# 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.

• Have you considered the domain convergence theorem? – MathWanderer Feb 27 at 20:58
• You can even show quite easily $X$ is dense in $\{0,1\}^S$, as every basic open subset intersects it (pick $0$'s outside the finitely many determined coordinates), so indeed far from closed. – Henno Brandsma Feb 27 at 22:44

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).

• $D_n$ is a subset of the index set $S$ (and we aren't concerned with any topology on it). $\{0, 1\}^I$ is compact for any index set $I$ (under the usual product topology). – Rob Arthan Feb 27 at 22:06
• Note that the result that the product of an arbitrary family of compact sets is compact is known as Tychonoff's theorem, which is not trivial to prove. – Math1000 Feb 27 at 22:10
• @Math1000: Thanks for that reminder. It's probably ignorance on my part, but I don't know of a way of proving that $\{0, 1\}^{\Bbb{N}}$ is sequentially compact that is simpler than the proof of Tychonoff's theorem (and I was assuming, possibly wrongly, that Tychonoff's theorem would be known to the OP, given the kind of problem he or she is looking at). Please do post a more direct proof of the sequential compactness of $\{0, 1\}^{\Bbb{N}}$ if you have one. – Rob Arthan Feb 27 at 22:19
• You could use (for the countable case) the homeomorphism of $\{0,1\}^\mathbb{N}$ with the standard Cantor set, and then you only rely on compactness of intervals, or Heine-Borel. – Henno Brandsma Feb 27 at 22:52
• @HennoBrandsma: that's true: arguably less direct, but simpler in not requiring Zorn's lemma. – Rob Arthan Feb 27 at 23:09

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 $$x_{n_k}\to z.$$