(sec:ms:compactness)=
# Compactness
The material in this section is primarily based on
{cite}`aliprantis1998principles,gopal2020introduction`.
```{div}
Recall from {prf:ref}`def-st-cover` that for a subset $A \subseteq X$,
a cover is a family $\{ A_i \}_{i \in I}$ of subsets of $X$ such that
$$
A \subseteq \bigcup_{i \in I} A_i.
$$
```
## Open Covers
```{index} Metric space; open cover
```
```{prf:definition} Open cover
:label: def-ms-open-cover
A family of open subsets $\{ A_i \}_{i \in I}$ of subsets of $(X,d)$
is an *open cover* of $A$ if it covers $A$.
```
```{prf:theorem} LindelĂ¶f
:label: res-ms-lindelof
Every open cover of a subset of $\RR^m$ can be reduced to
an at-most countable subcover.
```
```{prf:proof}
We call a point $a = (a_1, \dots, a_m) \in \RR^m$ a rational point
if every component of $a$ is a rational number.
1. Let $A$ be a subset of $\RR^m$.
1. Let $\{ \OOO_i \}_{i \in I}$ be an open cover of $A$ (possibly uncountable).
1. Thus, $A \subseteq \bigcup_{i \in I} \OOO_i$.
1. For each $x \in A$,
1. Choose an index $i_x \in I$ such that $x \in \OOO_{i_x}$.
1. Pick a rational point $a_x \in \RR^m$ and a rational positive number $r_x$
such that $x \in B(a_x, r_x) \subseteq \OOO_{i_x}$.
1. Consider the collection $C = \{ B(a_x, r_x) \ST x \in A \}$.
1. Since the set of rational points is countable and the set of rational numbers is countable,
hence the set of all open balls centered at rational points with rational radii is countable.
1. Hence, $C$ which is a subset of open balls with rational points as centers and
rational radii, is at most countable.
1. Thus, $C$ is an at most countable open cover of $A$.
1. Since each $B(a_x, r_x)$ is a subset of an $\OOO_{i_x}$, hence
there exists an at most countable subcover of $\{ \OOO_i\}_{i \in I}$ for $A$.
```
## Compact Sets
```{index} Compact set
```
```{prf:definition} Compact set
:label: def-ms-compact-set
Let $(X, d)$ be a metric space. A subset $A$ of $X$ is called
*compact* if every open cover of $A$ can be reduced to a
finite subcover.
```
```{index} Compact metric space
```
```{prf:definition} Compact metric space
:label: def-ms-compact-metric-space
Let $(X,d)$ be a metric space. If $X$ is itself a
compact set, then $(X,d)$ is called a *compact metric space*.
```
```{prf:example} $(0,1)$ is not compact
:label: ex-0-1-not-compact
The set $(0,1)$ is not compact in $\RR$.
We first show this the hard way by picking an
open cover for $(0,1)$ which cannot be
reduced to a finite subcover.
Later, we discuss how to verify compactness
through easy checks.
1. Consider the family of open intervals:
$$
C = \left \{ \left(\frac{1}{n}, 1 \right ) \ST n \geq 2 \right \}.
$$
1. For every $x \in (0,1)$, there is a natural number $n$ such that $x > \frac{1}{n}$.
1. Thus, $x \in \left(\frac{1}{n}, 1 \right )$.
1. Thus,
$$
(0, 1) \subseteq \bigcup_{n=2}^{\infty} \left(\frac{1}{n}, 1 \right )
$$
implying that $C$ is an open cover of $(0,1)$.
1. At the same time for every $n \geq 2$, we have:
$$
\left(\frac{1}{n}, 1 \right ) \subseteq (0, 1)
$$
since $\frac{1}{n} > 0$.
1. Thus,
$$
(0, 1) = \bigcup_{n=2}^{\infty} \left(\frac{1}{n}, 1 \right ).
$$
1. But there is no finite subcover of $(0,1)$ in $C$.
1. If there was a finite subcover, we could pick a maximum $n$ among those intervals.
1. But then $x = \frac{1}{n}$ won't belong to any of those intervals in the finite subcover.
1. Hence $(0,1)$ is not compact.
We later show in {prf:ref}`res-ms-compact-is-closed-bounded`
that every compact set is closed and bounded.
Hence, an easy way to say that $(0,1)$ is not
compact is by noticing that it is not closed.
```
## Characterization of Compactness
We have defined compactness as a property where every open
cover can be reduced to a finite subcover.
The characterization of a property involves identifying
other properties which are equivalent in the sense that
property A $\iff$ property B. If one is true then
the other must be true and vice versa.
We will see later that compactness of a set $A$ is equivalent
to the property that every sequence of $A$ has a subsequence
that converges to a point in $A$. However, before we go there,
let us examine some implications of this
property that every sequence of a set $A$ has a subsequence that
converges within the set $A$. These results will be
useful later in the characterization of compact sets.
```{prf:lemma} Lebesgue number
:label: res-ms-open-cover-lebesgue-number
Let $A \subseteq \bigcup_{i \in I} \OOO_i$ be an open cover of $A$.
If every sequence in $A$ has a subsequence which converges to a point of $A$,
then, there exists a number $\delta > 0$ such that for each $x \in A$, we have
$B(x, \delta) \subseteq \OOO_i$ for at least one $i \in I$.
