# Conditions needed to prove $|A|^{|P(A)|}=|P(P(A))|$

I'm trying to figure out if $${|A|}^{|P(A)|}=|P(P(A))|$$ (where $$A$$ is infinite) is provable without the Axiom of Choice.

I know unconditionally we have the lower bound: $$|A|^{|P(A)|}\geq 2^{|P(A)|}=|P(P(A))|$$

If we assume the Axiom of Choice we can use Tarski's theorem about choice, that is $$|X|=|X|^2$$ for all infinite sets $$|X|$$ as well as the trivial bound $$|X|<2^{|X|}=|P(X)|$$ to get an upper bound:

$${|A|}^{|P(A)|}\leq{(2^{|A|})}^{|P(A)|}=2^{|A||P(A)|}\leq 2^{{|P(A)|}^2} = 2^{|P(A)|}=|P(P(A))|$$

Then with these lower and upper bounds for $$|A|^{|P(A)|}$$ and the Schroder-Bernstein Theorem we get the equivalence. However, I'm wondering is this can be achieved without the Axiom of Choice since the bound $$|A|<2^{|A|}$$ is rather generous. Also, I thought I saw somewhere that $$|X|^{|Y|}=|P(Y)|$$ as long as $$|X|<|Y|$$ but that may be wrong. Thanks.

• You should add the condition that $A$ is infinite! Mar 28 at 14:46
• @AlexKruckman Thanks! Yeah, I should have specified that. Otherwise, even the Axiom of Choice cannot help.
– Ari
Mar 29 at 0:57
• "Also, I thought I saw somewhere that $|X|^{|Y|}=|P(Y)|$ as long as $|X|<|Y|$" this is definitely wrong without the axiom of choice. It is possible to have $X$ such that $|P(X)|=2^{|X|}<3^{|X|}<4^{|X|}<...<\aleph_0^{|X|}$
– ℋolo
Mar 29 at 18:08
• @ℋolo what if we only consider the case of $X$ and $Y$ are infinite?- similar to Alex's comment?
– Ari
Mar 29 at 20:50
• @Ari If you are asking for $|X|^{|Y|}=|Z|^{|Y|}$ for infinite $|X|,|Z|<|Y|$, the answer is still a "no", it is possible that you have a sequence of infinite cardinals $(X_i)_{i\in ℕ}$ such that $|X_i|>|X_{i+1}|$ for all $i$ and $|X_0|^{|X_0|}>|X_1|^{|X_0|}>|X_2|^{|X_0|}>\cdots>n^{|X_0|}>\cdots>3^{|X_0|}>2^{|X_0|}$, so we have infinitely many infinite cardinals, all smaller or equal to $|X_0|$ and their exponent is never equal
– ℋolo
Mar 29 at 22:40