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\title{Topology problems}
\maketitle

\section{Problems on topology}

\subsection{Basics of topologies}

\subsubsection{How are the topologies on $\mathbb{R}^{n}$ and on metric spaces
related to the general definition of topology?}

\begin{enumerate}
\setcounter{enumi}{\value{myenumi}}

\item What are some examples of topologies on $\mathbb{R}$, $\mathbb{R}^{n},$
finite sets, and other spaces? Which of these are homeomorphic?\ Which are not
homeomorphic? How would I\ construct more?

\item What is the relationship between topology of $\mathbb{R}$ from
calculus/analysis and from topology? An open set is in $\mathbb{R}$ a set $U$
such that for any $x\in U$ there is an $\varepsilon>0$ such that $\left(
x-\varepsilon,x+\varepsilon\right)  \subseteq U.$ A closed set is a set which
contains all of its limit points, i.e. if $x_{i}\in F$ for all $i,$ then
$\lim_{i\rightarrow\infty}x_{i}\in F.$

\begin{enumerate}
\item Show that every open subset of $\mathbb{R}$ is the complement of a
closed set and every closed subset is the complement of an open set.\newline

\item We say a function $f:D\mathbb{\rightarrow\mathbb{R}}$, where
$D\subset\mathbb{R}$, is continuous if for any sequence $\left\{
x_{i}\right\}  \subseteq D$ with $\lim_{i\rightarrow\infty}x_{i}\in D,$
\[
\lim_{i\rightarrow\infty}f\left(  x_{i}\right)  =f\left(  \lim_{i\rightarrow
\infty}x_{i}\right)  .
\]
Show that if $F\subseteq\mathbb{R}$ is closed, then $f^{-1}\left(  F\right)  $
is closed.\newline

\item Show that if $U\subseteq\mathbb{R}$ is open, then $f^{-1}\left(
U\right)  $ is open.\newline

\item Show that every open set is the union of intervals.\newline

\item If $U\subseteq\mathbb{R}$ is open, show that a function $f:U\rightarrow
\mathbb{R}$ is continuous (meaning the preimage of an open set is open) if and
only if for every $x\in U$ and for every $\varepsilon>0$ there exists a
$\delta>0$ such that $\left\vert f\left(  x\right)  -f\left(  y\right)
\right\vert <\varepsilon$ if $\left\vert x-y\right\vert <\delta.$
\end{enumerate}

\item Closed intervals are gotten by \textquotedblleft closing
up\textquotedblright\ intervals so that sequences must limit to points in the
interval itself. Does this notion generalize? How are closed sets related to
limits of sequences? In a topological space, a set $F$ is defined to be
\emph{closed} if $F^{C}=X\setminus F\in\mathcal{T}$, i.e. if $F^{C}$ is open.

\begin{enumerate}
\item Show that arbitrary intersections and finite unions of closed sets are
closed.\newline

\item Show that a map $f:X\rightarrow Y$ is continuous if and only if
$f^{-1}\left(  V\right)  $ is closed for any closed set $V\subseteq
Y.$\newline

\item A point $x\in X$ is a\emph{ limit point} of a set $A\subset X$ if every
open set $U$ containing $x$ also contains a point $y\in A\setminus\left\{
x\right\}  .$ The \emph{closure} of $A,$ denoted $cl\left(  A\right)  $ or
$\bar{A},$ is the intersection of all closed sets containing $A.$ Show that
$\bar{A}$ is closed.\newline

\item Show that $\bar{A}$ is equal to the union of $A$ and its limit
points.\newline

\item Show that if $A$ is closed, then $\bar{A}=A$ and $A$ contains its limit
points.\newline

\item (accumulation points) A point $x\in X$ is an \emph{accumulation point}
of $A$ if there exists a sequence in $A\setminus\{x\}$ that converges to $x$.
A point $x\in A$ is an \emph{isolated point} (of $A$) if there is an open set
$O$ such that $O\cap A=\{x\}$. Show that in a metric space, accumulation
points and limit points are the same.

\item Let $A\subseteq{\mathbb{R}}$, and $A^{\prime}$ denotes the set of all
the accumulation points of $A$. If $y\in A^{\prime}$ and $U\subseteq
{\mathbb{R}}$ is an open set containing $y$, show that there are infinitely
many distinct points in $A\cap U$.

\item Show that
\[
A^{\prime}=\bigcap_{x\in A}\overline{A\backslash\{x\}}.
\]
\newline

\item Give an example of a limit point that is not an accumulation point in a
topological space.
\end{enumerate}

\item Convince yourself that a metric space determines a topology. What is the
natural definition of continuous? How do these compare to both the definition
on $\mathbb{R}$ and for a general topological space? A \emph{metric space}
$(X,d)$ is a set $X$ and a function (called the metric) $d:X\times
X\rightarrow\mathbb{R}$ such that for all $x,y,z\in X,$ the metric satisfies:

\begin{itemize}
\item (positive definite) $d\left(  x,y\right)  \geq0$ with $d\left(
x,y\right)  =0$ if and only if $x=y$

\item (symmetric) $d\left(  x,y\right)  =d\left(  y,x\right)  $

\item (triangle inequality) $d\left(  x,z\right)  \leq d\left(  x,y\right)
+d\left(  y,z\right)  $
\end{itemize}

The ball of radius $r>0$ centered at $x\in X$ is defined to be $B\left(
x,r\right)  =\left\{  y\in X:d\left(  x,y\right)  <r\right\}  .$ We can
generalize the definition of open set on $\mathbb{R}$ to open set on a metric
space by saying a set $U\subseteq X$ is open if for every $x\in U$ there
exists an $r>0$ such that $B\left(  x,r\right)  \subseteq U.$

\begin{enumerate}
\item Show that $d\left(  x,y\right)  =\left\vert x-y\right\vert $ makes
$\mathbb{R}$ into a metric space.

