Conclusions on Totally Bounded metric spaces and Cauchy sequences

See also https://math.stackexchange.com/questions/9964/if-no-cauchy-subsequence-exists-must-a-uniformly-separated-subsequence-exist and https://math.stackexchange.com/questions/7210/which-metric-spaces-are-totally-bounded

 Let's list the conclusions from my recent posts. We'll use the same definitions and notation as before. These results are presumably all well known, though I haven't personally seen explicit statements of all of them before. I had some fun working out the details for myself!

Notes on terminology: I have seen both "uniformly separated" and "separated" used for sequences. But there is a completely different notion of separated sets for pairs of sets. So possibly my use of the term "separated set" is confusing. I have seen the term "uniformly discrete" used, but I don't think that I have yet found the most definitive standard terminology.

Theorem 1

Let $(X,d)$ be a metric space. Then  the following statements are equivalent.

(a) $X$ is totally bounded.

(b) $X$ has no infinite, uniformly separated subset.

(c) There is no uniformly separated sequence in $X$.

(d) Every sequence in $X$ has a Cauchy subsequence.

Taking negations, we have:

Theorem 1'

Let $(X,d)$ be a metric space. Then  the following statements are equivalent.

(a) $X$ is not totally bounded.

(b) $X$ has an infinite, uniformly separated subset.

(c) There is a uniformly separated sequence in $X$.

(d) There is a sequence in $X$ which has no Cauchy subsequence.

Then, for individual sequences, we have:

Theorem 2

Let $(X,d)$ be a metric space, let $(x_n)$ be a sequence in $X$, and set

$\qquad A=\{x_n:n \in \N\,\}.$

Then the following statements are equivalent.

(a) $(x_n)$ has no uniformly separated subsequence.

(b) Every subsequence of $(x_n)$ has a Cauchy subsequence.

(c) $A$ has no infinite, uniformly separated subset.

(d) $A$ is totally bounded.

Taking negations we have:

Theorem 2'

Let $(X,d)$ be a metric space, let $(x_n)$ be a sequence in $X$, and set

$\qquad A=\{x_n:n \in \N\,\}.$

Then the following statements are equivalent.

(a) $(x_n)$ has a uniformly separated subsequence.

(b)  $(x_n)$ has a subsequence which has no Cauchy subsequences.

(c) $A$ has an infinite, uniformly separated subset.

(d) $A$ is not totally bounded.

I wanted to use Theorem 2 to shorten the proof of Theorem 1, but (apart from being constructive) Theorem 2 isn't really any easier to prove than Theorem 1. So we might as well use Theorem 1 to prove Theorem 2. (We could note that whenever we have a well-order on $X$, then Theorem 1 becomes constructive too.)

Similarly we have the following pair of theorems.

Theorem 3

Let $(X,d)$ be a metric space and let $(x_n)$ be a sequence in $X$.

Then the following statements are equivalent.

(a) $(x_n)$ has a Cauchy subsequence.

(b) $(x_n)$ has a subsequence which has no uniformly separated subsequence.

Taking negations we have:

Theorem 3'

Let $(X,d)$ be a metric space and let $(x_n)$ be a sequence in $X$.

Then the following statements are equivalent.

(a) $(x_n)$ has no Cauchy subsequence.

(b) Every subsequence of  $(x_n)$ has a uniformly separated subsequence.

Some special cases to note:

• If the metric space $(X,d)$ is complete, then a sequence in $X$ is Cauchy if and only if it converges in $X$.
• In a finite-dimensional normed space over $\R$ or $\C$ (in particular in  $\R$ and  $\C$ with the usual metrics), the Heine-Borel Theorem and Bolzano-Weierstrass Theorem are both valid. As a result, in these special cases, you can check that the totally bounded subsets are just the bounded subsets.
• In particular, every bounded sequence in $\R$ has a Cauchy subsequence (which, of course, converges in $\R$), and no uniformly separated sequence in $\R$ can be bounded.

You can check for yourselves that the theorems above are correct for special cases such as when $X$ is a subset of $\R$ (with the usual metric).