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{Math 508 Fall 2014 \hfill Jerry Kazdan}
\centerline{\large \bf Compactness}
\vskip20pt
In these notes we will assume all sets are in a metric space $X$. These proofs are merely a rephrasing of this in Rudin -- but perhaps the differences in wording will help.
\smallskip
Intuitive remark: a set is compact if it can be guarded by a finite number of arbitrarily nearsighted policemen.
\bigskip
{\bf Theorem } \emph{A compact set $K$ is bounded}.
\smallskip
{\sc Proof } Pick any point $p\in K$ and let $B_n(p)=\{x\in K : d(x,p)< n\},\ n=1,2,\ldots$. These open balls cover $K$. By compactness, a finite number also cover $K$. The largest of these is a ball that contains $K$.
\bigskip
{\bf Theorem 2.34} \emph{A compact set $K$ is closed}.
\smallskip
{\sc Proof } We show that the complement $K^c=X-K$ is open. Pick a
point $p \not\in K$. If $q\in K$, let $V_q$ and $W_q$ be open balls
around $p$ and $q$ of radius $\frac{1}{2}d(p,q)$. Observe that if
$x\in W_q$ then
$$
d(q,p) \le d(q,x)+d(x,p) < \tfrac{1}{2}d(p,q) +d(x,p)
$$
so $d(x,p) >\tfrac{1}{2}d(p,q)$, that is, all the points in this ball
are at least $\tfrac{1}{2}d(p,q)$ from $p$ .
By compactness, a finite number of these balls, $W_{q_1},\ldots W_{q_N}$
cover $K$. Look at the corresponding balls $V_{q_1},\ldots V_{q_N}$.
They are all centered at $p$. The smallest (their intersection) is a
neighborhood of $p$ that contains no points of $K$.
\bigskip
{\bf Theorem 2.35} Closed subsets of compact sets are compact.
\smallskip
{\sc Proof } Say $F\subset K \subset X$ where $F$ is closed and $K$ is compact. Let $\{V_\alpha\}$ be an open cover of $F$. Then $F^c$ is a trivial open cover of $F^c$. Consequently $\{F^c\}\cup \{V_\alpha\}$ is an open cover of $K$. By compactness of $K$ it has a finite sub-cover -- which gives us a finite sub-cover of $F$.
\bigskip
{\bf Theorem 2.38} Let $I_n$ be a sequence of nested closed intervals in $\R$, so $I_n \supseteq I_{n+1},\,n=1,2,\ldots $. Then $\cap_{n=1}^\infty I_n$ is not empty.
\smallskip
{\sc Proof } Say $I_n=\{x\in\R : a_n\le x \le b_n\}$. The nested property means
$$
a_1\le a_n\le a_{n+1}\le b_{n+1}\le b_n\le b_1.
$$
Let $a=\sup a_n$ and $b=\inf b_n$. It is clear that
$\cap_{n=1}^\infty I_n= \{a\le x \le b\}$.
\bigskip
It is clear that this immediately extends to closed cells
(``rectangles'') in $\R^2$ and $\R^k$. We use it to show
\medskip
{\bf Theorem 2.40} Closed and bounded intervals $x\in \R :\{ a\le x
\le b\}$ are compact.
\smallskip
{\sc Proof } Idea: keep on dividing $a\le x \le b$ in half and use a microscope.
\smallskip
Say there is an open cover $\{G_\alpha\}$ that has no finite sub-cover. Divide the interval in half. Then one (or both) halves are closed sets with an open cover that has no finite cover. Keep on repeating this. At the $n^{\text{th}}$ step we have a closed interval $I_n$ of length $(b-a)/2^n$ where there is no finite sub-cover of our $\{G_\alpha\}$.
By the previous theorem, the intersection of these (nested) intervals
$\cap_{n=1}^\infty I_n$ has at point $p$. Since $p$ is contained in at least one of the
$\{G_\alpha\}$ so there is some interval around $p$. This shows that for $n$ large $I_n$ is covered by one of the sets $G_\alpha$. Contradiction.
\bigskip
{\bf Theorem 2.37} In any metric space, an infinite subset $E$ of a compact set $K$ has a limit point in $K$. [Bolzano-Weierstrass]
\smallskip
{\sc Proof } Say no point of $K$ is a limit point of $E$. Then each point of $K$ would have a neighborhood containing at most one point $q$ of $E$. A finite number of these neighborhoods cover $K$ -- so the set $E$ must be finite.
\bigskip
{\bf Theorem 2.41 } Let $\{E\in \R^k\}$. The following properties are equivalent:
(a) $E$ is closed and bounded.\\
(b) $E$ is compact.\\
(c)Every infinite subset of $E$ has a limit point in $E$. [Bolzano-Weierstrass Property]
\smallskip
{\sc Proof} We do this for sets $E\in\R^1$. The ore general case is then straightforward.
(a) implies (b): Since $E$ is bounded it is contained in some closed interval $I$. This interval is compact (Theorem 2.40). But then $E$ is a closed subset of a compact set so it is compact (Theorem 2.35).
\smallskip
(b) implies (c): Theorem 2.37.
\smallskip
(c) implies (a). If $E$ is not bounded, then for each $n=1,2,\ldots$ there is a point $x_n\in E$ with $\abs{x_n}>n$. This infinite set has no limit point, a contradiction.
\smallskip
If $E\subset \R$ is not closed then there is a point $p\in\R$ which is a limit point of $E$ but not in $E$. Thus, for each $n=1,2,3,\ldots$ there is a point $x_n\in E$ with $\abs{x_n-p}<1/n$. This set $S=\{x_1,\,x_2,\,\ldots\}$ has $p\not\in E$ as its only limit point. Contradiction.
\bigskip
{\sc Example } Let $K$ be a compact set in a metric space $X$ and let $p\in X$ but $p\not\in K$. Then there is a point $x_0$ in $K$ that is closest to $p$. In other words, let $\alpha = \inf_{x\in K} d(x,\,p)$. then there is at least one point $x_0\in K$ with $d(x_0,p)=\alpha$,
\smallskip
{\sc Remark: } There may be many such points, for example if $K$ is the unit circle $x^2 + y^2=1$ in the plane and $p=(0,0)$, then every point on the circle minimizes the distance to the origin.
\smallskip
{\sc Solution: } For any $n=1,2, \ldots$ there is at least one point $x_n\in K$ with $d(x_n,\,p)\le \alpha + \frac{1}{n}$. If this set $\{x_1, x_2,\ldots\}$ is finite (for instance if $K$ only has a finite number of points), pick the point closest to $p$. If the set has infinite many points, by the Bolzano-Weierstrass property it has a limit point $q$ in $K$. This is the desired point in $K$ that is closest to $p$.
\bigskip
{\sc Example } In $\ell_2$ the set of unit vectors $e_1=(1,0,0,\ldots),\, e_2=(0,1,0,0\ldots),\ \ldots$ is closed and bounded but not compact.
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