Is the set of sequences with values in a finite set separable?

Question

Let $\mathcal{A}$ be a finite set and consider the set of all sequences $\mathcal{A}^{\mathbb{Z}}$ on $\mathbb{Z}$ with values in $\mathcal{A}$. This set has a cardinality of $\mathcal{A}^{\mathbb{Z}}$ which is not countable, right? (is there actually a simple explanation?)

My actual question however is if there is at least a countable and dense subset (w.r.t. to the product topology when endowing each $\mathcal{A}$ with the discrete topology), i.e. is $\mathcal{A}^{\mathbb{Z}}$ separable?

Sadly, I couldn't come up with anything myself. I ask this to answer another question from ergodic theory, namely if the two-sided full shift over a finite alphabet is transitive.

Answer 1

The set of sequences $a_i$ that are constant for all but finitely many $i$ is countable and dense in $A^\Bbb{Z}$.

Answer 2

For the side question, as long as $|\mathcal{A}|\ge 2$ we have $2^{\aleph_0}\le|\mathcal{A}^{\mathbb{Z}}|$ so that it's uncountable.

For the actual question, I'll consider $\mathcal{A}^{\mathbb{N}}$, but the same idea works for $\mathbb{Z}$. We can see that the product topology on such a set has a basis of cones $\{x\in\mathcal{A}^{\mathbb{N}}:\sigma\text{ is contained in }x\}$ for $\sigma\in\mathcal{A}^{<\mathbb{N}}$. As a result, if we consider (say) the sequences which are almost always some $a\in\mathcal{A}$ (in the sense that there are only finitely many entries not $a$), then this set will be dense of size $\mathcal{A}^{<\mathbb{N}}=\aleph_0$.