Suppose that $\aleph_0\le |A|<|B|$. Compare $|\mathcal {P}(A)|$ and $|\mathcal {P}(B)|$, $|A^B|$ and $|B^A|$, $|B^A|$ and $|B|$

Question

Suppose that $\aleph_0\le |A|<|B|$. Compare below cardinalities.

1. $|\mathcal {P}(A)|$ and $|\mathcal {P}(B)|$

2. $|A^B|$ and $|B^A|$

3. $|B^A|$ and $|B|$

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My attempt:

From $|A|<|B|$, I can conclude $|\mathcal {P}(A)|\le |\mathcal {P}(B)|$. But I'm not sure if $|\mathcal {P}(A)|<|\mathcal {P}(B)|$.

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I have no idea how to compare $|A^B|$ and $|B^A|$, $|B^A|$ and $|B|$.

Please shed me some lights! Thank you so much!

Answer 1

For Part $1,$ you cannot prove in ZFC that $\kappa < \lambda$ implies that $2^\kappa < 2^\lambda,$ but you cannot come up with a counterexample using ZFC alone either, since it is true under GCH. It is consistent with ZFC that $2^{\aleph_1}=2^{\aleph_0}$ (this holds under Martin's axiom plus $\lnot$CH).

For part $3,$ similarly, the $\le$ result is clear, but you can't prove the strict inequality. This time a counterexample is available in ZFC alone. For $B=2^{\aleph_0}$ and $A=\aleph_0,$ we have $$ (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0\cdot\aleph_0} = 2^{\aleph_0}.$$

For part $2$, for $\kappa < \lambda,$ we have $$\lambda^\kappa\le 2^{\lambda\cdot \kappa}=2^\lambda \le \kappa^\lambda.$$ The rightmost inequality is actually an equality since $\kappa^\lambda \le 2^{\kappa\cdot \lambda}= 2^\lambda$. So we really want to compare $\lambda^\kappa$ to $2^\lambda.$ Instances of equality are again consistent with ZFC. For instance, under GCH, $\aleph_\omega^{\aleph_0} = \aleph_\omega^+= 2^{\aleph_\omega}.$ (And bof in the comments shows we don't need GCH for a counterexample. The beth equality that reduces to the aleph equality I got from cardinal arithmetic in GCH is provable in ZFC: $(\beth_\omega)^{\aleph_0} = 2^{\beth_\omega} = (\aleph_0)^{\beth_\omega}$. )