Sequentially compact but not compact Let X be a subset of product {0,1}$^S$ of uncountable class of {0,1} (S is an uncountable set), consisting of all those elements for which no more than a countable number of coordinates are nonzero. (The space {0,1} here is equipped with the discrete topology, and the space {0,1}$^S$ with the product topology). Prove that X is sequentially compact, but not closed in {0,1}$^S$ and therefore not compact.  
 A: Let $x_n$ be a sequence in $X$ and let $D_n = \{s : S \mid p_s(x_n) \neq 0\}$. Then each $D_n$ is countable and hence so is $D = \bigcup_{n \in \Bbb{N}} D_n$. So $x_n$ is a sequence in the compact, metrizable subspace $\{0, 1\}^D$ of $\{0, 1\}^S$ and therefore has a cluster point in $\{0, 1\}^D$, which is also a cluster point in $\{0, 1\}^S$. (Here, by abuse of notation, if $I \subset J$, I identify $\{0, 1\}^I$ with a subset of $\{0, 1\}^J$ by padding out with $0$s).
A: We have $X=\{x\in \prod_S \{0,1\}:x^{-1}(1)\ \text{is countable}\}$. For $s\in S,$ let $B_s=\{x\in X:x(s)=0\}.$ Then $\{B_s\}_{s\in S}$ is an open cover of $X$ with no finite subcover. 
Now let $(x_n)$ be a sequence in $X$. Let $S' = \bigcup_n x^{-1}_n (1).\ S'$ is countable since each $x^{-1}_n (1)$ is. Let $Y=\prod_{S'} \{0,1\}$ and define $f:Y\to X$ by $x\mapsto x|_{S'}$ so that $f(x_n)$ is a sequence in $Y.$ Since $S'$ is countable, $Y$ is sequentially compact, so some $(x_{n_k})\subseteq (x_n)$ satisfies $f(x_{n_k})\to y\in Y$.
To finish, notice that for each integer $n,\ x_n(S\setminus S')=\{0\}$ so that if we define $z\in X$ to be $z_s=y_s$ whenever $s\in S'$ and $z_s=0$ otherwise, then $x_{n_k}\to z.$
