I have a proof for the following problem but I am not sure if it is valid or generalizes to infinite number of sets. Please give your suggestions or a better proof.
$|pow(\mathbb{N})|$ is defined as cardinality of power set of natural numbers.
Now for infinite sequence of sets we have , $|\mathbb{N}| < |pow(\mathbb{N})| < |pow(pow(\mathbb{N}))| ....$
Here $pow^n(\mathbb{N})$ is the $n$ th set in the sequence.
$U := \bigcup\limits_{n=0}^\infty pow^n(\mathbb{N}) $
Prove that $ |pow^n(\mathbb{N})| < |U| $ for all natural number $n$
My proof:
We know that there exists no surjection from $pow^n(\mathbb{N})$ to $pow^{n+1}(\mathbb{N})$, namely from a set to its power set, for any natural number $n$.
So an element $e$ in $pow^{n+1}(\mathbb{N})$ does not have an inverse image in $pow^n(\mathbb{N})$. For the set $U$ this specific element $e$ is still in $U$.
So we can't form a surjection from $pow^n(\mathbb{N})$ to $U$, since we still can't find an inverse image for $e$, which implies stricly smaller than relation.