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Set $X:=\{0,1\}^\mathbb{R}$ equipped with the product topology, where $\{0,1\}$ is equipped with the discrete topology. By Tychonoff's Theorem, $X$ is compact. I want to show that $X$ is not sequentially compact.

The sequence $f_n:=_{\{n\}}$ is in $X$. Assume it has a convergent subsequence $f_{n_k}\to f$. In the discrete topology, only the eventually constant sequences converge. So there are two indices $i,j\in\mathbb{N}$ such that $_{\{n_i\}}=f_{n_i}=f_{n_j}=_{\{n_j\}}\implies n_i=n_j$. This is a contradiction since subsequences can't repeat the same index of $f_n$. Thus no convergent subsequence exists and $X$ is not sequentially compact.

I have two questions:

  1. Is the above correct?
  2. If yes, isn't this a contradiction to this example for sequentially compact $\not\Rightarrow$ compact? There the same $\{0,1\}^\mathbb{R}$ as above is considered and it is shown that $M:=\{f\in\{0,1\}^\mathbb{R}:f(x)=1\text{ for at most countably many }x\}$ is sequentially compact. The sequence from my proof lies in $M$ and should therefore have a convergent subsequence, contradicting my argument above. Where is my error?
Fatrago
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    It is not compact with the discrete topology, but it is with the product topology. – Plop Feb 27 '23 at 14:50
  • Moreover, in the product topology, the sequence you provide converges to the function that is zero everywhere. – Plop Feb 27 '23 at 14:56
  • @Plop: sorry for the inaccuracy, i mean ${0,1}$ with the discrete topology and ${0,1}^\mathbb{R}$ with the product topology. do the functions still converge to zero? – Fatrago Feb 27 '23 at 15:03
  • You might want to hit the edit button and correct that in the text of your post. – Lee Mosher Feb 27 '23 at 15:04
  • @Fatrago: Yes, yes. – Plop Feb 27 '23 at 15:06
  • Ah, I see where my problem is: The topology of the infinite product of discrete topologies is not necessarily discrete. – Fatrago Feb 27 '23 at 15:15

1 Answers1

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The mistake you made is believing that a product of discrete topologies is discrete.

Let $X$ and $Y$ be sets. The set $Y^X$ is endowed with a topology named the product of discrete topology, which has the property that for all sequence $(f_n)_n$ and $f \in Y^X$, $\lim_{n \to \infty} f_n = f$ if and only if $\forall x \in X,\quad \lim_{n \to \infty} f_n(x) = f(x)$.

Let now $F = \{0,1\}$, and $(x_n)_n$ be a sequence in $X$ that is injective (under some weak version of the axiom of choice or the definition of "infinite", the existence of such a sequence is equivalent to $X$ being infinite). Let us denote, for all $n$, $f_n := \textbf{1}_{\{x_n\}}$. Then the sequence $(f_n)_n$ converges to the null function. Indeed, let $x \in X$. For all $n$, if $f_n(x) = 1$, then for all $m >n$, $f_m(x) = 0$ by the injectivity assumption. Therefore, for all $x \in X$, the set $\{n \in \mathbb{N} \ \vert \ f_n(x) = 1\}$ is either empty or a singleton, so for all $x$, $\lim_n f_n(x) = 0$.

In particular, if $X$ is infinite, this topology on $Y^X$ is not the discrete topology, since it has a convergent sequence that is not eventually constant.

Plop
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