Heine-Borel implies Bolzano-Weierstrass theorem

I have to show that Heine-Borel theorem (closed and bounded sets in $\mathbb{R}$ are compact) implies Bolzano-Weierestrass theorem (BWT: Every bounded infinite subset in $\mathbb{R}$ has a limit point in $\mathbb{R}$). I want to understand several things before I can solve the problem, it does not seem easy, but I want to explain it correctly. Here are my thoughts until now:

1. Initially, I want to ask two things. The BWT implies existence of a limit point for all bounded, infinite subsets in $\mathbb{R}$, does it follow that $(0,1)$ has a limit point.
2. In my proofs related to compactness problems I usually use the fact that if an infinite set has a limit point, we can always choose a convergent subsequence from points in this set. Is it true? I feel that at least one of the above is wrong since this will make $(0,1)$ compact and it is not, not even complete.

At the end of the problem I should conclude that HB theorem is equivalent to completeness in $\mathbb{R}$.

Can you help me answering the above-stated questions, any further help on the problem will be also appreciated.

• Do you know that every infinite subset of a compact set has a limit point? Every infinite subset of a bounded set in $\Bbb{R}$ is an infinite subset of a closed interval, which is compact. Thus the BWT follows. May 1 '17 at 12:39
• Yes. $(0,1)$ has a limit point in itself, $0.5$ is one of them. May 1 '17 at 12:46
• $(0,1)$ is not even closed. Thus it is not compact. An infinite bounded set $S$ has a limit point $p$ in $\Bbb{R}$, but it is not necessary that $p \in S$. The set $S=\{1/n\}$ is one example. The only limit point is $0$, but $0 \notin S$. May 1 '17 at 12:49
• The HB theorem implies BWT, which is equivalent to completeness. Other equivalent statements include the monotonic convergence theorem, the least upper bound property and the nested interval theorem. May 1 '17 at 12:53
• Great, I am trying to visualize things on a graph, but I think I get the general idea. BWT says: A set is infinite and totally bounded then it has a limit point. Then it does not matter what sequence we choose, we always can find a convergent subsequence, since the set is infinite. When we add the bounds, we obtain a compact set. The limit point is 100% in the set since it becomes closed and it contains all of its limit points, hence whatever sequence we choose, we can find a convergent subsequence in the set. Is my idea a correct one? May 1 '17 at 12:56

Here is a short proof of Bolzano Weierstrass theorem based on Heine Borel theorem. Let $A$ be an infinite set which is bounded so that $A\subset [a, b]$ for some real numbers $a, b$. Also assume on the contrary that no point of $[a, b]$ is a limit point of $A$. Thus for each $x \in [a, b]$ there is a neighborhood of $I_{x}$ of $x$ which does not contain any point of $A$ different from $x$. By Heine Borel a finite number of these neighborhoods cover $[a, b]$ and hence $[a, b]$ contains at most a finite number of points $x_{1}, x_{2}, \dots, x_{n}$ (based on chosen finite number of neighborhoods $I_{x_{k}}$) of $A$. This is the contradiction we needed and hence there is a limit point of $A$ in $[a, b]$.
Nested Interval Principle: Let $\{I_{n}\}$ be a sequence of closed intervals such that $I_{n} \supseteq I_{n + 1}$ for all $n$. Then there is a point $c$ which lies in all the intervals $I_{n}$.