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\begin{center}
{\Large Intro to Topology -- HW 5.}
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\bigskip\noindent
\begin{enumerate}




\item Show that in a Hausdorff space $(X, \tau_X)$\newline
limits are unique. That is, if for a sequence of points $x_n\in X$, $n=1,2...$
we have $x_n\to a$ and $x_n\to b$, then $a=b$.


\item 
a.) In real analysis one often finds points of a closure using
convergence, since in $\R^n$ we have $x\in cl(A)$ if and only if
$\exists$ a sequence $x_n\to x$ with $x_n\in A$.
 Verify that the same is true in any metric space.

{\it (Hint: Think through the $\R^n$ case and generalize for any metric.
Recall that $x\in cl(A)$ if and only if $\forall U$ open sets s.t. $x\in U$ we have $U\cap A\not=\emptyset$.)}

b.) However, in general this  is not true! 
That is, $cl(A)$ and the set of limit points of $A$, for $A\subset X$ in a topological space $(X,\tau_X)$,  may be different.

Explain why by working out the following example:

let $X=\R$, and let $\tau_X=$ 
{\it the co-countable topology} on $X$, where the open sets are the $\emptyset$, $X$ and those sets whose complement is countable (possibly countably infinite).
Take $A=\R\setminus \{p\}$ for some $p\in\R$. 

What is $cl(A)$ and why?

    What is the set of limit points of $A$ and why? (A point $x\in X$ is a limit point if $\exists (x_n)$ sequence in $X$
 such that $x_n\rightarrow x$.)
 
     Conclude that the set of limit points of $A$ is not the same as $cl(A)$. 

\item Let $(X,d)$ be a metric space and $\tau_d=$the topology generated by $d$.

Prove that if $K\subset X$ is compact then
$K$ is closed.

Here is how you can start: 
if a set in a metric space $(X,d)$ is not closed, then it cannot be compact, because it has an open cover with no finite subcover. 

Specifically,
if a set $K$ is not closed, then $K$ is a proper subset of its closure, so $\exists x\in cl(K)$ such that $x\not\in K$. Then consider 
$$ A_n=\{y\in X\,|\, d(x,y)>{1\over n}\}\ \ \forall n\in\N
$$

Prove that the collection $\{A_n\}$ is a  cover of $K$ that has no finite subcover.

Also, $A_n$ must be open $\forall n\in\N$ - why?

Thus $K$ is not compact.


\item[EXTRA CREDIT]  
Let $X$ have the $\tau_X$=co-finite topology.

Prove that a sequence $(x_n)$ $n\in\N$ in  this space is divergent if and only if it has at least two eventually constant subsequences.

Prove that 
\end{enumerate}

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