\item Show that the open sets described above form a defines a topology. This
is called the \emph{metric topology}.

\item Show that a function $f:X\rightarrow Y$ between metric spaces $\left(
X,d_{X}\right)  $ and $\left(  Y,d_{Y}\right)  $ is continuous if and only if
for every $x\in X$ and for every $\varepsilon>0$ there exists a $\delta>0$
such that $d_{Y}\left(  f\left(  x\right)  ,f\left(  y\right)  \right)
<\varepsilon$ if $d_{X}\left(  x,y\right)  <\delta.$

\item Show that in a metric space limit points and accumulation points are the same.
\end{enumerate}

\item One way of producing metric spaces is with norms on vector spaces.
(Norms on vector spaces) Let $V$ be a vector space. A function $\left\Vert
\cdot\right\Vert :V\rightarrow\mathbb{R}$ on $V$ is a \emph{norm} if

\begin{itemize}
\item For all $x\in V$, $\left\Vert x\right\Vert \geq0$, and $\left\Vert
x\right\Vert =0$ if and only if $x=0$.

\item $\left\Vert ax\right\Vert =\left\vert a\right\vert \left\Vert
x\right\Vert $ for every $a\in\mathbb{R}$ and $x\in V$.

\item $\left\Vert x+y\right\Vert \leq\left\Vert x\right\Vert +\left\Vert
y\right\Vert $ for every $x,y\in V$.
\end{itemize}

\begin{enumerate}
\item Show that $d\left(  x,y\right)  =\left\Vert x-y\right\Vert $ defines a metric.

\item Let $C((0,1))$ be the set of bounded continuous functions on $(0,1)$ and
show that $\left\Vert f\right\Vert =\sup_{x\in\left(  0,1\right)  }\left\vert
f\left(  x\right)  \right\vert $ is a norm. Let $U=\{f:f(x)>0\quad\forall
x\in(0,1)\}$, and $V=\{f:f(x)\geq0\quad\forall x\in(0,1)\}$. For each of $U$
and $V$ determine if the set is open or closed or neither. You should prove
your answer.
\end{enumerate}

\item For $\mathbb{R}$, we notice that there are points whose distance is
arbitrarily large. Is this always true? Why or why not? Is this a property of
the topology of $\mathbb{R}$?

\item (Bounded metrics) For $x,y\in\mathbb{R}$ let
\[
d(x,y)=\frac{|x-y|}{1+|x-y|}%
\]


\begin{enumerate}
\item Show this is a metric.

\item Does this metric give $\mathbb{R}$ a different topology from the one
that comes from the usual metric on $\mathbb{R}$? You should prove your answer.

\item Generalize this to show that for any metric space $\left(  X,d\right)
,$ there is a bounded metric (i.e., one for which there exists $M>0$ such that
the distance between any two points is less than $C$) that generates the same topology.
\end{enumerate}

\setcounter{myenumi}{\value{enumi}}
\end{enumerate}

\subsubsection{Can we generalize the notion of a metric topology? What might
be a good way to do this?}

\begin{enumerate}
\setcounter{enumi}{\value{myenumi}}

\item (local base) Let $X$ be a set. A \emph{local base (or basis)}, or\emph{
neighborhood base},\emph{ }is a collection $\{\mathcal{N}(x)|x\in X\}$ of
subsets of $X$ satisfying

\begin{itemize}
\item $V\in\mathcal{N}(x)\implies x\in V$.

\item If $V_{1},V_{2}\in\mathcal{N}(x)$, then $\exists V_{3}\in\mathcal{N}(x)$
such that $V_{3}\subseteq V_{1}\cap V_{2}$.

\item If $V\in\mathcal{N}(x)$, then there exists a $W\in\mathcal{N}(x)$ such
that $W\subseteq V$ and the following holds: If $y\in W$, then there exists
$U\in\mathcal{N}(y)$ such that $U\subseteq W$.
\end{itemize}

\begin{enumerate}
\item Given a local base $\left\{  \mathcal{N}(x)\right\}  $, define a set
$\mathcal{T}$ by $U\in\mathcal{T}$ if and only if for any $x\in U,$ there
exists $V\in\mathcal{N}(x)$ such that $V\subseteq U.$ Show that $\mathcal{T}$
is a topology. We call $\left\{  \mathcal{N}(x)\right\}  $ a local base of the topology.

\item Suppose $\left\{  \mathcal{N}(x)\right\}  $ is a local base for a
topological space $(X,\mathcal{T})$. If $A\in\mathcal{N}(x)$, show that there
exists a $U\in\mathcal{T}$ such that $x\in U\subseteq A$. Hence neighborhoods
need not be open, but must contain an open set.

\item Show that open balls $B_{r}\left(  x\right)  =\left\{  y:d\left(
x,y\right)  <r\right\}  $ form a local base of a metric space $\left(
X,d\right)  $.

\item Show that closed disks $D_{r}\left(  x\right)  =\left\{  y:d\left(
x,y\right)  \leq r\right\}  $ (where $r>0$) form a local base of a metric
space $\left(  X,d\right)  $.

