# Complete and Totally Bounded Metric Space is Sequentially Compact

## Theorem

Let $M = \struct {A, d}$ be a metric space.

Let $M$ be complete and totally bounded.

Then $M$ is sequentially compact.

## Proof 1

Let $\left\langle{x_m}\right\rangle_{m \mathop \in \N}$ be an infinite sequence in $A$.

By the definition of a totally bounded metric space, we can use the axiom of countable choice to obtain a sequence $\left\langle{F_n}\right\rangle_{n \in \N}$ such that:

- For all $n \in \N$, $F_n$ is a finite $2^{-n}$-net for $M$.

For all $n \in \N$ and $y \in F_n$, define:

- $S_n \left({y}\right) = \left\{{m \in \N: d \left({x_m, y}\right) < 2^{-n}}\right\}$

It follows from the definition of a net that:

- $(1): \quad \displaystyle \N = \bigcup_{y \mathop \in F_n} S_n \left({y}\right)$

For all $n \in \N$, define:

- $G_n = \left\{{y \in F_n: S_n \left({y}\right)}\right.$ is infinite$\left.{}\right\}$

Since $F_n$ is finite by definition, it follows by $(1)$ that $G_n$ is non-empty.

For all $y \in G_n$, define:

- $\displaystyle T_n \left({y}\right) = \left\{{z \in G_{n+1}: S_n \left({y}\right) \cap S_{n+1} \left({z}\right)}\right.$ is infinite$\left.{}\right\}$

By $(1)$, it follows from the distributivity of intersection over union that:

- $\displaystyle S_n \left({y}\right) = \bigcup_{z \mathop \in F_{n+1}} \left({S_n \left({y}\right) \cap S_{n+1} \left({z}\right)}\right)$

Hence, by the definition of $G_n$, it follows that $T_n \left({y}\right)$ is non-empty.

From Countable Union of Countable Sets is Countable, it follows that the disjoint union $\displaystyle \bigsqcup_{n \mathop \in \N} G_n$ is countable.

Using the axiom of countable choice, there exists a sequence $\left\langle{\phi_n: G_n \to G_{n+1}}\right\rangle_{n \in \N}$ of mappings such that:

- $\forall n \in \N: \forall y \in G_n: \phi_n \left({y}\right) \in T_n \left({y}\right)$

Now, we use the Principle of Recursive Definition to construct a strictly increasing sequence $\left\langle{m_k}\right\rangle_{k \mathop \in \N}$ in $\N$.

Let $y_0 \in G_0$.

Let $m_0 \in S_0 \left({y_0}\right)$.

For all $k \in \N$, let:

- $y_{k+1} = \left({\phi_k \circ \cdots \circ \phi_1 \circ \phi_0}\right) \left({y_0}\right)$

where $\circ$ denotes composition of mappings.

Let $m_{k+1} > m_k$ be the smallest natural number such that:

- $m_{k+1} \in S_k \left({y_k}\right) \cap S_{k+1} \left({y_{k+1}}\right)$

Such an $m_{k+1}$ exists by:

and because:

- $S_k \left({y_k}\right) \cap S_{k+1} \left({y_{k+1}}\right)$ is infinite by the definitions of $T_k \left({y_k}\right)$ and $\phi_k$.

Note that:

- $m_k, m_{k+1} \in S_k \left({y_k}\right)$

Let $\hat x_k = x_{m_k}$.

Let $i < j$, where $i, j \in \N$.

Then, by Sum of Infinite Geometric Progression:

- $\displaystyle d \left({\hat x_i, \hat x_j}\right) \le \sum_{k \mathop = i}^{j-1} \left({d \left({\hat x_k, y_k}\right) + d \left({\hat x_{k+1}, y_k}\right)}\right) < \sum_{k \mathop = i}^{\infty} 2^{1-k} = 2^{2-i}$

Hence, by Sequence of Powers of Number less than One, the sequence $\left\langle{\hat x_k}\right\rangle_{k \in \N}$ is Cauchy.

By the assumption that $M$ is complete, the sequence $\left\langle{\hat x_k}\right\rangle$ converges in $M$.

