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I'm browsing the proof here on showing that if a metric space is totally bounded and complete then it is sequentially compact. I'm confused about two claims.

First, the proof says :

For each $n\in \mathbb{N}$ let $D_n$ be a finite subset of $X$ such that the open balls of radius $2^{−n}$ centred at the points of $D_n$ cover $X$.

Why does $D_n$ have to be finite? I know that's implied from the definition of totally boundedness. I guess my question is really why can't we let $D_n$ be infinite?

Second question (which is related to the first I think):

$D_0$ is finite, so there is a point $y \in D_0$ such that infinitely many terms of $\sigma$ are in $B(y_0,1)$

Why would $B(y_0,1)$ contain infinitely many terms? What if the metric space X is finite in the first place?

user1691278
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  • Think pigeonhole principle: you put infinitely many balls into a finite number of pots, so one pot has to have infinitely many. If $X$ is finite, then it is automatically sequentially compact. – Clayton Nov 13 '17 at 05:02

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If $D_n $ wasn't finite you couldn't conclude that infinitely many elements of $\sigma $ are contained in one element of $D_n $ (there could be one in each, for instance )...

Secondly $\sigma $ is a sequence, hence has infinitely many terms, some or all of which may be equal (such as when $X $ is finite)...