I have a question which may sound silly to you, but I'm confident that I don't understand Cantor's diagonal argument very well to use it. Any provided insight would be appreciated.
I was tasked with finding the cardinal number of the set $$A^{*} = \{\epsilon, a, b, c, aa, ab, ac ... \}$$
Which is the iterated set of set $A$, using Kleene's operator.
Now, at first I thought that the cardinality of this set is bigger than $\aleph_0$ because the set has infinitely many elements, but each string in the set can be up to infinite in length. I thought it was impossible to construct a bijection between $\mathbb{N}$ and $A^{*}$, but this simple observation did the trick: $1 \rightarrow \epsilon$, $2 \rightarrow a$, $3 \rightarrow b$, $4 \rightarrow c$, $5 \rightarrow aa$...
Question: Why can't we use Cantor's diagonal argument to prove that this set has a bigger cardinality than the set of natural numbers?
For example, if I have:
$s_1 = (A, a, c, b, a, c, ...)$
$s_2 = (c, B, b, c, c, a, ...)$
$s_3 = (a, a, C, a, c, a, ...)$
$s_4 = (b, c, a, A, c, c, ...)$
Each capital letter represents the letter at position $a_{ii}$. If I create a new string $d$ so that $d_1 \ne d_{1,1}$ etc, why isn't this a valid proof that the set's cardinality is bigger than $\aleph_0$?
This set is a countable union of finite sets $X_i$ where $X_i$ is the set of strings of length exactly $i$. Each $X_i$ has $|X_i| < \mathbb{N}$, so we have an inclusion into $\mathbb{N}$, and the countable union of countable sets is countable, by countable choice, so an upper bound for this set's cardinality is that of $\mathbb{N}$. On the other hand, the set is evidently infinite, so this is the cardinality.
Cantor's argument doesn't work because the string you obtain is not actually in the set.