Any such number $\delta > 0$ is called a *Lebesgue number* of $A$
for the open cover $\{\OOO_i \}_{i \in I}$.
```
```{prf:proof}
Assume the claim is false.
1. Then, for each $\delta > 0$, there exists some $x \in A$ such that
$B (x, \delta)$ is not a subset of any $\OOO_i$.
1. In particular, for each $n$, there exists some $x_n \in A$ such that
$B (x_n, \frac{1}{n}) \cap (X \setminus \OOO_i) \neq \EmptySet$ holds
for each $i \in I$.
1. Consider the sequence $\{ x_n \}$.
1. By the hypothesis, every
sequence has a subsequence that converges to a point of $A$.
1. Let $x \in A$ be the limit of such a subsequence of $\{ x_n \}$.
1. Since $x \in A$, $x \in \OOO_i$ for at least one $i \in I$.
1. Pick some $i \in I$ such that $x \in \OOO_i$.
1. Choose some $r > 0$ such that $x \in B(x, r) \subseteq \OOO_i$.
1. Recall that $\OOO_i$ is open and $x$ is its interior point. So such an $r > 0$
can be chosen.
1. Now, select some $n$ such that $\frac{1}{n} < \frac{r}{2}$ and $d(x, x_n) < \frac{r}{2}$.
1. It follows that $B(x_n, \frac{1}{n}) \subseteq B(x, r) \subseteq \OOO_i$.
1. But this contradicts with the selection of $x_n$ such that $B (x_n, \frac{1}{n})$ is
not contained in any $\OOO_i$.
1. Thus, a $\delta > 0$ must exist satisfying the condition that for each $x \in A$,
$B(x, \delta) \subseteq \OOO_i$ for at least one $i \in I$.
```
```{prf:lemma} Existence of finite cover of open balls
:label: res-ms-compact-set-existence-of-finite-cover
Let $A \subseteq X$ of a metric space $(X,d)$.
If every sequence in $A$ has a subsequence which converges to a point of $A$,
then,
for every $r > 0$ there exist $x_1, \dots, x_n \in A$ such that
$$
A \subseteq \bigcup_{j=1}^n B(x_j, r).
$$
```
This lemma simply claims that we can construct a finite open cover
of open balls for $A$ for every $r > 0$.
```{prf:proof}
Assume the claim to be false. Choose $r > 0$ such that it is not possible to
select a finite number of points from $A$ such that
$$
A \subseteq \bigcup_{j=1}^n B(x_j, r).
$$
1. Pick some $x_1 \in A$.
1. Choose some $x_2 \in A \setminus B(x_1, r)$. We can choose $x_2$ since
$A$ doesn't have a finite open ball cover.
1. Assuming that $x_1, x_2, \dots, x_n$ have been chosen inductively,
choose $x_{n+1}$ from the set $A \setminus \bigcup_{i=1}^n B(x_i, r)$.
1. By design, $d(x_n, x_m) \geq r$ holds for every $n \neq m$.
1. Then, no subsequence of $\{ x_n \}$ can converge.
1. This contradicts with the hypothesis that every sequence has a subsequence
that converges in $A$.
1. Hence, the claim must be true.
```
```{prf:theorem} Characterization of compactness
:label: def-ms-compact-characterization
Let $A$ be a subset of a metric space $(X,d)$.
The following statements are equivalent.
1. $A$ is compact.
1. Every infinite subset of $A$ has an accumulation point in $A$.
1. Every sequence in $A$ has a subsequence which converges to a point of $A$.
```
```{prf:proof}
(1) $\implies$ (2)
1. Let $S$ be an infinite subset of $A$.
1. Assume, by way of contradiction, that $S$ has no accumulation point in $A$.
1. Then for every $x \in A$, there exists a deleted neighborhood of $x$
that is disjoint with $S$.
1. That is, $\forall x \in A, \exists r_x > 0 \ST B (x, r_x) \cap ( S \setminus \{ x \}) = \EmptySet$.
1. Then, $B(x, r_x) \cap S$ can either be empty or it can at most contain $x$.
1. Thus, $B(x, r_x) \cap S \subseteq \{ x\}$.
1. Consider the open cover $C = \bigcup_{x \in A} B(x, r_x)$.
1. It is easy to see that $A \subseteq C$.
1. Since $A$ is compact (by hypothesis), there exists a finite set
$\{ x_1, \dots, x_n \} \subseteq A$ such that
$A \subseteq \bigcup_{i=1}^n B(x_i, r_{x_i})$.
1. But then
$$
S = A \cap S = \bigcup_{i=1}^n [B(x_i, r_{x_i}) \cap S] \subseteq \{ x_1, \dots, x_n \}.
$$
1. Thus, $S$ must be a finite set containing at most $n$ elements.
1. We arrive at a contradiction as $S$ is infinite.
1. Hence, $S$ has an accumulation point in $A$.
(2) $\implies$ (3)
1. We assume that every infinite subset of $A$ has an accumulation point in $A$.
1. Let $\{x_n \}$ be an arbitrary sequence of $A$.
1. If $\{x_n \}$ is constant, then every subsequence is constant and convergent.
1. If $\{x_n \}$ takes a finite number of distinct values, then there is at least
one value which must occur infinite times. Thus, at least one subsequence is
a constant sequence and hence convergent.
