# Bolzano-Weierstrass Theorem is false when S ⊂ Q

Show that the Bolzano–Weierstrass theorem is false when $S \subseteq \mathbb{Q}$. The Bolzano-Weierstrass theorem states that if a set $S\subseteq\mathbb{R}$ is infinite and bounded, it has an accumulation point.

I'm not really sure what to do for this problem, but this is what I have so far.

Assume for contradiction, there are no accumulation points in $\mathbb{Q}$

A set is closed if it contains its accumulation points

Since $S$ is closed and bounded, it is compact.

Consider $x\in S$, $x$ is not an accumulation point

Exists $\varepsilon>0$ such that $N^* (x,ε)\cap S≠∅$

Any help is greatly appreciated. Thanks in advance!

• A pedantic nitpick: The theorem is not false, what you are showing is that a bounded infinite subset of the rationals need not have an accumulation point. – copper.hat Oct 31 '17 at 3:27

HINT: Consider an increasing sequence tending to $\sqrt2$.
First, a closed and bounded subset of $\mathbb{R}$ is compact, not of $\mathbb{Q}$. In $\mathbb{Q}$, the only compact sets are the finite. An important distinction here is that $\mathbb{R}$ is what's called complete, where $\mathbb{Q}$ is not.
To answer your question, let $$E = \left\{ \sum_{k = 1}^{n} 10^{- k!} : n \in \mathbb{N} \right\} .$$ In $\mathbb{R}$, the limit point of $E$ would be $\sum_{k = 1}^{\infty} 10^{- k! }$, but that's an irrational number. So $E$ is an infinite bounded set, but in $\mathbb{Q}$ has no limit point.