\item $\mathcal{N}_{1}(x)$ and $\mathcal{N}_{2}(x)$ are local bases for a
space $X$. Show that the topology $\mathcal{T}_{1}$ generated by
$\mathcal{N}_{1}(x)$ is finer that the topology $\mathcal{T}_{2}$ generated by
$\mathcal{N}_{2}(x)$ if and only if for all $B\in\mathcal{N}_{2}(x)$, there is
a set $A\in\mathcal{N}_{1}(x)$ such that $x\in A\subseteq B$.
\end{enumerate}

\item Another idea for how to produce topologies: (basis for a topology) A
\emph{basis} $\mathcal{B}$ is a collection of subsets of $X$ such that

\begin{itemize}
\item For all $x\in X,$ there exists $U\in\mathcal{B}$ such that $x\in U$.

\item If $U,U^{\prime}\in\mathcal{B}$ and $x\in U\cap U^{\prime},$ then there
is a set $U^{\prime\prime}\in\mathcal{B}$ such that $x\in U^{\prime\prime}$
and $U^{\prime\prime}\subset U\cap U^{\prime}.$
\end{itemize}

\begin{enumerate}
\item Show that a basis generates a topology by taking the open sets to be all
sets we can form by taking a union of a collection of sets in $\mathcal{B}$.

\item Show this is equivalent to defining a set $V$ to be open if every point
$x\in V$ has a set $U\in\mathcal{B}$ such that $x\in U\subseteq V.$

\item Show that the collection of all balls forms a basis for the metric
topology. Show that not every open set is in the basis.
\end{enumerate}

\item How are topologies on the same space related to each other?
(finer/coarser) In this problem and the next, we explicitly write the topology
as a collection of open sets $\mathcal{T}$. Suppose $(X,\mathcal{T}_{1})$ and
$(X,\mathcal{T}_{2})$ are topological spaces and $\mathcal{T}_{2}%
\subseteq\mathcal{T}_{1}$. We say the topology $\mathcal{T}_{1}$ is
\emph{finer} than $\mathcal{T}_{2}$ and the topology $\mathcal{T}_{2}$ is
\emph{coarser} than $\mathcal{T}_{1}.$

\begin{enumerate}
\item Show that $\mathcal{T}_{1}$ is finer than $\mathcal{T}_{2}$ if and only
if for every $x\in X$ and every $U\in\mathcal{T}_{2}$ with $x\in U$, there is
a $V\in\mathcal{T}_{1}$ with $x\in V$ such that $V\subseteq U$.

\item Show that $\mathcal{T}_{1}$ is finer than $\mathcal{T}_{2}$ if and only
if the identity map $\mathrm{Id}:(X,\mathcal{T}_{1})\rightarrow(X,\mathcal{T}%
_{2})$ is continuous.

\item Can you find examples of two topologies that are comparable in this way,
and two that are not?
\end{enumerate}

\setcounter{myenumi}{\value{enumi}}
\end{enumerate}

\subsubsection{How can we construct new topologies?}

\begin{enumerate}
\setcounter{enumi}{\value{myenumi}}

\item (Product topology) Let $X_{\alpha}$ be a collection of topological
spaces indexed by a set $\mathcal{I}$. We define a topology on the Cartesian
product $\prod_{\alpha\in\mathcal{I}}X_{\alpha}$ as follows: a basis is given
by sets of the form $\prod_{\alpha\in\mathcal{I}}U_{\alpha}$, where
$U_{\alpha}\subseteq X_{\alpha}$ is open and for all but finitely many
$\alpha$ , $U_{\alpha}=X$ .

\begin{enumerate}
\item Prove that the above construction does indeed yield a basis.

\item Prove that this is the coarsest topology (with fewest open sets) such
that the projections $\pi_{\alpha}:\prod_{\alpha\in\mathcal{I}}X_{\alpha
}\rightarrow X_{\alpha}$ are continuous.

\item Compare the product topology and metric topology on $\mathbb{R}^{2}$.
\end{enumerate}

\item (Subspace topology) Suppose $(X,{\mathcal{T}})$ is a topological space.

\begin{enumerate}
\item If $W\subseteq X$ is a subset, show that the \emph{relative} or
\emph{subspace} topology of $W$ defined by $O\subseteq W$ is open only if
$O=W\bigcap U$ for some $U\in{\mathcal{T}}$ is indeed a topology.

\item Let $S_{2}\subset S_{1}\subset X$. Equip $S_{1}$ with the subspace
topology. There are two ways to define a topology on $S_{2}$. We can give it
the subspace topology it gets by thinking of it as a subset of $X$ or we can
give it the subspace topology it gets by thinking of it as a subset of $S_{1}%
$. Show that there two topologies on $S_{2}$ are the same.
\end{enumerate}

\item (Initial topology) Given a map $f:X\rightarrow Y$ and a topology on $Y$,
let ${\mathcal{T}}$ be the collection of subsets of $X$ of the form
$f^{-1}(U)$ where $U$ is open.

\begin{enumerate}
\item Show that this defines a topology on $X.$ This is called the
\emph{initial topology} or the \emph{weak topology}.

\item Suppose that the index $\alpha$ ranges over some index set $\mathcal{A}$
and for each $\alpha$ we have a topological space $(Y_{\alpha},\mathcal{S}%
_{\alpha})$ and a function $f_{\alpha}:X\rightarrow Y_{\alpha}$. In this case,
we cannot simply take the sets of the form $f_{\alpha}^{-1}\left(  U\right)  $
where $U$ is open. Instead, we need to take the coarsest topology (fewest open
sets) containing this set. We call this initial topology (or weak topology) as
well. Show that the initial topology constructed above is the weakest topology
on $X$ that makes all the functions $f_{\alpha}$ continuous.

