# Why is $A^\mathbb{N}$ with the discrete topology a polish space

I am currently preparing for a part of a seminar in topology/descriptive set theory and am working with the A. Kechris' book. I am confused about some results for one of the easier examples, the setting is the following:

The book wants me to show that "Any space $$A^{\mathbb{N}}$$ viewed with the discrete topology, where A is a countable set, is polish" (*) In the special case of A = {0,1} we would arrive at the Cantor space, which I know is separable as a second-countable space and also completely metrizable.

Here comes the confusion: I have shown that (1) the product of a sequence of polish spaces are polish. I have also shown that (2) the countable set $$A$$ with the discrete topology is polish by showing that (3) every set $$A$$ is equipped with the discrete topology iff $$A$$ is its only dense subset, thus separable (since $$A$$ also countable).

Now to show (*) I'd follow the path I assume the author had intended: Using (1) and (2) we conclude $$\Rightarrow$$ (*). But doesn't (3) contradict this claim, since this tells me that the only dense subset in $$A^{\mathbb{N}}$$ is $$A^{\mathbb{N}}$$ itself, because we are equipped with the discrete topology? And since $$A^{\mathbb{N}}$$ is uncountable, we arrive at $$A^{\mathbb{N}}$$ not separable, thus not polish?

Where is my mistake? Was (3) wrong? Did I misunderstand how the space $$A^{\mathbb{N}}$$ looks like?

• The book want you to assume that $A$ is discrete, not $A^\Bbb N$. Commented Nov 3, 2021 at 14:32
• Such a quick response, thanks! So do I gather correctly that none of the statements (1), (2), (3) or (*) are faulty and there is no contradiction because essentially I understood $A^{\mathbb{N}}$ wrong? Commented Nov 3, 2021 at 14:49
• Yes, that is the case. Commented Nov 3, 2021 at 15:15

The space $$A$$ is discrete ( and as it’s countable, even Polish). So your fact 1 applies and $$A^{\Bbb N}$$ is also Polish.
It’s quite far from being discrete and has in fact no isolated points at all unless $$A$$ is a singleton. It’s compact (and homeomorphic to the Cantor set) if $$A$$ is finite and not a singleton. If infinite it’s homeomorphic to the irrational numbers as a subspace of the reals. Both are standard Borel spaces.