0
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.