About $\kappa^\mu = \kappa$, where $\kappa,\mu$ are cardinals.

Question

The question is the following: Let $B$ a non-empty set, is there exists $\kappa$ an infinite cardinal such that $\kappa^{|x|} = \kappa$ for all $x \in B$?, where $|x|$ is the cardinal of $x$. I don't mind the use of Choice Axiom to find such cardinal.

Answer 1

If $B$ contains $\varnothing$, then no. For you would need $\kappa^{\varnothing}=\kappa$, and that is only true for $\kappa=1$, which is finite.

If $B$ does not contain $\varnothing$, then yes. (The fact that you assume every element of $B$ has a cardinality already assumes the Axiom of Choice, by the way). Since $B$ is a set, $\lambda = \sup\{|x|\mid x\in B\}$ exists.

If $\lambda$ is finite, then take $\kappa=\aleph_0$. If $\lambda$ is infinite, let $\kappa=2^{\lambda}$.

Then we have for each $x\in B$ that $1\leq |x|\leq \lambda$, so $$\kappa = \kappa^1 \leq \kappa^{|x|} \leq \kappa^{\lambda} =(2^{\lambda})^{\lambda} = 2^{\lambda\lambda} = 2^{\lambda}=\kappa.$$ Thus, $\kappa^{|x|}=\kappa$, as desired.

Answer 2

Yes, as long as all $x$ are nonempty.

Take some infinite cardinal $\lambda$ bigger than $|x|$ for all $x \in B$.

Then for all $x \in B$, we have $2^\lambda \leq (2^\lambda)^x = 2^{\lambda \cdot x} \leq 2^{\lambda \cdot \lambda} = 2^\lambda$. Therefore, $2^\lambda = (2^\lambda)^x$.