Conclusions on Totally Bounded metric spaces and Cauchy sequences

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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).

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