Recently I started reading up on some set theory and metric spaces. I just read about compact subsets and I thought I understood it but in the exercises I'm having difficulty with the following questions:
Which of the following, if any, are sequentially compact subsets of $l^{\infty}$ (the vector space of bounded sequences of real numbers)? Prove your answer, or provide a counterexample. You may assume the supremum norm is defined on $l^{\infty}$.
(a) $S_{1}=\{\mathbf{x}=(x_{n})\in l^{\infty}:|x_{i}|\leq1\forall i=1,\dots,r\}$ where $r$ is some fixed, positive integer.
(b) $S_{2}=\{\mathbf{x}=(x_{n})\in l^{\infty}:||\mathbf{x}||\leq1,x_{n}=0\forall n\geq r\}$ where $r$ is some fixed, positive integer.
I believe that (a) is not compact but I cannot come up with a counterexample for it. I have no idea for (b) though.
For the set $S_2$ in (b) if $(\mathbf{x_k})$ is any sequence in $S_2$ then by the fact that, every bounded sequence of real numbers has a convergent subsequence, for the first $r$ coordinates you can get a convergent subseqence with same indices and since the rest of the coordinates are zero, sequential compactness follows.
Another way to look at this is that $S_2$ is homeomorphic to the closed unit ball in $\mathbb{R}^r$ with the supremum norm. But the closed unit ball in $\mathbb{R}^r$ with the supremum norm is compact, whence it follows that $S_2$ is compact and in particular sequentially compact.
For the set $S_1$ in (a) consider the sequence $(\mathbf{x_k})$ in $S_1$ given by $\mathbf{x_k} = (1,0,\ldots,0, k,\ldots)$, here $k$ is at the $(r+1)$-th position. Clearly, $(\mathbf{x_k})$ has no convergent subsequence and hence $S_1$ is not sequentially compact.