# 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?
– user205011
Feb 27 '20 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. Feb 27 '20 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). Feb 27 '20 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. Feb 27 '20 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. Feb 27 '20 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. Feb 27 '20 at 22:52
• @HennoBrandsma: that's true: arguably less direct, but simpler in not requiring Zorn's lemma. Feb 27 '20 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 $$x_{n_k}\to z.$$