question about an infinite dimensional vector space Let $s$ denote the set of all complex sequences. Let $M \subseteq s$ be an infinite subset of $s$. If there exists numbers $\lambda_1, \lambda_2,.... $ such that for all $x =( r_k(x) ) \in M $ we have $| r_k(x) | \leq \lambda_k $, then $M$ is compact.
My attempt:
Take a sequence $(x_n^{(m)}) = ( r_1^{(m)},r_2^{(m)},.......) \subseteq M $ . By hypothesis, we know each $(r_k^{(m)} )$ is bounded, hence $M$ must be bounded. Still, we need to show $M$ is closed. But, how can I find an element $y$ in $M$ such that $(x_n^{(m)}) \to y $ thus showing $M$ is closed??
 A: There's a few issues with your proposed proof.
1) Sequential compactness is not in general equivalent to compactness. They are equivalent in a metric space, though. If you want to use the strategy you seem to be outlining, you'll need to produce a metric on $s$.
2) Until you supply a metric on $s$, you can't talk about $M$ being bounded or unbounded. Boundedness is not an intrinsic property of a space, unlike compactness- even if two spaces are homeomorphic, different metrics may say that the space is bounded or not. Consider $\mathbb{R}$ with the usual metric $d(x,y)=|x-y|$ versus the metric $d'(x,y)=\frac{|x-y|}{1+|x-y|}$.
Hint: Once you've produced a metric, try solving the problem first in the first coordinate, then in the first and second coordinate, then in the first, 2nd, and third, etc. You'll need to combine these in a bit of a clever way, though. Think about Cantor's diagonalization tricks.
Full Proof: We use the $\ell^\infty$ metric on $S$, ie $d(x,y)=\max_{n\in\mathbb{N}}(|x_n-y_n|)$. (Exercise: prove the triangle inequality for this metric). Now sequential compactness is equivalent to compactness for any subspace of $S$, as $S$ is a metric space.
To prove that $M$ is sequentially compact, take a subsequence which converges in the first coordinate, which we can do by the fact that the ball of radius $\lambda_1$ is compact in $\mathbb{C}$. Now, in this subsequence, pick a subsequence that converges in the second coordinate. Continue this process infinitely many times. Now, form a subsequence of our original sequence by taking the $n^{th}$ term to be the $n^{th}$ term of the $n^{th}$ subsequence. By construction, this subsequence converges to a limit in $M$. So $M$ is compact.
