My text book gives a proof that a seq. compact metric space is totally bounded, and I was wondering if someone could answer my question about it?
To start, a metric space is seq. compact if every sequence in some metric space $X$ has a convergent subseq.
A metric space is totally bounded if, $\forall \epsilon > 0, \exists$ a finite subset $A \subset X$ such that $X = \bigcup_{a \in A}S_{\epsilon}(a)$.
The proof given is:
Let $X$ be a seq. compact Metric Space, and let $\epsilon > 0$ be given. Choose a point $a_1 \in X$. If $X = S_{\epsilon}(a_1)$ we're done. Otherwise, choose $a_2 \in X-S_{\epsilon}(a_1)$. If $S_{\epsilon}(a_1) \cup S_{\epsilon}(a_2) = X$ we're done. Otherwise we continue.
Then the book says, "some union of the form $S_{\epsilon}(a_1)\cup S_{\epsilon}(a_2)\cup ... S_{\epsilon}(a_n)$ will necessarily contain every point of $X$; for if this process could be continued indefinitely, then the sequence $\{a_1,a_2,...,a_n\}$ would be a sequence with no convergent sub sequence.
I'm not sure I understand how the author is drawing the conclusion that $\{a_1,a_2,...,a_n\}$ would have no convergent sub sequence.
Is it because if $X = \bigcup_{x \in X}S_{\epsilon}(x)$, then the points of $X$ never get "close enough" to converge?
Thank you for your help.
You are right, $(a_n)$ can't have a converging subsequence because for all $m \neq n, d(a_m,a_n)\geqslant \epsilon$.