Given cardinal numbers $A$ and $B$. Does there hold

$$A>B \quad\Rightarrow\quad 2^A>2^B\quad?$$

In other words: is the cardinality of a set uniquely determined by the cardinality of its power set? My first guess was Yes. But then I thought about the continuum hypothesis and that without it there might be a cadinal $\aleph_1$ with $\aleph_0<\aleph_1<\mathfrak c$. I was not sure if $2^{\aleph_1}$ falls nicely between $\mathfrak c$ and $2^{\mathfrak c}$. So it might depend on CH? I have not much experience with cardinal arithmetic, so no clue how to think about this.

  • $\begingroup$ I seem to recall a theorem that $\aleph\mapsto2^\aleph$ is an increasing, but not necessarily strictly increasing "function". Unfortunately, I don't remember exactly the phrasing or any of the proof. $\endgroup$ – Arthur Sep 17 '17 at 11:34
  • $\begingroup$ @Arthur Very interesting! Would be nice if someone with a source or proof could verify this. $\endgroup$ – M. Winter Sep 17 '17 at 11:47
  • 1
    $\begingroup$ See this answer with a link to a detailed article. $\endgroup$ – Mikhail Katz Sep 17 '17 at 12:20
  • $\begingroup$ @Arthur. The axiom system ZFC, of set theory, if consistent, cannot disprove that there are infinite cardinals $A,B$ with $A>B$ and $2^A=2^B.$ This is shown by "Forcing", a method invented by Paul Cohen (...late 1960's .) Not an easy subject. In the 1930's Kurt Godel showed, by using his "constructible class" L, that if ZFC is consistent then it cannot disprove the Generalized Continuum Hypothesis (GCH), which is that $2^A=A^+$ for every infinite cardinal $A,$ where $A^+$ is the least cardinal greater than $A$. $\endgroup$ – DanielWainfleet Sep 17 '17 at 13:22
  • $\begingroup$ @DanielWainfleet Which is why I said "not necessarily strictly increasing" rather than "not strictly increasing". $\endgroup$ – Arthur Sep 17 '17 at 15:36

At "small" cardinals this can indeed fail when CH does not hold, so (in a model where $\text{MA}(\omega_2)$ holds for example), we have that $2^{\aleph_0} = 2^{\aleph_1} = \aleph_2$, while of course $\aleph_0 < \aleph_1$. So consistently it can fail right away, but if CH holds, then $2^{\aleph_0} = \aleph_1 < 2^{\aleph_1}$, and GCH (the generalised continuum hypethesis) will imply that we have $2^{\aleph_\alpha} = \aleph_{\alpha+1} < 2^{\aleph_{\alpha+1}}$, for all $\alpha$ so that it certainly holds for all cardinals $< \aleph_\omega$ in that case.

In fact, Easton's theorem gives us exactly what the only restrictions on the function $f:\kappa \to 2^\kappa$ are for cardinals $\kappa$. Consistently, we can have a lot of weirdness, but $f$ is always increasing, but not necesarily strictly increasing. E.g. even with GCH we can have $2^{\aleph_\omega} = \aleph_{\omega+2} = 2^{\aleph_{\omega+1}}$. This is a hard theorem to prove (but in Jech's set theory or some other books as well). The singular cardinals hypothesis is also relevant to the cardinality of the power set problem.

| cite | improve this answer | |
  • $\begingroup$ Easton's theorem only addresses regular cardinals. It ignores singular cardinals and does not consider large cardinals. We do not have yet a description of the possible behaviors of the exponential in full generality. In fact, in the models obtained by Easton's method the singular cardinal hypothesis holds. $\endgroup$ – Andrés E. Caicedo Sep 17 '17 at 12:38
  • $\begingroup$ @AndrésE.Caicedo Is $f$ being strictly increasing consistent with ZFC? $\endgroup$ – Henno Brandsma Sep 17 '17 at 12:40
  • $\begingroup$ Yes, of course. $\mathsf{GCH}$ gives us that. It is also consistent that the exponential is strictly increasing and $\mathsf{GCH}$ fails everywhere. $\endgroup$ – Andrés E. Caicedo Sep 17 '17 at 12:51

Not the answer you're looking for? Browse other questions tagged or ask your own question.