Does every bounded sequence have a Cauchy sub-sequence? In an answer to an earlier question it was explained why a bounded sequence is not guaranteed to be a Cauchy sequence. But does every bounded sequence have a Cauchy sub-sequence? I know that from any bounded sequence in $\mathbb{R}^n$ it is possible to pick an infinite sequence of the elements which will form a Cauchy sequence. Does this hold in other topologies as well? The proof I know for the case of $\mathbb{R}^n$ relies on using the ordering of $\mathbb{R}$ to find a Cauchy sub-sequence for one co-ordinate at a time. But the same proof could not be applied to a topology with no ordering.

No, it is not true in general that every bounded sequence has a Cauchy subsequence. Define a metric $d$ on $\Bbb R$ by $d(x,y)=\min\\{|x-y|,1\\}$, and consider the sequence $\sigma=\langle n:n\in\Bbb N\rangle$. Clearly $d(m,n)=1$ whenever $m,n\in\Bbb N$ and $m\
e n$, so $\sigma$ has no Cauchy subsequence, even though $\Bbb R$ itself is bounded in the metric $d$. (Note that $d$ generates the usual topology on $\Bbb R$.)