Bolzano-Weierstrass for metric spaces

Having previously proven that every Cauchy Sequence in the real numbers converges using the Bolzano-Weierstrass theorem, I am now attempting to prove that every Cauchy Sequence in a metric space converges. I have proven that every Cauchy Sequence in a metric space is bounded, but can I draw on Bolzano-Weierstrass to claim that every bounded sequence in a metric space has a convergent subsequence?

• It's not true, the metric space has to be complete for this to be true. For example you can find a sequence of rationals that converges to $\sqrt{2}\not\in\Bbb Q$. – Gregory Grant Feb 15 '16 at 2:00

In general what you want to prove is false.

Take for exemple the metric space $\mathbb Q$ with the distance induced from the usual distance on $\mathbb R$, and consider a Cauchy sequence of rational numbers converging to an irrational number.

• Thank you! So is it untrue that every Cauchy Sequence on a Metric Space converges? – Sarah H Feb 15 '16 at 2:04
• Right, as you say. – John B Feb 15 '16 at 2:05

It's not true, the metric space has to be complete for this to be true. For example you can find a sequence of rationals that converges to $\sqrt{2}\not\in\Bbb Q$.

Even in a locally compact complete metric space a bounded sequence may fail to have a Cauchy subsequence.

Consider the space of natural numbers with the discrete metric: $d(x,y)=1$ if $x\ne y,$ and the sequence $(x_n)_{n\in N}=(n)_{n\in N}.$

More generally, if $d$ is a metric on a set $S$ then the metric $e(x,y)=\min (1,d(x,y)$ is equivalent to $d$. That is, $d$ and $e$ generate the same topology. But $e$ is bounded, so every sequence is $e$-bounded:

For example, on the reals, let $e(x,y)=\min (1,|x-y|),$ and consider the sequence $(x_n)_{n\in N}=(n)_{n\in N}.$