# Proving a sequentially compact space is separable.

Is the following argument valid:

Let $X$ be the sequentially compact space. Divide the space into countable subspaces, and form sequences in each of those subspaces. If the space is uncountable, there will be an uncountable number of such countable subspaces.

Each of those sequences have convergent subsequences. Divide $X$ into two disjoint parts $A$ and $B$- $A$ contains points of every sequence until the $n^{th}$ term of each subsequence, and $B$ contains the rest of the terms. Clearly neither of $A$ or $B$ has a limit point of the other set.

Is this argument correct? Thanks!

The statement is false: the ordinal space $\omega_1$ is sequentially compact but not separable. I suspect that what you’re actually trying to prove is that every sequentially compact metric space is separable. Unfortunately, the argument that you present here makes no sense: the division into $A$ and $B$ that you describe is not possible.
Probably the most straightforward way to prove the theorem is to prove that $X$ is totally bounded. Then take a finite $2^{-n}$ net $D_n$ for each $n\in\Bbb N$, let $D=\bigcup_{n\in\Bbb N}D_n$, and show that $D$ is dense in $X$.
• What invalid assumption am I making? Why is such a division into $A$ and $B$ not possible? – fierydemon Aug 19 '13 at 7:08
• @Ayush: For each $x\in X$ and $n\in\Bbb Z^+$, $x$ is the $n$-th term of some sequence, so each point belongs in both $A$ and $B$. There are other problems. For instance, an uncountable space can be separable, $\Bbb R$ being an example, so the relevance of your assumption that $X$ is uncountable is far from clear. – Brian M. Scott Aug 19 '13 at 7:12
• Sorry I think I got my answer. There might a sequence in $A$, not totally contained within any of the countable subspaces, with a limit in $B$. – fierydemon Aug 19 '13 at 7:13
• @Ayush: Every point of the space is in $A$. – Brian M. Scott Aug 19 '13 at 7:13