Cardinality of powers of infinite sets

Question

How to prove that for infinite sets $S,S'$- (i) If $|S'|<|S|$, then $|S^{S'}|=|S|$? (ii) If $|S'|=|S|$, then $|S^{S'}|=|2^S|$? Also what is $|S^{S'}|$ when $|S'|>|S|$?

Answer 1

1. This one isn't true. Indeed, assuming Choice, for any infinite cardinal $\kappa$, we have $\kappa < \kappa^{\text{cf}(\kappa)}$ (this follows immediately from König's theorem), where $\text{cf}$ is cofinality. So let $S$ be any set of singular cardinality $\kappa$, and $S'$ be a set of cardinality equal to $\kappa$'s cofinality.
2. Under the axiom of choice, $$2^\kappa \leq \kappa^\kappa \leq (2^{\kappa})^\kappa = 2^{\kappa \times \kappa} = 2^\kappa$$ so equality holds throughout and $2^\kappa = \kappa^\kappa$. Note, however, that this is pretty much the best that we can say about $2^{\kappa}$: Easton's theorem states strongly that the operation $\kappa \mapsto 2^{\kappa}$ is basically not restricted at all in ZFC on the regular cardinals beyond the simple requirements of monotonicity, Cantor's theorem, and König's lemma.
3. When $\kappa > \lambda$, then $2^\kappa \leq \lambda^\kappa \leq \kappa^\kappa = 2^\kappa$, so $\lambda^\kappa = 2^\kappa$.

Note that in general, cardinal exponentiation is pretty much unspecified in ZFC. However, in the presence of the generalised continuum hypothesis, cardinal exponentiation is extremely precisely specified, and I'll give its behaviour here to demonstrate "the neatest thing you could possibly hope for". These are proved by much the same circular-inequality method as Point 2 above.

• If $\lambda < \text{cf}(\kappa)$, then $\kappa^\lambda = \kappa$. (Here, $\text{cf}$ is the cofinality.)
• If $\text{cf}(\kappa) \leq \lambda \leq \kappa$, then $\kappa^\lambda = \kappa^+$.
• If $\kappa < \lambda$, then $\kappa^\lambda = \lambda^+$.