\item Show that the subspace topology is the initial topology for the
inclusion map.

\item Show that the product topology is the initial topology for the
projection maps $\pi_{\alpha}.$
\end{enumerate}

\item (final topology) If $Y$ is a set, $(X,\mathcal{T})$ is a topological
space, and $f:X\rightarrow Y$ is a function, then we can define a
$\mathcal{T}^{\prime}$ on $Y$ by taking $\mathcal{T}^{\prime}$ to be all
subsets $U$ of $Y$ such that $f^{-1}(U)\in\mathcal{T}$.

\begin{enumerate}
\item Show this defines a topology. This is sometimes called the \emph{final
topology} or \emph{strong topology}. With this construction $f$ is a
continuous function. It is important to note that this construction works
because of the set identities
\begin{align}
f^{-1}(\cup_{\alpha}U_{\alpha})  &  =\cup_{\alpha}f^{-1}(U_{\alpha
})\nonumber\\
f^{-1}(\cap_{\alpha}U_{\alpha})  &  =\cap_{\alpha}f^{-1}(U_{\alpha})
\end{align}


\item Show that the final topology makes $f$ a continuous function and that it
is the finest topology (most open sets) that makes $f$ a continuous function.
\end{enumerate}

\item Let $X$ be a topological space and let $\sim$ be an equivalence
relation. Recall that an equivalence relation $\sim$ is a relation satisfying
the following properties:

\begin{itemize}
\item (reflexivity) $x\sim x.$

\item (symmetry) $x\sim y$ implies $y\sim x$

\item (transitivity) $x\sim y$ and $y\sim z$ implies $x\sim z.$
\end{itemize}

Then $Q=X/\sim$ denotes the set of equivalence classes of the relation. For
$x\in X$, we denote the equivalence class containing $x$ by $[x]$. There is a
natural quotient map $q:X\rightarrow Q$ given by $q\left(  x\right)  =\left[
x\right]  $. We now define the \emph{quotient topology} on $Q$ such that
$U\subseteq Q$ is open if $q^{-1}(U)$ is open in $X$. We call this the
\emph{quotient topology.}

\begin{enumerate}
\item Show that the quotient topology is, in fact, a topology.

\item Show that $q$ is continuous if $Q$ has the quotient topology.

\item Show that the quotient topology is the finest topology (most open sets)
such that $q$ is continuous.

\item Show that the quotient topology is the final topology for the quotient
map $q\left(  x\right)  =\left[  x\right]  .$(quotient topology)

\item \#The circle is a subset of $\mathbb{R}^{2}$ and so can be given the
subspace topology. We can also think of the circle as the interval $[0,2\pi]$
with the two endpoints identified. To be more precise, we define an
equivalence relation by defining $0\sim2\pi$, and no other distinct points are
equivalent. Since $[0,2\pi]$ has a topology, we can consider the quotient
topology on $[0,2\pi]/\sim$. Show that $[0,2\pi]/\sim$ is homeomorphic to the
circle $\left\{  \left(  x,y\right)  \in\mathbb{R}^{2}:x^{2}+y^{2}=1\right\}
$ with the subspace topology.
\end{enumerate}

\item (profinite topology) An arithmetic progress in $\mathbb{Z}$ is a set of
the form $\{k+nl:n\in\mathbb{Z}\}$ where $l$ is a positive integer and $k$ is
any integer. Define the \emph{profinite topology} on $\mathbb{Z}$ in which the
open sets are the empty set and unions of arithmetic progressions.

\begin{enumerate}
\item Show that an arithmetic progression is also a closed set in this topology.

\item Show that if there were only finitely many primes, then the set
$\{-1,1\}$ would be open.

\item Then show that this set is not open and conclude that there are
infinitely many primes.

\item Let $T^{\infty}$ be the product of countably infinitely many copies of
the unit circle with the product topology. Define the map $\phi:\mathbb{Z}%
\rightarrow T^{\infty}$ as follows:
\[
\phi(n)=\left(  \exp(2\pi in/2),\exp(2\pi in/3),\exp(2\pi in/4),\exp(2\pi
in/5),...\right)  .
\]
Show that this map is injective and the induced topology on $\mathbb{Z}$
coincides with the profinite topology.
\end{enumerate}

\setcounter{myenumi}{\value{enumi}}
\end{enumerate}

\section{Problems on compactness}

\begin{enumerate}
\setcounter{enumi}{\value{myenumi}}

\item Prove or disprove:

\begin{enumerate}
\item Show that $(0,1]$ is not compact.

\item $A$ is finite and $U$ is a open subset of ${\mathbb{R}}$. If $A\subseteq
U$, there exists an $\epsilon>0$ such that for all $x\in A$, $B(x,\epsilon
)\subseteq U$. ($B(x,\epsilon)$ denotes the open ball of radius $\epsilon$
centered at $x$.)

\item $P$ is countable and $U$ is a open subset of ${\mathbb{R}}$. If
$P\subseteq U$, there exists an $\epsilon>0$ such that for all $x\in P$,
$B(x,\epsilon)\subseteq U$.

\item $F$ is closed and $U$ is a open subset of ${\mathbb{R}}$. If $F\subseteq
U$, there exists an $\epsilon>0$ such that for all $x\in F$, $B(x,\epsilon
)\subseteq U$.