1. We are left with the case where $\{ x_n \}$ contains infinite distinct values.
1. Then, we can choose a subsequence $\{ y_n \}$ of $\{ x_n \}$ which consists of
all distinct values; i.e., $y_n \neq y_m$ whenever $n \neq m$.
1. Let $k_1 = 1$.
1. Assuming $k_1, \dots, k_n$ have been selected,
choose $k_{n+1} > k_n$ such that $x_{k_{n+1}} \neq x_{k_i}$ for $1 \leq i \leq n$.
1. This is possible since $\{x_n \}$ has infinite distinct values and
so far only $n$ distinct values have been chosen.
1. Now, let $y_n = x_{k_n}$.
1. It is clear that $\{y_n\}$ is a subsequence of $\{x_n\}$ consisting of
all distinct values.
1. Consider the set $Y = \{ y_1, y_2, \dots \}$ (not the sequence but the set).
1. By our hypothesis (2), since $Y$ is an infinite subset of $A$, hence it must
have an accumulation point in $A$.
1. Let $x$ be an accumulation point of $Y$.
1. Assume that $x \neq y_n$ for each $n$. If $x = y_k$ for some $k$, then drop the
first $k$ elements of $\{ y_n \}$.
1. This ensures that $d(x, y_n) > 0$ for every $n$.
1. We will now select a subsequence of $\{ y_n \}$ that converges to $x$.
1. Towards this, choose $m_1$ such that $d(y_{m_1}, x) < 1$.
1. Now, inductively, assuming that $m_1 < m_2, \dots < m_n$ have been chosen,
choose $m_{n+1}$ such that
$$
d(y_{m_{n+1}}, x) < r_{n+1} = \min \{\frac{1}{n+1}, d(y_1, x), d(y_2, x), \dots, d(y_{m_n}, x) \}.
$$
1. Since $d(x, y_n) > 0$, hence $r_{n+1} > 0$.
1. Since $x$ is an accumulation point of $Y$, hence for every $r > 0$, there exists a
point $y \in Y$ such that $d(x, y) < r$.
1. Thus, it is possible to pick a suitable $m_{n+1}$ such that $d(y_{m_{n+1}}, x) < r_{n+1}$.
1. Also, by design, $m_{n+1}$ must be greater than $m_n$
as $r_{n+1} < d(y_i, x) \Forall 1 \leq i \leq m_n$.
1. Thus, $\{ y_{m_n} \}$ is a subsequence of $\{ y_n \}$.
1. Hence, $\{y_{m_n}\}$ is a subsequence of $\{ x_n \}$ too.
1. Since $d(x, y_{m_n}) < \frac{1}{n}$, it follows that $\lim y_{m_n} = x$.
1. Thus, $\{ x_n \}$ has a convergent subsequence which converges to a point of $A$.
(3) $\implies$ (1)
1. Let $A \subseteq \bigcup_{i \in I} \OOO_i$ be an arbitrary open cover of $A$.
1. We can pick a Lebesgue number $\delta > 0$ such that for each $x \in A$,
we have $B(x, \delta) \subseteq \OOO_i$ for at least one $i \in I$
thanks to {prf:ref}`res-ms-open-cover-lebesgue-number`.
1. Now, thanks to {prf:ref}`res-ms-compact-set-existence-of-finite-cover`, we
can pick $x_1, \dots, x_n \in A$ such that:
$$
A \subseteq \bigcup_{j=1}^n B(x_j, \delta).
$$
1. Now, for each $j$ pick some $i_j \in I$ such that $B(x_j, \delta) \subseteq \OOO_{i_j}$.
1. Then,
$$
A \subseteq \bigcup_{j=1}^n B(x_j, \delta) \subseteq \bigcup_{j=1}^n \OOO_{i_j}.
$$
1. Thus, the open cover of $A$ has a finite subcover.
1. Thus, $A$ is compact.
```
```{index} Bolzano-Weierstrass property
```
```{prf:definition} Bolzano-Weierstrass property
:label: def-ms-bolzano-weierstrass-property
A set $A$ in a metric space has the *Bolzano-Weierstrass property* if
every sequence in $A$ has a convergent subsequence that converges
to a point in $A$.
```
```{prf:observation}
:label: res-ms-compact-bolzano-weierstrass-prop
A compact set has the Bolzano-Weierstrass property.
```
## Closedness and Boundedness
```{prf:theorem} Compact sets are closed and bounded
:label: res-ms-compact-is-closed-bounded
A compact set is closed and bounded.
```
```{prf:proof}
Assume $A$ to be compact. We shall show that $A$ must be bounded.
1. Choose an open cover for $A$ as $A \subseteq \bigcup_{x \in A}B(x, 1)$.
1. Since $A$ is compact, hence there exist finite set of points $\{ x_1, \dots x_n \} \subseteq A$
such that $A \subseteq \bigcup_{i=1}^n B(x_i, 1)$.
1. Let $M = \max \{ d(x_i, x_j) \ST 1 \leq i, j \leq n\}$.
1. For any $x, y \in A$, choose $i, j$ such that $x \in B(x_i, 1)$ and $y \in B(x_j, 1)$.
1. Then, by triangle inequality, we have:
$$
d(x, y) \leq d(x, x_i) + d(x_i, x_j) + d(x_j, y) < M + 2 < \infty.
$$
1. Thus, $\diam A = \sup d(x,y)$ is finite and $A$ is bounded.
We now show that if $A$ is compact then $A$ must be closed too.
