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\title{Topology problems}
\maketitle

\section{Problems on topology}

\begin{enumerate}
\item 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 $c=\sqrt{2}.$

\item 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 *Consider the following definitions for subsets of $\mathbb{R}$. An open
set is 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.

\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.

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

\item Show that every open set is the union of intervals.
\end{enumerate}

\item *(closed) Let $\left(  X,\mathcal{T}\right)  $ be a topological space. A
set $F\subseteq X$ is \emph{closed} if $F^{C}=X\setminus F\in\mathcal{T}$,
i.e. if $F^{C}$ is open. Show that arbitrary intersections and finite unions
of closed sets are closed.

\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 $V\subseteq Y.$

\item **In this problem we confirm that this definition of continuous is the
same as that we know from calculus.

\begin{enumerate}
\item A subset $U\subseteq\mathbb{R}$ is open if for each $x\in U$ there
exists $\varepsilon>0$ such that $\left(  x-\varepsilon,x+\varepsilon\right)
\subseteq U.$ Show that these open sets form a topology.

\item If $U\subseteq\mathbb{R}$ is open, show that a function $f:U\rightarrow
\mathbb{R}$ is continuous 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 *(limit points and closure) 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 $A.$

\begin{enumerate}
\item Show that $\bar{A}$ is closed.

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

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

\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 $O\cap A=\{x\}$. Let $A\subseteq{\mathbb{R}}$, and $A^{\prime}$
denotes the set of all the accumulation points of $A$.

\begin{enumerate}
\item 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
\[
A^{\prime}=\bigcap_{x\in A}c{l}(A\backslash\{x\}).
\]


\item Using this, or otherwise, show that $A^{\prime}$ is a closed set.

\item Show that $c{l}(A)=A\cup A^{\prime}$.
\end{enumerate}

\item *(interior) The interior of $A,$ denoted $int(A)$ or $\overset{\circ}%
{A}$ is the union of all open sets contained in $A.$

\begin{enumerate}
\item Show $int(A)$ is open.

\item Show that if $B$ is an open set contained in $A$, then $B\subseteq
int(A)$.
\end{enumerate}

\item **(metric topology) 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 $B_{r}\left(
x\right)  =\left\{  y\in X:d\left(  x,y\right)  <r\right\}  .$ 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 $B_{r}\left(  x\right)  \subseteq U.$

\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 $d_{X}\left(  x,y\right)  <\delta.$
\end{enumerate}

\item (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}

Show that $d\left(  x,y\right)  =\left\Vert x-y\right\Vert $ defines a metric.

\item 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.
\end{enumerate}

\item Let $C((0,1))$ be the set of bounded continuous functions on $(0,1)$
with the norm $\left\Vert f\right\Vert =\sup_{x\in\left(  0,1\right)
}\left\vert f\left(  x\right)  \right\vert $ (convince yourself this is a
norm!) so
\[
d(f,g)=\sup_{0<x<1}|f(x)-g(x)|.
\]
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.

\item ($l^{p}$) Let $l^{p}$ be the set of sequences $(x_{n})_{n=1}^{\infty}$
with $\sum_{n=1}^{\infty}|x_{n}|^{p}<\infty$. For such a sequence we define
\[
||(x_{n})_{n=1}^{\infty}||_{p}=\left[  \sum_{n=1}^{\infty}|x_{n}|^{p}\right]
^{1/p}%
\]


\begin{enumerate}
\item Show this is a norm and hence defines a metric space.

\item Consider the two sets
\begin{align*}
F &  =\{(x_{n})_{n=1}^{\infty}\in l^{p}:x_{n}\geq0\,\forall n\}\\
U &  =\{(x_{n})_{n=1}^{\infty}\in l^{p}:x_{n}>0\,\forall n\}
\end{align*}
Is $F$ closed? Is $U$ open?
\end{enumerate}

\item In a metric space limit points and accumulation points are the same.

\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 **(finer/coarser) 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.
\end{enumerate}

\item **(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 **(Product topology 1) Let $X$ and $Y$ be topological spaces. We define
a topology on the product $X\times Y$ bay taking as a basis sets of the form
$U\times V$ where $U$ is open in $X$ and $V$ is open in $Y.$

\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 $\pi_{1}:X\times Y\rightarrow X$ and $\pi_{2}:X\times
Y\rightarrow Y$ are continuous.

\item If $Z$ is also a topological space, show that $\left(  X\times Y\right)
\times Z$ is homeomorphic to $X\times\left(  Y\times Z\right)  .$
\end{enumerate}

\item **(Product topology 2) 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.
\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 the 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 **(quotient topology) 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.
\end{enumerate}

\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.

\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.

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

\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 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) 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
$x\in U,$ $y\in V$.

\begin{enumerate}
\item Show that a subspace of a Hausdorff space is Hausdorff but the quotient
of a Hausdorff space may not be Hausdorff.

\item Show that finite point sets in Hausdorff spaces are closed.
\end{enumerate}

\item (line with two origins) 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 (T1) 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 $y$. Prove that a space $X$ is a
T1-space if and only if any subset consisting of a single point is closed.
Also find an example of a topological space which is a T1-space, but not
Hausdorff (a T2-space).

\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.

\item 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.

\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 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 *(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}
\end{enumerate}

\section{Problems on compactness}

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

\item Show the following spaces are compact:

\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\}  $.

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

\item Prove or disprove the following:

\begin{enumerate}
\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 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 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.

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

\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 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
$X.\backslash$
\end{enumerate}
\end{enumerate}

\item *Show that if $X$ is compact and $F\subset X$ is closed, then $F$ is
compact (in the subspace topology).

\item **Show Theorem \ref{Heine-Borel1}.

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

\item *Prove Corollary \ref{Heine-Borel2}.

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

\item **Prove the Extreme Value Theorem (Theorem \ref{Extreme Value Theorem}).

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


\begin{enumerate}
\item Prove that with this metric, the entire space of $\mathbb{R}$ is not
compact even though it is closed and bounded.

\item Can you characterize all the compact sets in $(\mathbb{R},d)$ ?
\end{enumerate}

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

\item (Hausdorff/T2) 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
$x\in U,$ $y\in V$.

\begin{enumerate}
\item Show that a subspace of a Hausdorff space is Hausdorff but the quotient
of a Hausdorff space may not be Hausdorff.

\item Show that finite point sets in Hausdorff spaces are closed.

\item Show that compact subsets of a Hausdorff space are closed.
\end{enumerate}

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

\item (sequential compactness) 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 Let ${\mathbb{R}}\mathbb{P}^{n}$ denote the quotient space of
${\mathbb{R}}^{n+1}\setminus\{0\}$, by the equivalence relation $x\sim y$ if
and only if there exists $\lambda\neq0$ such that $x=\lambda y$. Show that
${\mathbb{R}}\mathbb{P}^{n}$ is compact.

\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.

\item (local compactness/one point compactification) 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 $x\in
U\subseteq N.$)

\begin{enumerate}
\item If $X$ is a compact space, then show that $X$ is locally compact.

\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.)

\item There is a canonical way to add one point to a locally compact Hausdorff
space to get a compact space. Namely, if $X$ is locally compact Hausdorff, 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 $X$. Prove that $\bar{X}$ , called the \emph{one point
compactification} of $X$, is a compact Hausdorff space.

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

\section{Problems on connectedness}

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

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

\item *Prove Proposition \ref{prop: image of ctd}.

\item *Prove the Intermediate Value Theorem.

\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. Show that if $X$ is path
connected, then it is 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 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.

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

\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 $X.$ 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.

\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 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.
\end{enumerate}


\end{document}