# Prove the Heine-Borel Thorem using Cauchy Sequence

Prove the Heine-Borel theorem using the Cauchy sequence property. If $A$ is closed and bounded then $A$ is compact.

Cauchy sequence definition: A sequence $<a_{n}:n\mathbb{N}>$ of real numbers is a Cauchy sequence if and only if for every $\epsilon > 0$ there exists a positive integer $n_{0}$ such that $n,m > n_{0} \Rightarrow |x_{n} - x_{m}| < \epsilon$.

Compact definition: $A\subset \mathbb{R}$ is compact means if $A\subset\cup O_{\alpha}$ where $O_{\alpha}$ is open then there exists $\alpha_{1},\ldots,\alpha_{n}$ such that $A\subset O_{\alpha_{1}}\cup O_{\alpha_{1}}\cup \ldots \cup O_{\alpha_{n}}$.

Proof: Let $A$ be closed and bounded, suppose $A\subset \bigcup O_\alpha$.

Case 1: $A = [a,b]$

$I_1 = [a,b]$, assume $I_1$ has no finite subcover. Consider $[a,(a+b)/2]$,$[(a+b)/2,b]$. If $O_{\alpha_{1}},O_{\alpha_{2}}$ cover $J_1$ and $O_{\beta_{1}},O_{\beta_{2}}$ cover $J_2$, then $$O_{\alpha_{1}},\ldots,O_{\alpha_{n}},O_{\beta_{1}},\ldots,O_{\beta_{n}}$$ cover $I$, contradiction! So, $J_1$ on $J_2$ has no finite subcover, call this $I_2$, proceed. Get, $$I_1\supset I_2\supset \ldots \supset I_n\supset\ldots$$ Now, let's define a subsequence $a_{j_{k}}$ by letting $a_{j_{k}}$ be the first term of the sequence $a_{j_{1}},\ldots,a_{j_{k-1}}$ and which is an element of $I_j$. We claim that $a_{j_{k}}$ is a Cauchy sequence.

This is where I am lost, I don't know how to continuge and prove that $A$ is compact, any suggestions would be greatly appreciated.

Note that each $I_j$ has the property that finitely many $O_{\alpha}$'s do not cover $I_j$. Furthermore, if $I_j = [a_j,b_j]$, then $$b_j - a_j = (b-a)/2^j$$ For any $n \in \mathbb{N}$, $a_m \in I_n$ for all $m\geq n$, and so $$|a_m - a_n| < (b-a)/2^n$$ so $(a_n)$ is a Cauchy sequence. Let $a$ be the limit of $(a_n)$, then $a \in I_1$, so $\exists \alpha_1$ such that $a\in O_{\alpha_1}$. Now choose $N \in \mathbb{N}$ such that $$(a-1/2^N,a+1/2^N) \subset O_{\alpha_1}$$ Now fing $n$ large enough so that $I_n \in (a-1/2^N,a+1/2^N)$. This contradicts the fact that each $I_n$ cannot be covered by finitely many $O_{\alpha}$'s.