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.
 A: 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 $m\ne n$, 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 $\Bbb R$.)
A: This happens for a metric space precisely when bounded is the same as totally bounded.
A subset $A$ of a metric space $X,d$ is 


*

*bounded if $\sup\{d(x,y):x,y\in A\}<\infty$

*totally bounded if, for every $\varepsilon>0$, there exist $x_1,x_2,\dots,x_n\in A$ such that $A\subseteq\bigcup_{i=1}^n B_d(x_i,\varepsilon)$


where $B_d(x,r)=\{y\in X:d(x,y)<r\}$.
It is clear that any totally bounded subset is bounded, but the converse is not true; for instance, for any metric $d$ we can define a metric $d'$ inducing the same topology as $d$ where every subset is bounded, namely
$$
d'(x,y)=\frac{d(x,y)}{1+d(x,y)}
$$
Now just take a metric space which is not totally bounded (for instance, complete but not compact) and you have an example of a bounded set which is not totally bounded.
Now suppose that every bounded set in $X,d$ is totally bounded. If a sequence is bounded, the set of its points is totally bounded, so it is precompact and its closure in the completion $\hat{X},\hat{d}$ is compact. A sequence in a compact metric space always has a convergent subsequence. (The result can also be proved without mentioning completions.)
Also the converse is true. If every bounded sequence has a Cauchy subsequence, then every bounded set is totally bounded.
Indeed, suppose $A\ne\emptyset$ is bounded, but not totally bounded. Then there is $\varepsilon>0$ such that no finite set of balls of radius $\varepsilon$ centered on points of $A$ covers $A$. In particular, chosen arbitrarily $x_0\in A$, there are points of $A$ outside of $B_d(x_0,\varepsilon)$. Suppose we have chosen $x_0,\dots,x_{n-1}$. Then we can also choose $x_n\in A$ such that
$$
x_n\notin\bigcup_{i=0}^{n-1}B_d(x_i,\varepsilon)
$$
Thus we have a sequence $(x_n)$ such that, for $m\ne n$, $d(x_m,x_n)\ge\varepsilon$. Such a sequence can have no Cauchy subsequence.
The standard metric on $\mathbb{R}^n$ has the property that every bounded subset is totally bounded.
The Heine-Borel theorem can be generalized to all complete metric spaces: a subset of a complete metric space is compact if and only if it is closed and totally bounded.
A: You are asking about boundness of a sequence, so I'm assuming we are talking about a metric space.
In a metric space, every Cauchy sequence is bounded. 
Let $(x_n)_{n \in \mathbb{N}} \subset X$ be a Cauchy sequence in a metric space, i.e., for each $\epsilon > 0$ there is a positive integer $n_0 \in \mathbb{N}$ such that for every $n, m \geq n_0$
$$ d(x_n,x_m) < \epsilon. $$
Take $\epsilon = 1$ and let $n_0$ be such integer. Then 
$$ d(0,x_n) \leq d(0,x_{n_0})+d(x_{n_0},x_n) < d(0,x_{n_0}) + 1, \; \forall n \geq n_0. $$
Now let $$\delta = \max_{1 \leq i \leq n_0} \{d(0,x_i),d(0,x_{n_0})+1\}$$ and we conclude that $d(0,x_n) \leq \delta$ for every $n \in \mathbb{N}$, so $(x_n) \subset B(0,\delta)$. 
