If $A = \{x_n:n\ge 1\}$ is totally bounded, then $(x_n)$ has a Cauchy subsequence (Carothers, Lemma 7.3)

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Let $(x_n)$ be a sequence in $(M,d)$, and let $A = \{x_n:n\ge 1\}$ be the range of the sequence. If $A$ is totally bounded, then $(x_n)$ has a Cauchy subsequence.

First, here are the definitions/characterizations of totally bounded I know:

$A\subset (M,d)$ is said to be totally bounded if for all $\epsilon>0$, there exist finitely many points $x_1,\ldots,x_n\in M$ such that $A\subset\bigcup_{i=1}^n B(x_i,\epsilon)$. Equivalently, $A\subset (M,d)$ is said to be totally bounded if for all $\epsilon>0$, there exist finitely many points $x_1,\ldots,x_n\in A$ such that $A\subset\bigcup_{i=1}^n B(x_i,\epsilon)$.

$A$ is totally bounded iff for all $\epsilon>0$, there are finitely many sets $A_1,A_2,…,A_n\subset A$ with $\text{diam}(A_i) < \epsilon$ for all $1\le i\le n$ such that $A\subset \bigcup_{i=1}^n A_i$. (The $A_i$'s might as well be subsets of $M$, we don't care – it is easy to see the equivalence of the two characterisations).

I need some help in understanding the proof:

  1. If $A$ is finite, we are done. Is the following reasoning correct: If $A$ is finite, then $x_n$ is eventually constant. Eventually constant sequences are Cauchy themselves – and so have a Cauchy subsequence. $\color{red}{\text{(Resolved).}}$
  2. Why is $A_1$ also totally bounded? We know that $A\subset \cup_{i=1}^n A_i$, where $A_1$ contains infinitely many points of $A$. I know that subsets of a totally bounded set are also totally bounded, but $A_1\subset A$ is not necessarily true! $\color{red}{\text{(Resolved).}}$
  3. Why does the author write (How?) in the second last line. Isn't it obvious that I can pick $x_{n_k}\in A_k$ since all the $A_k$'s are non-empty? What is there to justify? $\color{red}{\text{(Resolved).}}$
  4. $\text{diam}(A_k) < 1/k$ by construction, but why is $\text{diam}\{x_{n_j}:j\ge k\} \le \text{diam}(A_k)$? We need this to conclude that the chosen subsequence is indeed Cauchy. $\color{red}{\text{(Resolved).}}$
  5. I could simply have selected $1/2^k$ instead of $1/k$ right (or any other monotone decreasing function of $k$)? I don't find the choice of $1/k$ special. $\color{red}{\text{(Resolved).}}$

Here's the proof attached for reference:

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Thanks a lot for the help!

Best Answer

$A_1 \subseteq A$ is true! $A$ is a union by finitely many sets of diameter $<1$ (we can take $\varepsilon=\frac12$ in the definition of being totally bounded and know there are $x_1, \ldots x_n$ so that $A \subseteq \bigcup_{i=1}^n B(x_i, \frac12)$, we just take $A_1 =B(x_i, \frac12) \cap A$ for the ball that contains infinitely many points of $A$. So $_1$ is still totally bounded and its diameter is at most the diameter of $B(x_i, \frac12)$ which is $\le 1$ by the triangle inequality..

The above process we can keep on doing, using recursion.

Picking $x_{n_k}$ as a subsequence needs that each $A_k$ has infinitely many elements, not just non-emptyness: for $x_{n_1}$ we use non-emptyness, but then we need some $x_n$ with $n > n_1$ to be in $A_2$ and this is what the infinity of points gives us. This keeps on: having $n_k$ we need elements with larger index than $n_k$ to be in $A_{k+1}$ to keep the subsequence going (which must have a monotonically increasing subset $n_1 < n_2 < \ldots$ as indices).

And yes any other decreasing sequence of diameters would have done just as well, as long as it goes below any given $\varepsilon>0$.

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