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.

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

$|A^B|$ and $|B^A|$

$|B^A|$ and $|B|$

My attempt:

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

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!

@LeAnhDung You cannot conclude $|\mathcal P(A)| < |\mathcal P(B)|.$ For instance, it is consistent with ZFC that $2^{\aleph_1} = 2^{\aleph_0}.$ (However, it is also consistent with ZFC that $|A| < |B|\to |\mathcal P(A)| < |\mathcal P(B)|,$ e.g. this holds under the generalized continuum hypothesis.) — spaceisdarkgreen, Oct 13 '18 at 23:12
Hi @spaceisdarkgreen! You meant: 1. $|X|<|Y| \implies |X^A|\le|Y^A|$. Under ZFC, we're unable to conclude $|X^A|<|Y^A|$ from $|X|<|Y|$. 2. $|X|<|Y| \implies |\mathcal {P}(A)|\le |\mathcal {P}(B)|$. Under ZFC, we're unable to conclude $|\mathcal {P}(A)|< |\mathcal {P}(B)|$ from $|X|<|Y|$. Is my understanding of your previous comment correct? — Akira, Oct 13 '18 at 23:15
@LeAnhDung I'm not sure what $X$ and $Y$ are here, so I think we're not quite understanding each other. I was saying that while $$|A|<|B| \implies |P(A)| \le |P(B)|$$ is provable in ZFC, $$ |A| < |B| \implies |P(A)| < |P(B)|$$ is undecidable in ZFC (so in particular you cannot prove it). — spaceisdarkgreen, Oct 13 '18 at 23:19
@LeAnhDung Well, would assume they are. I'm just saying that I was only talking about part one of your problem here and I wasn't sure where you read the first part in what I wrote. It also happens to be true that $|X| < |Y|$ does not imply the strict $|X^A| < |Y^A|$ — spaceisdarkgreen, Oct 13 '18 at 23:24
@spaceisdarkgreen In ZFC, can we conclude 1. $|X| < |Y|\implies |A^X| < |A^Y|$ in case $0<|A|$ and 2. $|B| < |B^A|$ in case $1<|A|$? — Akira, Oct 13 '18 at 23:54
@Akira No, neither of those hold in general. For the first one, what I just said about power sets is just $A=2$ and of course $A=1$ is never true. The second one also fails. For finite $|A|,$ this fails for any infinite $B.$ There are also infinite counterexamples. — spaceisdarkgreen, Oct 14 '18 at 00:03
@spaceisdarkgreen I ask 2. since in Alvin Lepik's answer, he asserted that $|B| = |B|^{1} < |B|^{|A|} = |B^A|$. I'm not sure if this statement is correct or not. — Akira, Oct 14 '18 at 00:09
@Akira It is not in general. Let $B=2^{\aleph_0}$ and $A=\aleph_0.$ — spaceisdarkgreen, Oct 14 '18 at 00:17
Thank you so much for your patience @spaceisdarkgreen! As a result, Alvin Lepik's answer is incorrect. — Akira, Oct 14 '18 at 00:21

spaceisdarkgreen · Accepted Answer · 2018-10-14T03:16:26.997

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: $(\beth_\omega)^{\aleph_0} = 2^{\beth_\omega} = (\aleph_0)^{\beth_\omega}$. )

Thank you so much! Everything gets clearer with your answer. — Akira, Oct 14 '18 at 00:40
In ZFC alone $\beth_\omega^{\aleph_0}=2^{\beth_\omega}$ where $\beth_0=\aleph_0$, $\beth_{n+1}=2^{\beth_n}$, $\beth_\omega=\sum_{n\lt\omega}\beth_n$. https://en.wikipedia.org/wiki/Beth_number — bof, Oct 14 '18 at 02:43
$$\beth_\omega^{\aleph_0}\le\left(2^{\beth_\omega}\right)^{\aleph_0}=2^{\beth_\omega\aleph_0}=2^{\beth_\omega}$$ $$2^{\beth_\omega}=2^{\beth_0+\beth_1+\beth_2+\cdots}=2^{\beth_0}2^{\beth_1}2^{\beth_2}\cdots=\beth_1\beth_2\beth_3\cdots\le\beth_\omega\beth_\omega\beth_\omega\cdots=\beth_\omega^{\aleph_0}$$ — bof, Oct 14 '18 at 03:06

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

1 Answers1