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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.

Ingrid
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  • Have you considered the domain convergence theorem? –  Feb 27 '20 at 20:58
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    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 '20 at 22:44

2 Answers2

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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).

Rob Arthan
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    $D_n$ 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 '20 at 22:06
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    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 '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. – Rob Arthan 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. – Henno Brandsma Feb 27 '20 at 22:52
  • @HennoBrandsma: that's true: arguably less direct, but simpler in not requiring Zorn's lemma. – Rob Arthan Feb 27 '20 at 23:09
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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 $x_{n_k}\to z.$

Matematleta
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