Talk:Sequentially Compact Metric Space is Totally Bounded/Proof 2

What if $x_N=x_M=x$?

Whow can we assure that there exist $x_N$,$x_M$ such as $x_N \ne x_M$ in order to get the contradiction?


 * This follows from the assumption that we have a contradicting $\epsilon$, as the first few lines of the proof (up to 'Now, as $X$ is sequenntially compact [...]') demonstrate. It may appear a bit strange since one is tempted initially to reason from both (contradicting) sides at once. --Lord_Farin 11:55, 20 May 2012 (EDT)
 * By the way, note that $\epsilon/2$ suffices instead of $\epsilon/4$ since we have a strict inequality present; minute detail, though. --Lord_Farin 11:57, 20 May 2012 (EDT)