Towards this, we show that $A$ contains all its closure points.
1. Let $x \in \closure A$.
1. Due to {prf:ref}`res-ms-closure-point-as-limit`, there exists
a sequence $\{ x_n \}$ of $A$ with $\lim x_n = x$.
1. Due to {prf:ref}`def-ms-compact-characterization`, $\{x_n\}$ has
a subsequence that converges to a point of $A$ (since $A$ is compact by hypothesis).
1. As per {prf:ref}`res-ms-subsequence-convergence`, subsequences of
a convergent sequence converge to the same limit.
1. Thus, $x$ must be in $A$.
1. Thus, $\closure A \subseteq A$. Thus, $\closure A = A$. Thus, $A$ is closed.
```
Although every compact set is closed and bounded, the converse need
not be true. See {prf:ref}`ex-discrete-space-closed-bounded-not-compact`
for an example of closed and bounded set (in discrete space)
which is not compact.
In fact, discrete space is a
{prf:ref}`complete ` metric space.
Yet, it has closed and bounded sets which are not compact.
In the specific case of Euclidean spaces, all closed and
bounded sets are compact too.
See {prf:ref}`Heine-Borel theorem ` below.
## Nested Sequence of Compact Sets
```{prf:theorem} Nonempty nested intersection property
:label: res-ms-nested-seq-compact-nonempty-intersection
Let $\{ A_n \}$ be a nested sequence of nonempty compact subsets
of $(X, d)$ such that
$$
S_1 \supseteq S_2 \supseteq S_3 \supseteq \dots
$$
Then their intersection is nonempty; i.e.,
$$
\bigcap_{n=1}^{\infty} S_n \neq \EmptySet.
$$
```
```{prf:proof}
We prove this result by way of contradiction.
1. Assume for contradiction that $\bigcap_{n=1}^{\infty} S_n = \EmptySet$.
1. Let $V_n = X \setminus S_n$ for every $n \in \Nat$.
1. Since $S_{n +1 } \subseteq S_n$ for every $n$
hence $V_n \subseteq V_{n+1}$ for every $n$.
1. Then $V = \{ V_n \ST n \in \Nat \}$ is a collection of open sets.
1. Since $\bigcap_{n=1}^{\infty} S_n = \EmptySet$, hence
$$
X &= X \setminus \EmptySet \\
&= X \setminus \left ( \bigcap_{n=1}^{\infty} S_n \right ) \\
&= \bigcup_{n=1}^{\infty}(X \setminus S_n) \\
&= \bigcup_{n=1}^{\infty} V_n.
$$
1. Hence $V$ is an open cover for every $S_n$
as $S_n \subseteq X = \bigcup_{n=1}^{\infty} V_n$.
1. In particular, $V$ is an open cover for $S_1$.
1. Since $S_1$ is compact, hence there exists a
finite subcover of $S_1$ indexed by a finite
set of natural numbers $I$.
1. Let $I = \{ i_1, i_2, \dots, i_k \}$ and the
finite subcover be $F_k = \{ V_i \in V \ST i \in I \}$.
1. Since $I$ is a finite set, it has a maximum element.
1. Let $m = \max I = \max \{ i_1, i_2, \dots, i_k \}$.
1. Then $F_m = \{ V_1, \dots, V_m \}$ is also a finite
subcover of $S_1$ since $F_k \subseteq F_m$.
1. But then $\bigcup_{k=1}^m V_k = V_m$
since $V_n \subseteq V_{n+1}$ for every $n$.
1. Hence $S_1 \subseteq V_m$.
1. Then $S_m \subseteq S_1 \subseteq V_m$.
1. But $V_m = X \setminus S_m$, hence $V_m \cap S_m = \EmptySet$
a contradiction.
1. Hence, the intersection of the nested compact sets must be nonempty.
```
## Continuity
```{prf:theorem} Continuous images of compact sets are compact
:label: res-ms-compact-continuous-map
Let $f: (X, d) \to (Y, \rho)$ be a continuous function.
Let $A$ be a compact subset of $X$ with $A \subseteq \dom f$.
Then $f(A)$ is a compact subset of $Y$.
```
```{prf:proof}
We prove this by showing that any open cover of $f(A)$
can be reduced to a finite subcover.
1. Let $f(A) \subseteq \bigcup_{i \in I} \OOO_i$ be an open cover for $f(A)$.
1. Then $A \subset \bigcup_{i \in I} f^{-1} (\OOO_i)$.
1. Since $f$ is continuous,
hence $f^{-1} (\OOO_i)$ is an open subset of $X$ for every $i \in I$.
1. Since $A$ is compact, there exist indices $i_1, \dots, i_n$ such that
$A \subseteq \bigcup_{j=1}^n f^{-1}(\OOO_{i_j})$.
1. Then
$$
f(A) \subseteq f\left ( \bigcup_{j=1}^n f^{-1}(\OOO_{i_j}) \right )
= \bigcup_{j=1}^n f(f^{-1}(\OOO_{i_j}))
\subseteq \bigcup_{j=1}^n \OOO_{i_j}.
$$
1. Thus, $A$ is compact.
```
## Lipschitz Continuity
```{prf:theorem}
:label: res-ms-compact-llc-lipschitz
Let $f : (X,d) \to (Y, \rho)$ be a (partial) function with
$S = \dom f$.
