# 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. Commented May 1, 2017 at 12:39
• Yes. $(0,1)$ has a limit point in itself, $0.5$ is one of them. Commented May 1, 2017 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 $0 \notin S$. Commented May 1, 2017 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. Commented May 1, 2017 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? Commented May 1, 2017 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 $I_{n}$.