\item $K$ is compact and $U$ is a open subset of ${\mathbb{R}}$. If
$K\subseteq U$, there exists an $\epsilon>0$ such that for all $x\in K$,
$B(x,\epsilon)\subseteq U$.
\end{enumerate}

\item What kinds of applications of the compactness property can you imagine?
Keep in mind both definitions of compactness (regular and sequential) and
think about different kinds of topological spaces (e.g., subsets of
$\mathbb{R}$, function spaces, etc.) It might help to know that closed
intervals are compact, the sphere is compact, as is the set of continously
differentiable functions $f:\left[  0,1\right]  \rightarrow\left[  0,1\right]
$ (where the topology on the function space is generated by the uniform
distance $d\left(  f,g\right)  =\sup\left\{  \left\vert f\left(  x\right)
-g\left(  x\right)  \right\vert :x\in\left[  0,1\right]  \right\}  $).

\item What are some examples of spaces that are compact and of spaces that are
not compact? Can you make a noncompact space from compact? Can you make a
compact space from a noncompact space? Consider both examples and
constructions. Are finite spaces compact? Are any countable spaces compact?
What are the compact subspaces of $\mathbb{R}$?

\begin{enumerate}
\item Any finite topological space is compact.

\item The finite-dimensional sphere $\left\{  x\in\mathbb{R}^{n}:\left\vert
x\right\vert ^{2}=1\right\}  $ is compact.

\item The Cantor set $K$ defined by $K=\left[  0,1\right]  \setminus%
%TCIMACRO{\dbigcup \limits_{m=1}^{\infty}}%
%BeginExpansion
{\displaystyle\bigcup\limits_{m=1}^{\infty}}
%EndExpansion%
%TCIMACRO{\dbigcup _{k=0}^{3^{m-1}-1}}%
%BeginExpansion
{\displaystyle\bigcup_{k=0}^{3^{m-1}-1}}
%EndExpansion
\left(  \frac{3k+1}{3^{m}},\frac{3k+2}{3^{m}}\right)  $ is compact.

\item A set $X$ is has the discrete topology if for each $x\in X,$ the set
$\left\{  x\right\}  $ is open. Show that there is only one such topology and
that it is compact if and only if $X$ is finite

\item Let ${\mathbb{R}}\mathbb{P}^{n}$ denote the quotient space of
${\mathbb{R}}^{n+1}\setminus\{0\}$, by the equivalence relation $x\sim y$ iff
$\exists\lambda\neq0$, s.t. $x=\lambda y$. Show that ${\mathbb{R}}%
\mathbb{P}^{n}$ is compact.

\item Given a set $X,$ the finite complement topology is the topology where
open sets are sets whose complement are finite. Show this is a topology and
that $X$ is compact with this topology.
\end{enumerate}

\item Suppose $(X,{\mathcal{T}})$ is a topological space. When are subspaces compact?

\begin{enumerate}
\item We can define compactness of a subset as follows: $F\subseteq X$ is
compact if every collection $\left\{  U_{i}\right\}  _{i\in I}$ of open sets
in $X$ such that $F\subseteq%
%TCIMACRO{\dbigcup \limits_{i\in I}}%
%BeginExpansion
{\displaystyle\bigcup\limits_{i\in I}}
%EndExpansion
U_{i},$ there is a finite subcover. Show that the topological space $F$ with
the subspace topology is compact if and only if $F$ is compact as a subset of
$X.$

\item Is the definition of compact subspace consistent? Why am I concerned?
Let $S_{2}\subset S_{1}\subset X$. We give $S_{1}$ the subspace topology.

\begin{enumerate}
\item Show that if $S_{2}$ is open in $S_{1}$ then it need not be open in $X$.
Show that it is if $S_{1}$ is open in $X$.

\item Show that if $S_{2}$ is compact in $S_{1}$, then $S_{2}$ is compact in
$X.$
\end{enumerate}
\end{enumerate}

\item Proof of Extreme Value Theorem

\begin{enumerate}
\item Prove $\left[  0,1\right]  $ is compact.\newline

\item Prove that if $X$ and $Y$ are compact, then $X\times Y$ is compact.
\newline

\item Prove the Heine Borel Theorem (for compact subsets of $\mathbb{R}^{n}%
$).\newline

\item If $f:X\rightarrow Y$ is continuous and $X$ is compact, then $f\left(
X\right)  $ is compact (with the subspace topology).\newline

\item Prove the Extreme Value Theorem.
\end{enumerate}

\item How are compactness and closedness related?

\begin{enumerate}
\item Show that if $X$ is compact and $F\subset X$ is closed, then $F$ is
compact (in the subspace topology).\newline

\item Show that compact subsets need not be closed. Hint: consider the
indiscrete topology that has only two open sets.

\item What properties of a topology might ensure that compact subsets are
closed? Hint: Look at the complement of the compact set and see if one can
find an open set around each point.

\item Show that if $X$ is compact and $Y$ is Hausdorff and $f:X\rightarrow Y$
is a continuous bijection, then $f$ is a homeomorphism. Why does this work?
\end{enumerate}

\item (sequential compactness) Compact subsets of $\mathbb{R}$ are often
determined in terms of sequences. A sequence $\left\{  x_{n}\right\}  $ in a
topological space $X$ \emph{converges} to $x_{\infty}\in X$ if for every open
set $U$ containing $x_{\infty},$ there exists $N$ such that $x_{n}\in U$ if
$n>N.$ We say a space is \emph{sequentially compact} if every sequence has a
convergent subsequence. Prove that in a metric space, a set is compact if and
only if it is sequentially compact

\item (one point compactification) How can we make a noncompact space into a
subset of a compact space?

\begin{enumerate}
\item Use stereographic projection to show $\mathbb{R}^{n}$ is homeomorphic to
the sphere $S^{n}\setminus\left\{  pt\right\}  \subset\mathbb{R}^{n+1}.$

\item Can we add one point to any space and find a topology that makes it
compact? Given a topological space $X,$ let $\bar{X}=X\cup\{\infty\}$. The
open sets of $\bar{X}$ are the open sets of $X$ together with the sets
$(X\setminus K)\cup\{\infty\}$, where $K$ is a compact subset of $X$.