Since $\left\langle{\hat x_k}\right\rangle$ is a convergent subsequence of $\left\langle{x_m}\right\rangle$, it follows that $M$ is sequentially compact by definition.

$\blacksquare$

## Proof 2

The results:

show that it suffices to prove that $M$ is compact.

Aiming for a contradiction, suppose that $M$ is not compact.

Let $\CC$ be an open cover for $A$ such that $\CC$ does not have a finite subcover for $A$.

By the definition of a totally bounded metric space, we can use the axiom of countable choice to obtain a sequence $\sequence {F_n}_{n \mathop \in \N}$ such that:

- For all $n \in \N$, $F_n$ is a finite $2^{-n}$-net for $M$.

For the sake of notational brevity, let $\map {B'_n} a$ denote the open $2^{-n}$-ball of $a$ in $M$.

For all $n \in \N$, define:

Since $F_n$ is finite by definition, it follows by the definition of a net that $G_n$ is non-empty.

For all $y \in G_n$, define:

- $\map {T_n} y = \leftset {z \in G_{n + 1}: \map {B'_n} y \cap \map {B'_{n + 1} } z}$ is not covered by any finite subset of $\rightset \CC$

By the definition of a net, it follows from the distributivity of intersection over union that:

- $\ds \map {B'_n} y = \bigcup_{z \mathop \in F_{n + 1} } \paren {\map {B'_n} y \cap \map {B'_{n + 1} } z}$

Hence, by the definition of $G_n$, it follows that $\map {T_n} y$ is non-empty.

From Countable Union of Countable Sets is Countable, it follows that the disjoint union $\ds \bigsqcup_{n \mathop \in \N} G_n$ is countable.

Using the axiom of countable choice, we can obtain a sequence $\sequence {\phi_n: G_n \to G_{n + 1} }_{n \mathop \in \N}$ of mappings such that:

- $\forall n \in \N: \forall y \in G_n: \map {\phi_n} y \in \map {T_n} y$

Let $x_0 \in G_0$.

For all $n \in \N$, define:

- $x_{n + 1} = \map {\paren {\phi_n \circ \cdots \circ \phi_1 \circ \phi_0} } {x_0}$

where $\circ$ denotes composition of mappings.

For all $n \in \N$, define:

- $A_n = \map {B'_n} {x_n} \cap \map {B'_{n + 1} } {x_{n + 1} }$

Note that $A_n$ is non-empty; otherwise, by Union of Empty Set, $\O$ would be a cover for $A_n$.

Let $y \in A_n$.

Then:

\(\ds \map d {x_n, x_{n + 1} }\) | \(\le\) | \(\ds \map d {x_n, y} + \map d {x_{n + 1}, y}\) | ||||||||||||

\(\ds \) | \(<\) | \(\ds \frac 1 {2^n} + \frac 1 {2^{n + 1} }\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \frac 3 {2^{n + 1} }\) |

Let $i < j$, where $i, j \in \N$.

Then, by Sum of Infinite Geometric Progression:

\(\ds \map d {x_i, x_j}\) | \(\le\) | \(\ds \sum_{k \mathop = i}^{j - 1} \map d {x_k, x_{k + 1} }\) | ||||||||||||

\(\ds \) | \(<\) | \(\ds \sum_{k \mathop = i}^\infty \frac 3 {2^{k + 1} }\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \frac 3 {2^i}\) |

Hence, by Sequence of Powers of Number less than One, the sequence $\sequence {x_k}$ is Cauchy.

By the assumption that $M$ is complete, the sequence $\sequence {x_k}$ converges to some limit $x \in A$.

Choose $U \in \CC$ such that $x \in U$ (which can be done because $\CC$ covers $A$).

By the definition of an open set, we can choose a strictly positive real number $\epsilon$ such that $\map {B_\epsilon} x \subseteq U$.

