.

\vspace{0.05in}

Is it {\sc True}, \fbox{\sc Possible}, or {\sc False} \quad that $|\mathcal{T}|=5$? 

\vspace{0.05in}

There are many examples of topologies on $\mathbb{Z}$ which consist of more or less than 5 elements. For example: $\mathcal{T} = \mathcal{P}(\mathbb{Z})$ the discrete topology has infinitely many elements or $\mathcal{T} = \{ \varnothing, \mathbb{Z}\}$ the trivial topology only has 2 elements. So a topology on $\mathbb{Z}$ certainly does not have to have 5 elements.

On the other hand, there are many topolgies on $\mathbb{Z}$ which contain exactly 5 elements. For example: $\mathcal{T} = \{\varnothing, \mathbb{Z}, \{0\}, \{1\}, \{0,1\} \}$ is a topology (it contains the empty set, the whole set, is closed under unions, and finite intersections). %Another example would be $\mathcal{T} = \{ \varnothing, \mathbb{Z}, 3\mathbb{Z}, 3\mathbb{Z}+1, 3\mathbb{Z}+2 \}$ where $3\mathbb{Z} = \{ \dots,-6,-3,0,3,6,\dots \}$, $3\mathbb{Z}+1 = \{ \dots,-5,-2,1,4,7,\dots \}$, and $3\mathbb{Z}+2 = \{ \dots,-4,-1,2,5,8,\dots \}$. 

\vspace{0.1in}

\item Let $X$ be a metrizable topological space.

\vspace{0.05in}

Is it {\sc True}, {\sc Possible}, or \fbox{\sc False} \quad that $X$ has a non-metrizable subspace? 

\vspace{0.05in}

This is false since every subspace of a metrizable topological space is itself metrizable. 

Let $d$ be a metric which induces the topology on $X$ ($d$ must exist since $X$ is metrizable). Let $A \subseteq X$. Consider the metric $d$ restricted to $A$. This gives us a metric on $A$ ($d(a,b) \geq 0$ for all $a,b \in A$ since this is true for all elements of $X$, $d(a,b)=0$ iff $a=b$ for all $a,b \in A$ since this is true for all elements of $X$, $d(a,b)=d(b,a)$ for all $a,b \in A$ since this is true for all elements of $X$, and $d(a,b)+d(b,c) \geq d(a,c)$ for all $a,b,c \in A$ since this is true for all elements of $X$).

So the metric on $X$ gives us a metric on $A$. We need to show that this metric gives us the same topology as the subspace topology. Let's show that the basic open sets for both topologies match. Let $B^A_\epsilon(a) = \{ x \in A \;|\; d(x,a) < \epsilon \}$ and $B^X_\epsilon(a) = \{ x \in X \;|\; d(x,a) < \epsilon \}$. We have that sets of the form $B^A_\epsilon(a)$ (where $a \in A$ and $\epsilon >0$) form a basis for the metric topology on $A$. On the other hand, $B^X_\epsilon(x)$ (where $x \in X$ and $\epsilon >0$) form a basis for the metric topology on $X$. If the balls in $X$ are intersected with $A$, we get a basis for the subspace topology on $A$ (remember open sets in the subspace topology are intersections of open sets in $X$ intersected with the subspace). Let $a \in A$ and $\epsilon>0$, then  $B^X_\epsilon(a) \cap A = \{x \in X \;|\; d(x,a)<\epsilon \} \cap A = \{ x \in A \;|\; d(x,a)<\epsilon \} = B^A_\epsilon(a)$.

Let $U$ be open in $A$ (in the metric topology). For every $u \in U$ we have an open ball $B^A_\epsilon(u) \subseteq U$ but this is the same as $B^X_\epsilon(u) \cap A \subseteq U$. So $U$ is open in $A$ (in the subspace topology). Conversely, let $U$ be open in $A$ (in the subspace topology). For every $u \in U$ there is some open ball in $X$ whose intersection with $A$ is contained in $U$. However, we can assume that this ball is centered at $u$ itself (the radius may need to be shrunk, but this is always possible). So we get $B^X_\epsilon(u) \cap A \subseteq U$. But this is the same as $B^A_\epsilon(u) \subseteq U$. So $U$ is open in $A$ (in the metric topology).  