\item There is one problem with this; what is it? A space is \emph{locally
compact} if every point has a compact neighborhood. (A neighborhood of a point
$x$ is an set $N$ such that there exists and open set $U$ such that $x\in
U\subseteq N.$)

\item If $X$ is locally compact Hausdorff space, then prove that $\bar{X}$ ,
called the \emph{one point compactification} of $X$, is a compact Hausdorff space.

\item Are compact spaces locally compact? Show this is indeed true.

\item Show that the one point compactification of $\mathbb{R}^{n}$ is
homeomorphic to the sphere $S^{n}.$

\item Show that the rational numbers $\mathbb{Q}$ is not locally compact.
(Hint: Given a neighborhood of $q\in\mathbb{Q}$, it must contain a closed
interval $\left[  q-r,q+r\right]  \cap\mathbb{Q}$ which is compact if $N$ is
compact. Then show that $\left[  q-r,q+r\right]  \cap\mathbb{Q}$ has a cover
with no finite subcover.)
\end{enumerate}

\item Can we generalize Heine-Borel to metric spaces, where both closed and
bounded make sense? Show that it is not true for metric spaces in general.

\begin{enumerate}
\item For $x,y\in\mathbb{R}$ let
\[
d(x,y)=\frac{|x-y|}{1+|x-y|}.
\]
Prove that with this metric, the entire space of $\mathbb{R}$ is not compact
even though it is closed and bounded.

\item A sequence $\left\{  x_{n}\right\}  $ in a metric space $\left(
X,d\right)  $ is said to be \emph{Cauchy} if for every $\varepsilon>0$ there
exists $N>0$ such that $d\left(  x_{n},x_{m}\right)  <\varepsilon$ whenever
$n,m\geq N.$ A metric space is \emph{complete} if every Cauchy sequence converges.

\item Show that every convergent sequence is Cauchy. (The converse is only
true if the metric space is complete.)

\item Give an example of a metric space that is not complete.

\item Show that a compact metric space can be covered by finitely many balls
of any given radius.

\item Show that every compact metric space is complete.

\item A metric space $\left(  X,d\right)  $ is totally bounded if and only if
for every real number $\varepsilon>0$, there exists a finite collection of
open balls in $X$ of radius $\varepsilon$ whose union contains $X$. Show that
a complete, totally bounded metric space is compact. Explain why the previous
example fails to satisfy these assumptions.
\end{enumerate}
\end{enumerate}

\section{Other Problems}

\setcounter{enumi}{\value{myenumi}}

\subsection{Basic questions on the theorems:}

\begin{enumerate}
\item Intermediate Value Theorem:\ What is it useful for? Use the Intermediate
Value Theorem to show that there is a number $c\in\lbrack0,\infty)$ such that
$c^{2}=2.$ We call this number $c=\sqrt{2}.$\newline

\item Extreme Value Theorem. Give an example of applying it to a function.
What other applications are there? Use the Extreme Value Theorem to show
Rolle's theorem: If $f:\left[  a,b\right]  \rightarrow\mathbb{R}$ is
differentiable and $f\left(  a\right)  =f\left(  b\right)  $ then there is a
$c\in\left[  a,b\right]  $ such that $f^{\prime}\left(  c\right)  =0.$

\item Jordan Curve Theorem: Are there examples for which this is difficult to
tell if it is true?\ What happens if the curve is not simple?\ What
applications are there? Use the Jordan Curve Theorem to show there is no
continuous injective map of the complete graph $K_{5}$ into the plane. 

\item What are some properties that are true for real spaces that are only
true for some topologies?
\end{enumerate}

\subsubsection{What are some properties based on how many open sets generate
the topology?}

\begin{enumerate}
\setcounter{enumi}{\value{myenumi}}

\item (first countable) $(X,\mathcal{T})$ is \emph{first countable} if it has
a local base $\mathcal{N}(x)$ such that at every point $x\in X$, the
collection of neighborhoods $\mathcal{N}(x)$ is countable.

\begin{enumerate}
\item Show that $\mathbb{R}$ is first countable.

\item Show that the metric topology on a metric space $(X,d)$ is first countable.

\item Is $(X,\mathcal{T})$ is first countable, show that every point $x\in X$
has a countable collection of open neighborhoods $U_{n}\ni x$ such that
$U_{n+1}\subseteq U_{n}$, and $x$ is an interior point for an open set $O$ if
and only if there is an index $n$ such that $x\in U_{n}\subseteq O$.

\item With the same definitions as the previous part, show that if you
construct a sequence by picking arbitrary points $y_{n}\in U_{n}$, it follows
that the sequence $\{y_{n}\}$ converges.

\item If $(X,\mathcal{T})$ is first countable, and $(Y,\mathcal{S})$ is any
topological space, show that $f:X\rightarrow Y$ is continuous if and only if
it is sequentially continuous.
\end{enumerate}

\item (second countable) A topological space is \emph{second countable} if it
has a countable basis.

\begin{enumerate}
\item *Show that $\mathbb{R}$ with the usual topology is second countable.

