Complete and Totally Bounded Metric Space is Sequentially Compact/Proof 2

== Theorem == 
== Proof ==
The results:
:Compact Space is Countably Compact
:Countably Compact Metric Space is Sequentially Compact
show that it suffices to prove that $M$ is compact.
 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:
:$G_n = \leftset {y \in F_n: \map {B'_n} y}$ is not covered by any finite subset of $\rightset \CC$
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:
Let $i < j$, where $i, j \in \N$.
Then, by Sum of Infinite Geometric Progression:
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:
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.
Category:Complete and Totally Bounded Metric Space is Sequentially Compact