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\begin{center}{\bf Homework 3, due Thursday, September 13, 2012}\end{center}
Let $X$ be a metric space with metric $d$. Let $S\subseteq X$. We say that
$S$ is \emph{bounded} if for some $r>0$ and $x\in X$ we have $S\subseteq D_X(x,r)$.
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If $X$ is a topological space and $S\subseteq X$, note that
$\operatorname{Cl}(S)=S$ if $S$ is closed.
I.e., the closure of a closed set is the set itself.
[We had two equivalent definitions of the closure $\operatorname{Cl}(S)$ of a set $S$.
You can use whichever one you prefer.
The first was that $\operatorname{Cl}(S)$ consists of all $x\in X$ such that
every open neighborhood of $x$ meets $S$. The second was that
$\operatorname{Cl}(S)$ is the intersection of all closed subsets of $X$ that contain $S$.
Here is a proof that $\operatorname{Cl}(S)=S$ if $S$ is closed, using definition 1:
if $x\in S$ then every open neighborhood $U$ of $x$
clearly meets $S$ (since $x$ is in both $U$ and $S$), and hence (by definition 1)
$x\in \operatorname{Cl}(S)$. Thus $S\subseteq \operatorname{Cl}(S)$.
And if $x\in \operatorname{Cl}(S)$, then every open neighborhood of $x$ meets
$S$ (by definition 1). If $x\not\in S$, then $x\in S^c$ and $S^c$ is open
(since $S$ is closed), so $S^c$ is an open neighborhood of $x$ that does not meet $S$, which is
a contradiction. Thus we must have $x\in S$, and hence $\operatorname{Cl}(S)\subseteq S$.
Now here is a proof that $\operatorname{Cl}(S)=S$ if $S$ is closed, using definition 2:
by definition 2, $\operatorname{Cl}(S)=\cap_{C\in A}C$, where $A$ is the set of all closed subsets of $X$ that
contain $S$, so $S\subseteq \operatorname{Cl}(S)$ since $\operatorname{Cl}(S)$ is an intersection of sets $C$
each of which contains $S$. And $\operatorname{Cl}(S)\subseteq S$ since one of the sets $C$
whose intersection is $\operatorname{Cl}(S)$ is $S$ itself, since by hypothesis $S$ is a closed subset of $X$
which contains $S$.]
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Do any 5 of the 6 problems.
Each problem is worth 20 points. Solutions will be graded for correctness, clarity and style.
\begin{itemize}
\item[(1)] Let $X$ be a topological space and let $S\subseteq X$.
Show that $\operatorname{Fr}(S)$ is closed.
(Recall that $\operatorname{Fr}(S)$ consists of every $x\in X$ such that
every open neighborhood of $x$ meets both $S$ and $S^c$.)
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\item[(2)]
Let $X$ be a topological space and let $S\subseteq X$.
Show that $\operatorname{Cl}(S)=\operatorname{Int}(S)\cup \operatorname{Fr}(S)$.
(Recall that $\operatorname{Int}(S)$ consists of every $x\in X$ such that
some open neighborhood of $x$ is contained in $S$.)
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\item[(3)] Let $X$ be a topological space and $S$ a connected subset.
Show that $\operatorname{Cl}(S)$ is also connected.
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\item[(4)] Let $X$ be a nonempty metric space with metric $d$. Let $S\subseteq X$.
Show that $S$ is bounded if and only if for every $y\in X$, there is an $r_y>0$ such
that $S\subseteq D_X(y,r_y)$.
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\item[(5)] Let $f:X\to Y$ be a continuous map of topological spaces.
Let $S\subseteq Y$. Show that
$\operatorname{Cl}(f^{-1}(S))\subseteq f^{-1}(\operatorname{Cl}(S))$.
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\item[(6)] Let $f:X\to Y$ be a continuous map of topological spaces.
Let $S\subseteq Y$. Show that $\operatorname{Cl}(f^{-1}(S))=f^{-1}(\operatorname{Cl}(S))$
need not hold (give a specific example).
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\end{itemize}
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