\vspace{0.1in}

\item $X$ is a path connected topological space.

\vspace{0.05in}

Is it \fbox{\sc True}, {\sc Possible}, or {\sc False} \quad that $X$ is still path connected when given a coarser topology? 

\vspace{0.05in}

Consider a continuous map between two topological spaces: $f:X_1 \to X_2$. This means that the inverse image of an open set in $X_2$ under the map $f$ is an open set in $X_1$. If we add more open sets to the topology of $X_1$, the continuity of $f$ is not damaged. If we remove open sets from the topology of $X_2$, again the continuity of $f$ is not damaged. In summary, given a continuous map $f:X_1 \to X_2$, $f$ will remain continuous even if we give a finer topology to $X_1$ or a coarser topology to $X_2$.

Now consider a path connected space $X$. Let $a,b \in X$ and let $f:[0,1] \to X$ be a path from $a$ to $b$ (i.e. $f$ is continuous, $f(0)=a$, and $f(1)=b$). Suppose that $X$ is given a coarser topology. Then $f$ will still be continuous. This means we still have a path from $a$ to $b$. So, yes, $X$ is still path connected.

\vspace{0.1in}

\newpage

\item Let $X$ be a non-empty simply ordered set. Give $X$ the order topology. Let $a,b \in X$ with $a<b$.

\vspace{0.05in}

Is it {\sc True}, \fbox{\sc Possible}, or {\sc False} \quad that $\overline{(a,b)}=[a,b]$?

\vspace{0.05in}

This may hold, but it doesn't have to. The issue at hand is whether $a$ has a immediate predecessor and whether $b$ has an immediate successor. Elements in $\mathbb{Z}$ have immediate predecessors and successors, so intervals in $\mathbb{Z}$ will fail to have this closure property. On the other hand, elements in $\mathbb{R}$ do not have immediate predecessors or successors, so intervals in $\mathbb{R}$ will have this closure property.

Consider $(0,1)$ in $\mathbb{R}$. We have that $\overline{(0,1)} = [0,1]$. 

On the other hand, consider $(0,1)$ in $\mathbb{Z}$. Then $(0,1) = \{ n \in \mathbb{Z} \;|\; 0<n<1 \} = \varnothing$. Whereas, $[0,1] =$\\ $\{ n \in \mathbb{Z} \;|\; 0 \leq n \leq 1 \} = \{0,1\}$. So $\overline{(0,1)}=\overline{\varnothing} = \varnothing \not= [0,1]$.

\vspace{0.1in}

\item Let $X$ be a finite topological space. Let $f:X \to Y$ be continuous.

\vspace{0.05in}

Is it \fbox{\sc True}, {\sc Possible}, or {\sc False} \quad that $f(X)$ is compact?

\vspace{0.05in}

Any finite space is automatically compact: If $|X|=n$, then the power set of $X$ has $2^n$ elements. Thus any topology on $X$ has at most finitely many open sets (since a topology on $X$ is a subset of the power set of $X$). So any cover is already a finite (sub)cover. Now $X$ is finite, so $f(X)$ (the range of $f$) must be finite (the range can be smaller than $X$ but not bigger). Thus $f(X)$ is compact. 

Notice that we didn't even need to use the assumption that $f$ is continuous! We could have (uneccesarily) used this assumption if we had argued: $X$ is finite so it's compact. $f$ is continuous so it sends compact sets to compact sets. Therefore, $f(X)$ is compact.

\vspace{0.1in}

\item Let $f:\mathbb{R} \to \mathbb{R}_\ell$ be a continuous map. 

\vspace{0.05in}

Is it {\sc True}, {\sc Possible}, or \fbox{\sc False} \quad that $f(0)=0$ and $f(1)=1$? 

