I understand that for some set of letters $S$, its Kleen star is of cardinality $\aleph_0$, as you can sort the words created and enumerate them. However, what I don't understand is why the following is incorrect:
Suppose we activate the Kleen star on some countable $S, S^*$ is also countable. Now, also activate powerset on $S$, so $P(S)$ is of cadrdinality $\aleph$.
However why can't I just build an injective function $f$ from $P(S) \text{ to }S^*$ like this:
$\forall X \in P(S). f(X) =$ lexicographically sort the set, then concatenate all of the items in it.
we get that for 2 different sets, the output will be different, so cororally from this is that $|S^*| \geq |P(S)| = \aleph$ which is a contradiction to the countability of $S^*$.
Where is the error here?
We note that all the words in $S^*$ are finite. If $S$ is not finite, then there exist infinite subsets in $P(S)$. Concatenation of the elements in an infinite set of $P(S)$ will not yield a strong in $S^*$.