3

This question is motivated by an answer that I recently gave here.

Basically I wanted to prove that $|\operatorname{End}V|<|\operatorname{End}V^*|$ as soon as a vector space $V$ is big enough.

I boiled it down to showing that for any two cardinals $a,b$, one has $$a<b\implies a^a<b^b$$ As much as this seems reasonable, I know that exponentiation can be tricky so I searched references and wasn't able to find anything. So I'd like to ask:

On what reasonable conditions on cardinals $a,b,c,d$ can one claim that $a^b<c^d$?

And if that's too broad,

Does anyone have a reference to prove that $a<b\implies a^a<b^b$?

Arnaud Mortier
  • 27,276

1 Answers1

5

Well, off the bat this is not provable in general. It is consistent (e.g. assuming various forcing axioms) that $$\aleph_0^{\aleph_0}=\aleph_1^{\aleph_1}=\aleph_2,$$ so here we have $a=\aleph_0$ and $b=\aleph_1$. But it can consistently happen with many other cardinals as well.

There is no easy condition to ensure that this inequality happens, except more or less tautological ones, it is increasing when it is increasing. Specifically, note that $\kappa^\kappa=2^\kappa$ for all infinite cardinals. So we can simply recast the question as when does $\kappa<\lambda$ imply that $2^\kappa<2^\lambda$?

This is an easy consequence of the Generalized Continuum Hypothesis, but can still hold without it. There is no standard name for this axiom, but I like "Injective Continuum Function". Others might prefer "weak GCH".

(Note that if $2^\kappa\leq\lambda$, then it is necessarily the case that $2^\kappa<2^\lambda$, simply by Cantor's theorem. The above point, however, shows that it is not necessary in some cases for this to be true. So the example on the first paragraph cannot work assuming the Continuum Hypothesis, regardless to the value of $2^{\aleph_1}$.)

Asaf Karagila
  • 393,674
  • Thanks for the answer. I feared that when $a$ and $b$ are too close this could happen. It might be that in the case of $|V|$ and $|V^*|$ they are always too close. – Arnaud Mortier Aug 10 '18 at 15:12
  • Well. If $\dim V=\kappa$ then $\dim V^=2^\kappa$. So not that close. But indeed, depending on the cardinality of the field, it might not change the cardinality of $V$ and $V^$ all that much. – Asaf Karagila Aug 10 '18 at 15:13
  • I thought like this first, but then again it is consistent that $2^{\aleph_0}=\aleph_1$, and your example shows that the property fails there. – Arnaud Mortier Aug 10 '18 at 15:14
  • Well, if $2^{\aleph_0}=\aleph_1$, then $\aleph_0^{\aleph_0}=\aleph_1$, and $\aleph_1^{\aleph_1}>\aleph_1$. – Asaf Karagila Aug 10 '18 at 15:15
  • Oh nice, I see - both properties are consistent but not simultaneously. One really needs to be careful. It's good to learn things. – Arnaud Mortier Aug 10 '18 at 15:16
  • Yes. I've added a remark. – Asaf Karagila Aug 10 '18 at 15:19