\vspace{0.05in}

Let $C$ be a connected subset of $\mathbb{R}_\ell$. Suppose $a,b \in C$ and $a<b$. Then $U=(-\infty,b)$ and $V=[b,\infty)$ separate $C$ ($U$ and $V$ are open and disjoint. Their union contains $C$ and $a \in U$, $b \in V$ so they're non-empty). Therefore, $C$ is not connected (contradiction). Therefore, if $C$ is connected in $\mathbb{R}_\ell$, it must either be empty or a singleton set. 

Now $\mathbb{R}$ is connected. Continuous maps send connected sets to connected sets. Therefore, the range of $f$ is connected. Using the paragraph above, we conclude that the range of $f$ is a singleton set (it can't be empty). This means that $f$ is a constant map. So the only continuous maps from $\mathbb{R}$ to $\mathbb{R}_\ell$ are constant maps. Thus the proposed map sending $0$ to $0$ and $1$ to $1$ cannot exist.
\vspace{0.1in}

\item Let $A$ be a countable closed subset of $X=[0,1] \times [0,1]$ (the unit square in $\mathbb{R}^2$). Let $f:\mathbb{R} \to A$ be continuous.

\vspace{0.05in}

Is it \fbox{\sc True}, {\sc Possible}, or {\sc False} \quad that there is an open interval $I=(a,b) \subseteq \mathbb{R}$ such that $f$ restricted to $I$ is constant? 

\vspace{0.05in}

I had intended to add in the assumption that $f$ is {\it onto}. Oops! Let's prove this with and without that assumption. 

Now $[0,1]$ is compact, path connected, and Hausdorff so $X$ is too. $A$ is a closed subset of $X$ so it is compact. Thus $A$ is a countable compact Hausdorff space. Therefore$A$ must have an isolated point ($A$ may or may not be connected).

Let $a \in A$ be an isolated point (we just established that one must exist). So $\{ a\}$ is open, thus $f^{-1}\{ a\}$ is open in $\mathbb{R}$. Supposing that $f$ is onto, something must map to $a \in A$ so $f^{-1}\{ a \}$ is a non-empty open subset of $\mathbb{R}$. Thus it must contain some element $x$ (since it's non-empty) and some $\epsilon$-neighborhood of $x$ (since it's open). So we have $(x-\epsilon,x+\epsilon) \subseteq f^{-1}\{a\}$. By definition for all $x-\epsilon < y < x+\epsilon$, $f(y)=a$ ($f$ is constant on the interval $(x-\epsilon,x+\epsilon)$). Note: If $f$ isn't onto, it could be that $f^{-1}\{a\} = \varnothing$. So this argument won't work without assuming that $f$ is onto.

Now let's prove this statement without the onto assumption. Notice that $\mathbb{R}$ is connected, so $f(\mathbb{R}) \subseteq A$ must be connected. We know that $A$ is countable compact and Hausdorff (as argued above). So $f(\mathbb{R})$ is countable Hausdorff and also connected (we can't immedaitely conclude it's compact since we don't know if the range is closed or not). Let $C = \overline{f(\mathbb{R})}$ (the closure in $A$). Then $C$ is a closed subset of a compact space, so it's compact. Since $A$ is countable and Hausdorff and $C \subseteq A$, $C$ must be countable and Hausdorff. Finally, $C$ is connected since the closure of a connected set is still connected. Thus $C$ is a countable compact connect Hausdorff space. It must have an isolated point, say $y \in C$. Then $\{ y \}$ is closed (because $C$ is Hausdorff) and $\{ y \}$ is open (since $y$ is isolated). Thus $\{ y\}$ is a non-trivial clopen set. But $C$ is connected, so $\overline{f(\mathbb{R})} = C = \{ y \}$.  Now $f(\mathbb{R}) \not= \varnothing$ (the range cannot be empty). Thus $f(\mathbb{R}) = \{y\}$. This means $f$ is a constant map. So $f$ isn't just constant on some interval, it's constant everywhere!

\vspace{0.1in}

\end{enumerate}


\end{document}