Assume that $f$ is locally Lipschitz continuous
at every $x \in S$.
Let $A \subseteq S$ be a compact subset of $S$.
Then, $f$ is Lipschitz function on $A$.
In other words, there exists a constant $L > 0$
such that
$$
\rho(f(x),f(y)) \leq L d(x, y)
$$
for every $x, y \in A$.
```
```{prf:proof}
We proceed as follows.
1. Since $f$ is locally Lipschitz continuous
on $S$ hence $f$ is continuous
by {prf:ref}`res-ms-lipschitz-to-uniform-cont`.
1. Thus, by {prf:ref}`res-ms-compact-continuous-map`,
$f(A)$ is compact.
1. Hence, $f(A)$ is closed and bounded.
1. Thus, there exists $M > 0$ such that
for any $x, y \in A$,
$\rho(f(x), f(y)) \leq M$.
1. For contradiction, assume that $f$ is
not Lipschitz on $A$.
1. Then, there is no $L > 0$ such that
$$
\rho(f(x), f(y)) \leq L d(x,y) \Forall x, y \in A.
$$
1. Then, there exist two sequences $\{ x_n \}$
and $\{ y_n \}$ of $A$ such that
$$
\lim_{n \to \infty} \frac{\rho(f(x_n), f(y_n)) }{d(x_n,y_n)} = \infty.
$$
1. But $\rho(f(x_n), f(y_n)) \leq M$.
1. Thus,
$$
\lim d(x_n,y_n) = 0.
$$
1. Since $A$ is compact, hence $\{ x_n \}$
has a convergent subsequence.
1. Let $\{x_{n_k} \}$ be a convergent subsequence
with $x = \lim_{k \to \infty} x_{n_k}$.
1. By compactness of $A$, $x \in A$.
1. Then, $f$ cannot be not locally Lipschitz continuous at $x$.
1. This contradicts our hypothesis.
1. Thus, $f$ must be Lipschitz continuous on $A$.
```
## Homeomorphism
```{prf:theorem} Homeomorphism preserves compactness
:label: res-ms-compact-homeomorphism-pres
Let $f: (X, d) \to (Y, \rho)$ be a homeomorphism.
Then, $A$ is a compact subset of $(X,d)$ is and only if
$f(A)$ is a compact subset of $(Y, \rho)$.
```
```{prf:proof}
Let $A \subseteq X$ be compact.
Then, $f(A)$ is compact since $f$ is continuous
due to {prf:ref}`res-ms-compact-continuous-map`.
Let $f(A)$ be compact.
Since $f$ is a homeomorphism, hence $f^{-1}$ is continuous
and bijective.
Hence, $f^{-1}(f(A)) = A$ is compact
due to {prf:ref}`res-ms-compact-continuous-map`.
```
## Compact Spaces
```{prf:theorem}
:label: res-ms-compact-closed-subset
Every closed subset of a compact space is compact.
```
```{prf:proof}
Let $(X, d)$ be a compact metric space and let $A$ be a closed subset of $X$.
1. Let $\{\OOO_i\}_{i \in I}$ be an open cover of $A$.
We have, $A \subseteq \bigcup_{i \in I}\OOO_i$.
1. $X = A \cup (X \setminus A)$.
1. Then, $X \subseteq (X \setminus A) \cup \bigcup_{i \in I} \OOO_i$.
1. Since all $\OOO_i$ are subsets of $X$, hence we can write it as:
$X = (X \setminus A) \cup \bigcup_{i \in I} \OOO_i$.
1. Since $A$ is closed, hence $X \setminus A$ is open.
1. Thus, $(X \setminus A) \cup \bigcup_{i \in I} \OOO_i$ is an open cover of $X$.
1. But $X$ is compact. Hence, there exist finite indices $i_1, \dots, i_n$ such that
$X = (X \setminus A) \cup \OOO_{i_1} \cup \dots \cup \OOO_{i_n}$.
1. But then $A \subseteq X$ and $A \cap (X \setminus A) = \EmptySet$ imply that:
$A \subseteq \OOO_{i_1} \cup \dots \cup \OOO_{i_n}$.
1. Thus, $A$ is compact.
```
```{prf:theorem} Continuous maps are closed
:label: res-ms-compact-continuous-closed-map
Let $(X, d)$ be a compact metric space and suppose that
$f : (X, d) \to (Y, \rho)$ is a (total) continuous function.
Then $f$ is a {prf:ref}`closed mapping `.
If $f$ is bijective, then $f$ is a
{prf:ref}`homeomorphism `.
```
```{prf:proof}
Let $C$ be a closed subset of $X$.
1. Due to {prf:ref}`res-ms-compact-closed-subset`, $C$ is compact.
1. Since $f$ is continuous, hence, due to {prf:Ref}`res-ms-compact-continuous-map`,
$f(A)$ is compact.
1. As per {prf:ref}`res-ms-compact-is-closed-bounded`,
since $f(A)$ is compact, hence $f(A)$ is closed.
1. Thus, $f$ maps every closed set to a closed set.
1. Thus, $f$ is a closed mapping.
Now, assume that $f$ is bijective too.
