Proof Using the Heine-Borel Theorem How can I use the Heine-Borel Theorem to prove that every bounded infinite subset of the real numbers has a limit point that is in the real numbers?
 A: A bounded infinite subset of the real numbers can be used to create an infinite sequence of distinct elements $a_n$ contained in the bounded set. The set can be contained in a closed interval since it's bounded, and the closed interval is compact by Heine-Borel, and therefore the sequence $a_n$ has a convergent subsequence.
A: Let $S$ be a bounded infinite set and assume that it has no limit points. 
Then $\overline{S}\backslash S=\emptyset$ or equivalently $S$ is
closed. This because every element of $\overline{S}\backslash S$
is a limit point. 
Now Heine-Borel tells us that $S$ is compact.
For every $x\in S$ we can find an open set $U_{x}$ with $S\cap U_{x}=\left\{ x\right\} $.
This because $x$ is not a limit point of $S$. Then the $\left\{ U_{x}\right\} _{x\in S}$ form an open subcover
of $S$. If $F\subset S$ is finite then $S\backslash F\neq\emptyset$
and for $z\in S\backslash F$ we have $z\notin\cup_{x\in F}U_{x}$.
This tells us that no finite subcover can be found hence contradicts
that $S$ is compact. 
We conclude that the assumption that $S$ has no limit points must be a false one. 