By Sequence of Powers of Number less than One, we can choose a natural number $n$ such that:

- $\dfrac 1 {2^n} < \dfrac \epsilon 2$

By the definition of a limit, we can choose a natural number $m > n$ such that:

- $\map d {x_m, x} < \dfrac 1 {2^n}$

For all $y \in A_m$, we have:

\(\ds \map d {y, x}\) | \(\le\) | \(\ds \map d {y, x_m} + \map d {x_m, x}\) | ||||||||||||

\(\ds \) | \(<\) | \(\ds \dfrac 1 {2^m} + \dfrac 1 {2^n}\) | ||||||||||||

\(\ds \) | \(<\) | \(\ds \epsilon\) |

That is:

- $A_m \subseteq \map {B_\epsilon} x \subseteq U$

Since $\subseteq$ is a transitive relation, it follows that $A_m \subseteq U$.

That is, $A_m$ is covered by the singleton $\set U \subseteq \CC$.

But this contradicts the definitions of $\map {T_m} {x_m}$ and $\phi_m$.

Hence our initial assumption that $M$ is not compact was false.

Hence the result.

$\blacksquare$

## Proof 3

Let $M$ be both complete and totally bounded.

Let $\left \langle{a_k}\right \rangle$ be any infinite sequence in $A$.

Let $\epsilon \in \R_{>0}$.

Let $x_1, \ldots, x_n \in X$ be a finite set of points such that:

- $\displaystyle A = \bigcup_{i \mathop = 1}^n B_\epsilon \left({x_i}\right)$

where $B_\epsilon \left({x_i}\right)$ represents the open $\epsilon$-ball of $x_i$.

This is known to exist as $M$ is totally bounded.

Then for every $k \in \N$, there is some $j_k \in \left\{{0, \dots, n}\right\}$ such that $d \left({a_k, x_{j_k}}\right) \le \epsilon$.

For some $j \in \left\{{0, \dots, n}\right\}$, we must have $j_k = j$ for infinitely many $k$, and it follows by setting $x := x_{j_k}$.

Setting $x := x_{j_k}$, we see that:

- $(1): \quad$ There is some $x \in X$ such that $d \left({a_k, x}\right) \le \epsilon$ for infinitely many $k$.

Now let $\left \langle{a_k}\right \rangle$ be any infinite sequence in $A$.

By $(1)$, there is some $x_1 \in X$ such that $d \left({a_k, x_1}\right) \le 1/2$ for infinitely many $k$.

Now we can apply $(1)$ to the subsequence of $\left \langle{a_k}\right \rangle$ which consisting of those elements for which $d \left({a_k, x_1}\right) \le 1/2$.

Thus we can find $x_2 \in A$ such that infinitely many $k$ satisfy both $d \left({a_k, x_2}\right) \le 1/4$ and $d \left({a_k, x_1}\right) \le 1/2$.

Now we proceed inductively, to obtain a sequence $\left \langle {x_m}\right \rangle$ with the property that there exist infinitely many $k$ such that, for $1 \le j \le m$:

- $(2) \quad d \left({a_k, x_j}\right) \le 2^{-j}$

Now define a subsequence $\left \langle {a_{k_m}}\right \rangle$ inductively by letting $k_0$ be arbitrary, and choosing $k_{m+1}$ minimal such that $k_{m+1} > k_m$ and such that $(2)$ holds for $k = k_m$ and all $1 \le j \le m$.

Let $\epsilon > 0$, and choose $n$ sufficiently large that $1/2^{n-1} < \epsilon$.

Then:

- $d \left({a_{k_r}, a_{k_s}}\right) \le d \left({a_{k_r}, x_n}\right) + d \left({a_{k_s}, x_n}\right) \le 2 \cdot 2^{-n} < \epsilon$

whenever $r, s \ge n$.

So this subsequence is a Cauchy sequence and hence, because $M = \left({A, d}\right)$ is complete by assumption, it is convergent.

Thus we see that $\left \langle{a_k}\right \rangle$ has a convergent subsequence.

Hence, by definition, $M$ is sequentially compact.

$\blacksquare$

## Axiom of Countable Choice

This theorem depends on the Axiom of Countable Choice.

Although not as strong as the Axiom of Choice, the Axiom of Countable Choice is similarly independent of the Zermelo-Fraenkel axioms.

As such, mathematicians are generally convinced of its truth and believe that it should be generally accepted.

## Sources

- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*(2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $5$: Metric Spaces: Complete Metric Spaces