1. Thus, $f^{-1}$ exists. Let $g = f^{-1}$.
1. Then, due to bijection property $g^{-1}(A) = f(A)$ holds for every subset $A$ of $X$.
1. Thus, $g^{-1}(A) = f(A)$ is a closed subset of $Y$ whenever $A$ is a closed subset of $X$.
1. Thus, as per {prf:ref}`res-ms-continuous-function-characterization` $[(5) \implies (1)]$, $g$ is continuous.
1. Thus, both $f$ and $f^{-1} = g$ are continuous.
1. Thus, $f$ is a homeomorphism.
```
```{prf:theorem} Compact domain + Continuity = Uniform continuity
:label: res-ms-compact-dom-continuity-uniform
Let $f: (X, d) \to (Y, \rho)$ be {prf:ref}`continuous ` on $X$.
If $X$ is compact, then $f$ is
{prf:ref}`uniformly continuous `.
```
```{prf:proof}
We proceed as follows.
1. Let $\epsilon > 0$.
1. Since $f$ is continuous on $X$, hence for every $x \in X$,
there exists $r_x > 0$ such that
$\rho(f(y), f(x)) < \epsilon$ holds whenever $d(x, y) < 2 r_x$.
1. The collection of open balls $B(x, r_x)$ covers $X$;
i.e., $X = \bigcup_{x \in X}B(x, r_x)$.
1. Since $X$ is compact, there exists a set of finite number of points $x_1, \dots, x_n$
such that $X = \bigcup_{i=1}^n B(x_i, r_{x_i})$.
1. Now, let $\delta = \min \{r_{x_1}, \dots, r_{x_n} \}$.
1. Since $\delta$ is the minimum of a finite number of positive numbers, hence $\delta > 0$.
1. Now, pick any $x, y \in X$ that satisfy $d(x, y) < \delta$.
1. There exists an integer $i$ such that $d(x, x_i) < r_{x_i}$
(due to the finite open cover).
1. Therefore, $\rho(f(x), f(x_i)) < \epsilon$.
1. Now, by triangle inequality:
$$
d(y, x_i) \leq d(y, x) + d(x, x_i) < \delta + r_{x_i} \leq 2 r_{x_i}
$$
holds true.
1. Thus, $\rho(f(x_i), f(y)) < \epsilon$, since $d(y, x_i) < 2 r_{x_i}$.
1. Thus,
$$
\rho(f(x), f(y)) \leq \rho(f(x), f(x_i)) + \rho(f(x_i), f(y)) < \epsilon + \epsilon = 2\epsilon.
$$
1. Thus, $f$ is uniformly continuous.
```
## Euclidean Spaces
Recall that $\RR^m$ are called Euclidean spaces with the standard metric:
$$
d (x, y) \triangleq \left ( \sum_{i=1}^m |x_i - y_i|^2 \right )^{\frac{1}{2}}.
$$
By $0 \in \RR^m$ we shall mean the vector $(0, \dots, 0)$.
The compact subsets of a Euclidean space are precisely those
sets which are closed and bounded.
```{prf:theorem} Heine-Borel theorem
:label: res-ms-heine-borel-euclidean
A subset of a Euclidean space is compact if and
only if it is closed and bounded.
```
Compare this to {prf:ref}`Heine-Borel theorem ` for the real line.
We established there that for a closed and bounded subset of $\RR$,
any open cover can be reduced to a finite subcover. There, we defined
closed and bounded sets as compact sets. Our treatment of compactness
in this section is more general.
We start here with the definition that a compact set is one for which
any open cover can be reduced to a finite subcover.
We then show in this theorem that in the special case of
Euclidean spaces $\RR^m$, the compact subsets are identical to the
closed and bounded subsets of $\RR^m$.
```{prf:proof}
We have shown in {prf:ref}`res-ms-compact-is-closed-bounded` that
every compact set is closed and bounded.
For the converse, we assume $A$ is a closed and bounded subset of $\RR^m$.
We will show that every sequence of $A$ has a subsequence converging in $A$.
1. Since $A$ is bounded, we can pick $M > 0$ such that $d(x, y) \leq M$ for all $x, y \in A$.
1. Fix an element $y \in A$.
1. Let $a = (a_1, \dots, a_m) \in A$ be some arbitrary point of $A$. Then
$$
|a_i | \leq d(a, 0) \leq d(a, y) + d(y, 0) \leq M + d(y, 0)
$$
holds for every $1 \leq i \leq m$.
1. Thus, the set of real numbers consisting of the $i$-th coordinates
of the elements of $A$ is a bounded set.
1. Choose an arbitrary sequence $\{ x_n \}$ of $A$.
1. Recall from {prf:ref}`Bolzano Weierstrass theorem ` for real numbers
that every bounded sequence of real numbers has a convergent subsequence.
1. Note that if we form the sequence of real numbers from
any particular coordinate of $\{x_n\}$, (say first coordinates or second coordinates)
then the sequence is bounded by $M + d(y, 0)$.
1. Every such sequence of real numbers (from a fixed coordinate of $\{x_n\}$) will
have a convergent subsequence.
1. Thus, there is a subsequence $\{x^1_n\}$ of $\{x_n\}$ whose first coordinates
form a sequence in $\RR$ that converges in $\RR$.
1. Now, we choose $\{x^2_n\}$ as a subsequence of $\{x^1_n\}$ so that the
corresponding sequence of second coordinates of $\{x^2_n\}$ converge in $\RR$.
