# Heine Borel theorem

Heine–Borel theorem

For a subset $S$ of the real numbers $\mathbb{R}$, the following two statements are equivalent:

a) $S$ is closed and bounded.

b) for any sequence $(a_n)$ in $S$ there exist a subsequence which converges to some $L$ belonging to $\Bbb R$.

## I am ok with the first implication (a $\Rightarrow$ b)

For the second implication, I am assuming that if b holds. And I suppose for a contradiction that $X$ is not bounded i.e for all positive real $M$ there exist $x \in X$ such that $x\le M$ or $x>M$. Then I am getting stuck with finding a sequence such that every subsequence of it will not converge.

Same thing for proving that $X$ is closed. I suppose for the contrary that $X$ is not closed. Does that mean $X$ has an isolated point since it is not closed? If so where can I find the contradiction here?

I would appreciate any help.

• "Not closed" means that $X$ has an accumulation point not in $X$. – Robert Z Oct 26 '17 at 6:34
• This is not Heine Borel, but Bolzano-Weierstrass. Heine Borel deals with open covers. – Paramanand Singh Oct 26 '17 at 6:51
• My edit ea to change the :$\Bbb R$ " to "$X$" in the phrase " ... converging to some $L$ in $\Bbb R$" and to fix the grammar of it. – DanielWainfleet Oct 26 '17 at 7:52
• My second edit was to undo the first edit, in view of the Answer given, but to change $X$ to $S$ in (b), – DanielWainfleet Oct 26 '17 at 7:56

b) for any sequence $a_n$ in $S$ there exist a subsecuence which is convergent to some $L\in S$.
You use that if a sequence converges all it's subsequences converges to the same. This means that if $S$ is not closed there exists an accumulation point $a$ not in $S$ that is there is a sequence $a_n$ in $S$ such that $a_n\to a$ which in turn means that any subsequence to $a_n$ converges to $a$.
For the boundedness you use the opposite property. That is if a sequence diverges then all subsequence does that too. If $S$ is unbounded there exists a sequence $a_n$ such that $|a_n|>n$, this means that a subsequence $s_n$ to $a_n$ also must fulfil the property that $|s_n|>n$ which means that $s_n$ can't converge to an element of $S$.