Cardinal Exponentiation Lemma

Question

The following lemma is stated in Kunen's Set Theory:

$\kappa^{\lambda}= 2^{\lambda} = |\mathcal{P}(\lambda)|$ whenever $\kappa,\lambda$ are cardinals with $\lambda$ infinite and $2 \le \kappa \le 2^{\lambda}$. Also, $2^{\aleph_\alpha} \ge \aleph_{\alpha + 1}$ for every ordinal $\alpha$.

I've seen proofs here on stack exchange that show $\kappa^{\lambda}= 2^{\lambda}$ when $2 \le \kappa \le \lambda$ with the following reasoning:

$$2^\lambda \le \kappa^{\lambda} \le (2^\kappa)^\lambda = 2^{\kappa \cdot \lambda} = 2^\lambda.$$ The last equality makes sense because $\kappa \le \lambda$ and $\kappa \cdot \lambda = \text{max}(\kappa, \lambda).$

However, I want to know how to prove this theorem with, instead, the condition that $\kappa \le 2^\lambda.$ Can we use the same chain or inequalities and equalities above? Is the last equation still justified, or do we need a new proof?

Answer 1

Yes, the same thing works: $$2^\lambda \le \kappa^\lambda\le (2^\lambda)^\lambda = 2^\lambda. $$

Only difference is in the middle inequality, you use the assumption $\kappa\le 2^\lambda$ combined with monotonicity rather than Cantor's theorem $\kappa <2^\kappa$ combined with monotonicity.

Answer 2

If we put $\kappa = 2^\lambda$, then the equation $2^{\kappa \cdot \lambda} = 2^\lambda$ becomes $2^{2^\lambda} = 2^\lambda$, which is false by Cantor's Theorem. So the final equation is not justified in this situation.