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\font\bigbf=cmbx12 at 24pt
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\def\endtf{\egroup\medskip}
\def\false{\centerline{TRUE\qquad{\color{blue} FALSE}}}
\def\true{\centerline{{\color{blue} TRUE}\qquad FALSE}}
\newcount\qno
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\def\intro#1{\medskip\noindent{\it #1.}\quad}
\def\proof{\intro{Proof}}
\def\endproof{\leavevmode\hskip\parfillskip\vrule height 6pt width 6pt
depth 0pt{\parfillskip0pt\medskip}}
\let\bu\bullet
{\bigbf\centerline{Holden Rohrer}\bigskip\centerline{MATH 4317\qquad
HOMEWORK \#7}}
\question{(3 points each) Answer True or False. No justification
needed.}
\item{a)} The arbitrary intersection of compact sets is compact.
\tf
\true
\endtf
\item{b)} The arbitrary union of compact sets is compact.
\tf
\false
\endtf
\item{c)} Let $A$ be arbitrary, and let $K$ be compact. Then, the
intersection $A\cap K$ is compact.
\tf
\false % open set \subset closed bounded set
\endtf
\item{d)} If $F_1\supset F_2\supset F_3\supset F_4\supset\cdots$ is a
nested sequence of nonempty closed sets, then the intersection $\cap_n
F_n$ is not empty.
\tf
\false
\endtf
\item{e)} The intersection of a perfect set and a compact set is
perfect.
\tf
\false
\endtf
\item{f)} The rationals contain a perfect set.
\tf
\true % empty set
\endtf
\question{(3.3.10) Let $K = [a,b]$ be a closed interval. Let ${\cal O} =
\{O_\lambda | \lambda\in \Lambda\}$ be an open cover for $[a,b].$ Show
that there is a finite subcover using this strategy: define $S$ to be
the set of all $x\in [a,b]$ such that $[a,x]$ has a finite subcover from
${\cal O}.$}
\item{a)} Argue that $S$ is nonempty and bounded, and thus $s = \sup S$
exists.
\proof
Let $x = a.$
We know that $[a,a] = \{a\}\subseteq [a,b],$ so there must exist, in
${\cal O},$ an $O_\lambda$ such that $a\in O_\lambda.$
Taking $\{O_\lambda\}$ as the subcover, we have shown that $S$ is
nonempty.
From the definition, $S\subseteq [a,b],$ so $|S| \leq \max\{|a|,|b|\},$
making $S$ bounded.
Therefore, $s = \sup S$ exists.
\endproof
\item{b)} Show that $s = b.$ Explain why this implies ${\cal O}$ admits
a finite subcover. That is, every closed interval is compact.
\proof
From the definition $S\subseteq [a,b],$ and for all $y\in[a,b],$ we know
$y \leq b,$ so $s\leq b.$
For the sake of contradiction, assume $s < b.$
From the open cover, we know there must be an open interval
$V_\epsilon(s)$ containing $s,$ with some size $\epsilon,$ and from the
definition of supremum, there must be $x\in [s-\epsilon,s)\cap S,$ so
by taking the finite cover of $x$ and appending $V_\epsilon(s),$ we
construct a finite cover which includes $y\in (s,s+\epsilon)\cap[a,b],$
and this is nonempty since $s < b,$ and $y > s,$ giving us a
contradiction of our assumption about the supremum.
This conclusively tells us that $s = b,$ so an open cover of $[a,b]$
must admit a finite subcover.
This is the definition of compactness.
\endproof
\item{c)} Use the fact that closed intervals are compact to conclude
that a set $F$ that is closed and bounded is compact.
\proof
Let ${\cal O}$ be an open cover for $F.$
We are given that $F$ is bounded by some $M,$ so we construct
$${\cal O}\cup\left\{\left(\bigcup_{\lambda\in\Lambda}
O_\lambda\right)^c\right\},$$
The appended set is an open set because the complement of a closed set
is an open set.
We thus have an open cover for $[-M,M]\supseteq F,$
and we have shown that $[-M, M]$ admits a finite subcover, so taking the
obtained finite subcover sans the appended set, we have a finite
subcover for $F.$
\endproof
\question{We proved: the countable intersection of compact nonempty
$K_1\supset K_2\supset \cdots$ is not empty. We did so using a
diagonalization and the limit definition of compact sets. Prove this
theorem using the open cover definition of a compact set. (Assume
otherwise, and construct an open cover of $K_1$ without finite
subcover.)}
\proof
Assume for the sake of contradiction that the intersection of these sets
is empty.
Since $K_i$ is closed (assuming we are working in a Hausdorff space),
$K_i^c$ is open, and $C = \{K_2^c,K_3^c,\ldots\}$ is an open cover,
since $\bigcup_{2\leq i} K_i^c = \left(\bigcap_{2\leq i} K_i\right)^c =
X,$ where $X$ is the space.
This set is therefore an open cover of $K_1,$ but any finite subcover
${\cal O}$ must omit, for some $N,$ all $K_n$ where $n>N.$
Let $x\in K_n.$
For $m\leq n,$ $K_n\subseteq K_m,$ so $x\in K_m,$ and $x\not\in K_m^c,$
so $x\not\in \bigcup_{o\in\cal O} o.$
But $x\in K_1,$ contradicting the existence of a finite subcover from
this open cover.
Therefore, $K_1$ is not compact, but that is a contradiction.
\endproof
\bye
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