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

Question

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?

Answer 1

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.