pre-compactness, total boundedness and "Cauchy sequential compactness" From Wikipedia:

A metric space is called pre-compact or totally bounded if any
  sequence has a Cauchy subsequence; this can be generalised to uniform
  spaces.

Alternatively, pre-compactness and total boundedness can be defined differently for a uniform space (note that a metric space is a uniform space):

Pre-compact subspace is a subset whose closure is compact.
A subset $S$ of a uniform space $X$ is totally bounded if and
  only if, given any entourage $E$ in $X$, there exists a finite cover
  of $S$ by subsets of $X$ each of whose Cartesian squares is a subset
  of $E$.

Let me call a uniform space to be Cauchy sequential compact, if any
sequence in it has a Cauchy subsequence.
I was wondering if


*

*For a metric space, pre-compactness, total boundedness and Cauchy
sequential compactness are all equivalent?

*Same question for a uniform space?


Added: 


*

*Pre-compactness in the first quote is defined differently from the
one in the second quote. So now my question is narrowed down to
whether total boundedness and Cauchy sequential compactness are
equivalent in both metric spaces and uniform spaces.

*Pete's reply says yes for metric spaces, and now  what can we say
about uniform spaces?
Thanks and regards!
 A: It’s a straightforward result that a uniform space $X$ is totally bounded iff every net in $X$ has a Cauchy subnet, and the usual equivalence between nets and filters shows that this is in turn equivalent to the statement that for each filter in $X$ there is a finer Cauchy filter. However, this does not guarantee that every sequence has a Cauchy subsequence. One counterexample is $\beta\omega$: it’s a compact Hausdorff space, so it has a unique compatible uniformity and is both complete and totally bounded in that uniformity, but the sequence $\langle n:n\in\omega\rangle$ has no convergent subsequence.
I’ve not worked much with uniform spaces, but in that context I’ve generally seen precompact used as a synonym of totally bounded. The situation with respect to the other meaning of the term is the same as in metric spaces. If $Y$ is a subspace of a complete uniform space $X$, then $\operatorname{cl}_XY$ is complete; if $Y$ is also totally bounded $-$ which is an inherent property, just as it is for metric spaces $-$ then $\operatorname{cl}_XY$ is totally bounded and therefore compact. Thus, a subset of a complete uniform space is precompact (in this other sense) iff it is totally bounded. This will not be the case in arbitrary uniform spaces, however.
A: I'm afraid I don't know the answer about the uniform spaces part offhand.
Concerning metric spaces, here are some relevant facts:

*

*A metric space is compact iff it is sequentially compact.


*A metric space is totally bounded iff for all $\epsilon > 0$, it can be written as a finite union of subsets each of diameter at most $\epsilon$.


*Every subspace of a totally bounded metric space is totally bounded.


*A metric space is totally bounded iff each sequence admits a Cauchy subsequence.


*A metric space is sequentially compact iff it is complete and totally bounded.
(For proofs see for instance pages 14 to 18 of this expository note of mine.)
Note that 2) and 3) show that total boundedness is an "intrinsic" property of a metric space: i.e., a subset $Y$ of a metric space $(X,d)$ is totally bounded as a subset iff  $(Y,d)$ is totally bounded as a metric space in its own right.  Since total boundedness is equivalent to what you call Cauchy sequential compactness, Cauchy sequential compactness is also intrinsic.  But precompactness is not intrinsic, so it can't be equivalent to the other two.
Note also that total boundedness is in general weaker than compactness.  For instance, the rational numbers intersect $[0,1]$ are totally bounded but not compact.
