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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!

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  • $\begingroup$ 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. $\endgroup$ – copper.hat Oct 31 '17 at 3:27
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Use the fact that rational numbers are dense in real, so pick out a irrational number and then approximate it with rational numbers, such rational numbers can be taken as mutually distinct, so they constitute an infinite set, since those rational numbers are convergent, they are bounded, but fail to have an accumulation point as by assumption the accumulation point is exactly the irrational number.

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HINT: Consider an increasing sequence tending to $\sqrt2$.

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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.

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