\item Give an example of a space that is first countable but not second countable.

\item Show that every second countable space is also first countable.
\end{enumerate}

\item (dense/separable) A subspace $A$ of $X$ is called \emph{dense} if the
closure of $A$ is $X$. A topological space $X$ is called \emph{separable}, if
there exists a countable dense subset.

\begin{enumerate}
\item Show that $\mathbb{R}^{n}$ is separable.

\item \#Show that if $X$ is second countable, then $X$ is separable.
\end{enumerate}

\item (Hausdorff/T2)

\begin{enumerate}
\item Are points closed in every topology? Show that they may not be.

\item What assumptions can we make to ensure point sets are closed? A space is
\emph{Hausdorff} (or \emph{T2}) if for every two points $x,y\in X,$ there are
disjoint open sets $U$ and $V$ such that $x\in U,$ $y\in V$. Show that finite
point sets in Hausdorff spaces are closed.

\item Show that $\mathbb{R}$ is Hausdorff. Is $\mathbb{Q}$ Hausdorff? What
about a general metric space?

\item (line with two origins) Are there interesting examples of non-Hausdorff
spaces? The line with two origins is defined to be the quotient of
$\mathbb{R\times}\left\{  1\right\}  \cup\mathbb{R\times}\left\{  2\right\}  $
by the equivalence relation $\left(  a,1\right)  \sim\left(  a,2\right)  $ if
$a\neq0.$ Show that the line with two origins is not Hausdorff.

\item How can we construct Hausdorff spaces? Show that a subspace of a
Hausdorff space is Hausdorff but the quotient of a Hausdorff space may not be Hausdorff.

\item (Zariski topology) Consider the topology on ${\mathbb{R}}^{n}$ in which
the open sets are the empty set and the complements of the common zero levels
sets of finitely many polynomials. Show that this is indeed a topology on
$R^{n}$. This is called the \emph{Zariski topology}. Show also that the
Zariski topology is not Hausdorff.
\end{enumerate}

\item (T1) Consider a weaker assumption than Hausdorff. A topological space
$X$ is called a \emph{T1-space} (or a Tychonoff space) if for any two
different points $x,y\in X$ there exists an open set $U$ which contains $x$
but does not contain $y$.

\begin{enumerate}
\item Prove that a space $X$ is a T1-space if and only if any subset
consisting of a single point is closed.

\item Can a space be T1 but not Hausdorff? Let the group ${\mathbb{R}}$ act on
${\mathbb{R}}^{2}$ by
\[
t.(x,y)=(x,y+tx).
\]
Prove that the quotient space with the quotient topology is not Hausdorff, but
is the union of two disjoint Hausdorff subspaces. Also show that the quotient
space is a T1-space.
\end{enumerate}

\item Find a topological space $X$ and a sequence $x_{n}$ in $X$ which
converges but has more than one limit. What additional property on $X$ implies
that limits of sequences are unique? 

\item A function $f:{\mathbb{R}}\rightarrow{\mathbb{R}}$ is \emph{lower semi
continuous} if for all $x\in{\mathbb{R}},\epsilon>0$, there exists a
$\delta>0$ such that $|y-x|<\delta$ implies that $f(y)>f(x)-\epsilon$.

\begin{enumerate}
\item Show that the collection $\mathcal{B}=\{(\alpha,\infty)\,|\,\alpha
\in{\mathbb{R}}\}$ is a basis. Let $T^{\prime}$ denote the topology generated
by $\mathcal{B}$. Show that $T^{\prime}\subset T_{metric}$ and the containment
is strict. (Hint: One idea is to show that $T^{\prime}$ is not Hausdorff.)

\item Show that the collection $\mathcal{B}$ along with the empty set and all
of ${\mathbb{R}}$ is the topology generated by the base $\mathcal{B}$,
\emph{i.e.} $T^{\prime}=\mathcal{B}\cup\{\emptyset,{\mathbb{R}}\}$.

\item Show that $T^{\prime}$ is second countable.

\item Show that a function $f:({\mathbb{R}},T_{metric})\rightarrow
({\mathbb{R}},T^{\prime})$ is continuous, if and only if it is lower
semi-continuous by the earlier definition.

\item Show that a function $f:({\mathbb{R}},T^{\prime})\rightarrow
({\mathbb{R}},T_{metric})$ is continuous, if and only if it is a constant function.
\end{enumerate}

\item Let $X=\mathbb{N}\cup\{e\}$. Define a collection $\mathcal{T}$ by
$A\subseteq X$ is in $\mathcal{T}$ if and only if $A$ does not contain $e$
(this includes the empty set) or both $e\in A$ and $A^{c}$ is finite (this
includes $X$).

\begin{enumerate}
\item Show that $\mathcal{T}$ is a topology on $X$.

\item Show that $\mathcal{T}$ is second countable.

\item Show that $\mathbb{N}$ is dense in $(X,\mathcal{T})$.

\item Is $(X,\mathcal{T})$ compact? (see later section)

\item A function $f:\mathbb{N}\rightarrow{\mathbb{R}}$ is the same thing as a
sequence $x_{n}$. We will say that $g:X\rightarrow{\mathbb{R}}$ is a
continuous extension of $f$ if $g(n)=f(n)\,\forall n\in{\mathbb{N}}$. Show
that $f$ has a continuous extension iff $x_{n}=f(n)$ is a convergent sequence.
Further, the continuous extension is given by $g(e)=\lim_{n\rightarrow\infty
}f(n)$.