1. Proceeding in this manner, after $m$ steps, we have a subsequence
$\{x^m_n\}$ of $\{x_n\}$ with the property that for each $1 \leq i \leq m$,
the sequence of its $i$-th coordinates forms a convergent subsequence in $\RR$.
1. Since each of the coordinates of $\{x^m_n\}$ converges in $\RR$,
hence, $\{x^m_n\}$ converges in $\RR^m$.
1. But since $A$ is closed, hence $\{x^m_n\}$ converges to a point of $A$.
1. Thus, every sequence of $A$ has a convergent subsequence in $A$.
1. Thus, by {prf:ref}`def-ms-compact-characterization`, $A$ is compact.
```
Note that the property *convergence in individual coordinates
implies convergence in $\RR^m$* is due to the specific choice
of Euclidean metric.
```{prf:theorem} Attainment of minimum and maximum values
:label: res-ms-compact-real-valued-min-max-attain
Let $f : (X, d) \to \RR$ be a real valued function.
If $f$ is continuous then $f$ attains a maximum and minimum value on
every compact subset of $\dom f$.
```
```{prf:proof}
Let $A$ be a compact subset of $\dom f$.
1. Then, $f(A)$ is compact in $\RR$ due to {prf:ref}`res-ms-compact-continuous-map`.
1. Then, $f(A)$ is closed and bounded due to {prf:ref}`Heine-Borel theorem `.
1. Since $f(A)$ is bounded, hence it has an infimum and supremum.
1. Since $f(A)$ is closed, hence its infimum and supremum lie inside $f(A)$ itself.
1. Thus, $f$ attains a maximum and minimum value in $A$.
```
```{prf:theorem} Bolzano Weierstrass theorem for bounded subsets of $\RR^m$
:label: res-ms-bw-theorem-rn-closed-bounded-set
Every sequence in a closed and bounded set in $\RR^m$
has a convergent subsequence.
```
```{prf:proof}
Let $A$ be a closed and bounded subset of $\RR^n$.
1. By {prf:ref}`res-ms-heine-borel-euclidean`, $A$ is compact.
1. By {prf:ref}`def-ms-compact-characterization`,
every sequence in $A$ has a subsequence which converges to a point of $A$.
```
```{prf:theorem} Bolzano Weierstrass theorem for bounded sequences of $\RR^m$
:label: res-ms-bw-theorem-rn-bounded-sequence
Every bounded sequence in $\RR^m$ has a convergent subsequence.
```
```{prf:proof}
Let $\{ x_m \}$ be a bounded sequence of $\RR^m$.
Then there exists a closed ball $B[0, M]$ such that
$\{ x_m \} \subset B[0, M]$.
$B[0, M]$ is closed and bounded.
From {prf:ref}`res-ms-bw-theorem-rn-closed-bounded-set`,
every sequence in a closed and bounded subset of $\RR^n$
has a convergent subsequence.
```
## Totally Bounded Metric Spaces
```{index} Totally bounded metric space
```
```{prf:definition} Totally bounded space
:label: def-ms-totally-bounded-metric-space
A metric space $(X,d)$ is called *totally bounded* if for each
$r > 0$, there exists a finite number of points
$x_1, \dots, x_n \in X$ such that
$$
X = \bigcup_{i=1}^n B(x_i, r).
$$
```
```{prf:theorem} Compactness implies total boundedness
:label: res-ms-compact-is-totally-bounded
A compact metric space is totally bounded.
```
```{prf:proof}
Let $(X,d)$ be a compact metric space.
1. Let $r > 0$.
1. Consider the family of open balls $\{ B(x, r) \}_{x \in X}$.
1. Then $X = \bigcup_{x \in X} B(x, r)$.
1. Since $X$ is compact, there is a finite subcover of open balls.
1. Thus, for every $r > 0$, there $X$ is a union of finite open balls.
1. Thus, $X$ is totally bounded.
```
```{prf:example}
:label: ex-ms-totally-bounded-but-not-compact-space
We showed earlier in {prf:ref}`ex-0-1-not-compact`
that $(0,1)$ is not compact.
However $(0,1)$ is totally bounded.
1. Let $r > 0$.
1. Pick any $n> \frac{1}{r}$. Thus, $r > \frac{1}{n}$.
1. Consider the points $\frac{1}{n}, \frac{2}{n}, \dots, \frac{n-1}{n}$.
1. $\frac{k}{n} + r > \frac{k}{n} + \frac{1}{n} = \frac{k+1}{n}$.
1. $\frac{k}{n} - r < \frac{k}{n} - \frac{1}{n} = \frac{k-1}{n}$.
1. Thus,
$$
\left (\frac{k-1}{n}, \frac{k+1}{n}\right ) \subseteq B\left (\frac{k}{n}, r \right )
$$
1. Thus,
$$
(0, 1) = \bigcup_{k=1}^{n-1}B\left (\frac{k}{n}, r \right ).
$$
with the caveat that the first and last balls are restricted within the set $(0,1)$.
1. Thus, we have a finite union of open balls.
1. Thus, $(0,1)$ is totally bounded.
```
```{prf:theorem} Completeness and Compactness
:label: res-ms-complete-totally-bounded-compact
A metric space is compact if and only if it is complete and totally bounded.
```
```{prf:proof}
Let $(X,d)$ be a metric space. Assume that $(X,d)$ is
compact.
1. $(X,d)$ is totally bounded ({prf:ref}`res-ms-compact-is-totally-bounded`).
1. Since $(X,d)$ is compact, hence every sequence has a convergent
subsequence that converges to a point in $X$ ({prf:ref}`def-ms-compact-characterization` (3)).
1. Thus, if $\{x_n\}$ is a Cauchy subsequence of $X$, then it has
a subsequence with limit $x \in X$ leading to $\lim x_n = x$.
1. Thus, every Cauchy sequence of $X$ converges in $X$.
1. Thus, $(X,d)$ is complete.
For the converse, assume that $(X,d)$ is complete and totally bounded.
To show that $(X,d)$ is compact, we will show that
every infinite subset of $X$ has an accumulation point in $X$
({prf:ref}`def-ms-compact-characterization` (2)).
1. Let $A$ be an infinite subset of $X$.
1. Since $X$ is totally bounded,
there exists a finite subset $F_1 \subset X$
such that
$$
X = \bigcup_{x \in F_1} B(x, 1).
$$
1. We can extend the open balls with closed balls without problem:
$$
X = \bigcup_{x \in F_1} B[x, 1].
$$
1. Thus:
$$
A = A \cap X = \bigcup_{x \in F_1}( A \cap B[x, 1]).
$$
1. Since $A$ is infinite, there is some $x_1 \in F_1$ such that $A_1 = A \cap B[x_1, 1]$
is an infinite set.
1. If $A \cap B[x, 1]$ were finite for each $x \in F$, then $A$ would be finite
as a finite union of finite sets.
1. Since $X$ is totally bounded, we can again find a finite subset $F_2 \subset X$ such that
$$
X = \bigcup_{x \in F_2} B\left [x, \frac{1}{2}\right].
$$
1. Since $A_1$ is infinite, there is some $x_2 \in F_2$ such that $A_2 = A_1 \cap B\left[x_2, \frac{1}{2} \right]$
is an infinite set.
1. Proceeding inductively, if $x_1, x_2, \dots, x_n$ have been chosen, we can
choose $x_{n+1}$ such that the set
$$
A \cap B[x_1, 1] \cap B\left [x_2, \frac{1}{2} \right]
\cap \dots \cap B\left [x_n, \frac{1}{n}\right]
\cap B\left [x_{n+1}, \frac{1}{n+1} \right]
$$
is infinite.
1. Define
$$
E_n = B[x_1, 1] \cap B\left [x_2, \frac{1}{2} \right]
\cap \dots \cap B\left [x_n, \frac{1}{n}\right]
$$
for each $n$.
1. Then for each $n$:
1. $E_n$ is nonempty and closed.
1. $A \cap E_n$ is infinite.
1. $E_{n+1} \subseteq E_{n}$.
1. $\diam E_n \leq \frac{2}{n}$. $\lim \diam E_n = 0$.
1. Due to {prf:ref}`res-ms-complete-nested-closed-nonempty`,
there exists $a \in X$, such that $a \in E_n$ for each $n$.
1. Now, if $y \in A \cap E_n$, then
$$
d(a, y) \leq d(a, x_n) + d(x_n, y) < \frac{1}{n} + \frac{1}{n} = \frac{2}{n}.
$$
1. Thus, for every $r > 0$, we can pick $n > \frac{2}{r}$ such that
for every $y \in A \cap E_n$, $y \in B(a, r)$.
1. Thus, $a$ is an accumulation point of $A$.
1. Thus, every infinite subset of $X$ has an accumulation point in $X$.
1. Thus, $X$ is compact.
```
## Equivalent Metrics
```{prf:theorem} Metric equivalence and compactness
:label: res-ms-eq-metric-compactness
Let $d_a$ and $d_b$ be two different metrics on the same set $X$
that are equivalent.
Then, a set $A \subseteq X$ is compact in $(X, d_a)$
if and only if $A$ is compact in $(X, d_b)$.
In other words, the compact sets in the two metric spaces
are identical.
```
```{prf:proof}
Assume $A$ to be compact in $(X, d_a)$.
1. Let $A \subseteq \bigcup_{i \in I} \OOO_i$ be an open cover of $A$
in $(X, d_b)$; i.e. $\OOO_i$ are open in $(X, d_b)$.
1. Since $d_a$ and $d_b$ are equivalent, hence $\OOO_i$ are open in $(X, d_a)$ too.
1. Thus, $\bigcup_{i \in I} \OOO_i$ is an open cover for $A$ in $(X, d_a)$ too.
1. Since $A$ is compact in $(X, d_a)$, hence, there exist finite indices
$i_1, \dots, i_n$ such that
$A \subseteq \OOO_{i_1} \cup \dots \cup \OOO_{i_n}$.
1. But then, $\OOO_{i_1}, \dots, \OOO_{i_n}$ are open in $(X, d_b)$ too.
1. Thus, $A \subseteq \OOO_{i_1} \cup \dots \cup \OOO_{i_n}$ is a finite
open subcover of $A$ in $(X, d_b)$.
1. Thus, every open cover of $A$ in $(X, d_b)$ can be reduced to
a finite subcover.
1. Thus, $A$ is compact in $(X, d_b)$.
A similar reasoning establishes that if $A$ is compact in $(X, d_b)$
then $A$ is compact in $(X, d_a)$ too.
```