\item Every element $a=(l,a_{1},a_{2},a_{3},\ldots)\in{\mathbb{R}}%
\times{\mathbb{R}}^{N}$ defines a function $f_{a}:X\rightarrow{\mathbb{R}}$ by
$f_{a}(n)=a_{n},f_{a}(e)=l$. Let $Y\subset{\mathbb{R}}\times{\mathbb{R}}^{N}$
denote the set of all the convergent sequences with their associated limits,
\emph{i.e.} $(l,a_{1},a_{2},a_{3},\ldots)\in Y\implies a_{n}\rightarrow l$.
Find the weakest topology on $X$ such that for all $a\in Y$, $f_{a}%
:X\rightarrow{\mathbb{R}}$ is continuous.

\item Can you find a metric on $X$ such that the metric topology is identical
to the topology $\mathcal{T}$ above? Any topology with this property is said
to be \emph{metrizable}.
\end{enumerate}

\item What are some basic examples of connected and disconnected spaces?

\begin{enumerate}
\item Show $\left(  0,1\right)  $ is connected.

\item Show $\left(  0,2\right)  \setminus\left\{  1\right\}  $ is disconnected.

\item Show that ${\mathbb{R}}$ and ${\mathbb{R}}^{2}$ are not homeomorphic.
Hint: use the notion of a connected set.

\item Find all the different topologies, up to homeomorphism, on a 4-element
set, which make it a connected topological space.

\item Prove that the closure of a connected subspace is connected, but if the
closure of a space is connected the space may not be connected.
\end{enumerate}

\item Intermediate Value Theorem.

\begin{enumerate}
\item Prove that if $X$ is connected and $f:X\rightarrow Y$ is continuous then
$f\left(  X\right)  $ is connected.\newline

\item Prove the Intermediate Value Theorem.

\item Explain the proof of the Intermediate Value Theorem. What are the tricky parts?

\item Use the Intermediate Value Theorem to prove a special case of Brouwer's
Fixed Point Theorem: every continuous map $f:\left[  -1,1\right]
\rightarrow\left[  -1,1\right]  $ has a fixed point.
\end{enumerate}

\item (path connected). A \emph{path} in $X$ is a continuous map
$\gamma:\left[  a,b\right]  \rightarrow X.$ A space $X$ is \emph{path
connected} if any two points can be joined by a path.

\begin{enumerate}
\item Show that if $f:X\rightarrow Y$ is continuous and $X$ is path connected
then $f\left(  X\right)  $ is path connected.

\item Let $X$ be the union of the origin in ${\mathbb{R}}^{2}$ and the graph
of $\sin(1/x)$ on $(0,\infty)$. Show $X$ is connected but not path connected.

\item A space $X$ is called locally path-connected, if for each $x\in X$ and
every neighborhood $U$ of $x$, there exists a path-connected neighborhood $V$
of $x$ contained in $U$. Show that if $X$ is connected and locally
path-connected, then it is path-connected.

\item Show that every open subset of $\mathbb{R}^{n}$ is locally path connected.
\end{enumerate}

\item Are connected spaces and path connected spaces the same?

\begin{enumerate}
\item Show that if $X$ is path connected, then it is connected.

\item Let $K=\left\{  1/n:n=1,2,3,\ldots\right\}  .$ Define comb space to be
$C=\left(  [0,1]\times0\right)  \cup\left(  K\times\left[  0,1\right]
\right)  \cup\left(  0\times\left[  0,1\right]  \right)  $ with the subspace
topology ($C$ is a subset of $\left[  0,1\right]  \times\left[  0,1\right]
$). Define the deleted comb space to be $C^{\prime}=C\setminus0\times\left(
0,1\right)  .$ Show that the deleted comb space is connected but not path connected.
\end{enumerate}

\item Can we find interesting examples of path connected spaces%
%TCIMACRO{\TEXTsymbol{>}}%
%BeginExpansion
$>$%
%EndExpansion
?\ Recall the Cantor set $K$ is defined by
\[
K=\left[  0,1\right]  \setminus%
%TCIMACRO{\dbigcup \limits_{m=1}^{\infty}}%
%BeginExpansion
{\displaystyle\bigcup\limits_{m=1}^{\infty}}
%EndExpansion%
%TCIMACRO{\dbigcup _{k=0}^{3^{m-1}-1}}%
%BeginExpansion
{\displaystyle\bigcup_{k=0}^{3^{m-1}-1}}
%EndExpansion
\left(  \frac{3k+1}{3^{m}},\frac{3k+2}{3^{m}}\right)  .
\]
Show that the complement of $K\times K$ in the unit square $[0,1]\times
\lbrack0,1]$ is path-connected.

\item (totally disconnected) A space is \emph{totally disconnected} if the
only connected subsets are one-point sets. Show that the set of rational
numbers between $0$ and $1$ (i.e. $\mathbb{Q\cap}\left[  0,1\right]  $) with
the subspace topology is totally disconnected. Show that the Cantor set is
totally disconnected.

\item (connected component) Given $X$, we can define an equivalence relation
on $X$ by setting $x\sim y$ if there is a connected subset containing both $x$
and $y.$ The equivalence classes are called \emph{components} or
\emph{connected components} of $X.$

\begin{enumerate}
\item Prove that each connected component of a topological space $X$ is closed.

\item Show that if $A$ is a both open and closed, non-empty, connected subset
of a topological space $X$, then $A$ is a connected component.

\item Show that if a topological space has finitely many connected components,
then each of them is open and closed.
\end{enumerate}

\setcounter{myenumi}{\value{enumi}}
\end{enumerate}


\end{document}