# When $2^\alpha = 2^\beta$ implies $\alpha=\beta$ ($\alpha,\beta$ cardinals)

Sorry if this is a silly question. I was wondering, under what axioms of set theory is it true that if $\alpha$,$\beta$ are cardinals, and $2^\alpha=2^\beta$, then $\alpha=\beta$? Do people use these conditions to prove interesting results?

This question is prompted from a recent perusing of Johnson's "Topics in the Theory of Group Presentations", where in the first few pages he "proves" free groups of different rank are non-isomorphic: the number of mappings from a free group of rank $\omega$ to the group $\mathbb{Z}/2\mathbb{Z}$ is $2^\omega$, which would be invariant under isomorphism; and then he assumes the topic of my question: $2^\alpha=2^\beta$ implies $\alpha=\beta$.

But I remember reading something a few years ago about a student of R.L. Moore (I only remember his last names was Jones) "proving" the Moore Space conjecture, and using that $\alpha > \beta$ implied $2^\alpha > 2^\beta$, but that this was incorrect.

Anyway, I realized I don't know anything about when this is true or false, so I thought I'd ask.

• I don't know the answer to the set theory question. But if you want to prove that isomorphic free groups have the same rank, you can just pass to the abelianization, then tensor to Q, and reduce to the well-known fact that any two bases of the same vector space have the same cardinality. – Pete L. Clark Mar 5 '10 at 2:14
• Yeah, in fact the proof in Johnson's newer book is not flawed like this one is. I personally don't tensor with Q; If A is the abelianization, I look at A/2A :) – Steve D Mar 5 '10 at 2:16
• I am better with Google than with cardinal arithmetic :-) math.niu.edu/~rusin/known-math/99/luzin_easton – Carsten S Mar 5 '10 at 2:18
• See also this answer mathoverflow.net/questions/1924/#6594 and this related question mathoverflow.net/questions/12943, on the symmeric groups on sets of different sizes, when 2^a = 2^b but a not = b. – Joel David Hamkins Mar 5 '10 at 4:31
• See also math.stackexchange.com/a/420484/462. – Andrés E. Caicedo Jun 15 '13 at 3:23

François gives the correct affirmative answer. For the negative side, the usual method of proving that the negation of the Continuum Hypothesis is consistent with ZFC is to use the method of forcing to add Aleph2 many Cohen reals, so that 2ω = ω2 in the forcing extension V[G]. In this model V[G], it is also true that 2ω1 = ω2. Thus, this model shows that it is not necessarily true that different-sized sets have different sized power sets. The case of symmetric groups is likely more interesting than free groups, because in this model, the symmetric groups Sω and Sω1 have the same cardinality ω2. Nevertheless, these two groups are not isomorphic, as explained in this MO question.

The general answer about what can be true for the continuum function κ --> 2κ is exactly provided by Easton's Theorem. This remarkable theorem states that if you have any class function E, defined on the regular cardinals κ, with the properties that

• κ ≤ λ implies E(κ) ≤ E(λ)
• κ < E(κ)
• κ < Cof(E(κ))

then there is a forcing extension V[G] in which 2κ = E(κ) for all regular cardinal κ. In particular, this shows that the sizes of the power sets (on regular cardinals) are restricted to obey only and exactly the hypotheses listed explicitly above. Each of these properties corresponds to a well-known fact about cardinal exponententiation.

Using Easton's theorem, we can build models of set theory where 2κ = κ++ for all regular κ. The added power of the Woodin/Foreman result mentioned by François is that they also get this for singular cardinals.

The point now is that there are innumerable examples provided by Easton's theorem that satisfy your hypothesis that the continuum function is one-to-one. If one begins with a model of GCH and selects any injective Easton function E, then the resulting model of set theory V[G] will have E as it's continuum function κ --> 2κ = E(κ) for regular κ, and the GCH will continue to hold at singular κ, preserving injectivity. So one is quite free to satisfy your hypothesis while having any kind of crazy failures of GCH.

• +1: Excellent answer! I was going to fill in some of the gaps in my answer later, but I'm happy you did it instead since your answers are always much richer than mine. – François G. Dorais Mar 5 '10 at 3:17
• Well, I don't know about that, François, since your answers are usually excellent. In this case, you gave the positive answer, and I spoke about the negative side, where the hypothesis fails. But I suppose Easton's theorem goes both ways, doesn't it? You can use it either to make positive or negative instances of the phenomenon. – Joel David Hamkins 0 secs ago. – Joel David Hamkins Mar 5 '10 at 3:45
• And you completely avoid the large cardinal hypothesis in Woodin's Theorem, which is much more satisfactory. – François G. Dorais Mar 5 '10 at 3:56
• I have to say, I will probably pick this as the answer. I will wait for a little bit longer to see if anything drastic changes; I also wish I could pick all 4 answers as best! – Steve D Mar 5 '10 at 5:06

The Generalized Continuum Hypothesis, i.e. $2^\kappa = \kappa^+$ for every infinite cardinal $\kappa$, implies the desired result, but it can hold in other circumstances. For example, Woodin has shown that it is relatively consistent with the existence of a supercompact cardinal that $2^\kappa = \kappa^{++}$ for every infinite cardinal $\kappa$. (See Foreman and Woodin, The generalized continuum hypothesis can fail everywhere, Annals of Mathematics 133, 1991, 1-35.)

The statement $2^{\aleph_0} < 2^{\aleph_1}$ is known as the Weak Diamond Principle. This principle was introduced by Devlin and Shelah (A weak version of $\diamondsuit$ which follows from $2^{\aleph_{0}}<2^{\aleph_{1}}$, Israel J. Math. 29, 1978, 239-247). This principle has been very useful as a substitute for stronger enumeration principles such as the Continuum Hypothesis and Jensen's Diamond Principle $\diamondsuit$. I think the Weak Diamond Principle was inspired by Shelah's work on the Whitehead Problem (which may be related to your group theoretic application).

• Note that Woodin's Theorem, remarkable as it may be, is overkill here. As Joel Hamkins pointed out in his answer, the classical forcings of Easton and Cohen can be used to force that $2^\kappa=\kappa^{++}$ for $\aleph_0 \leq \kappa < \aleph_\omega$ and $2^\kappa = \kappa^{+}$ when $\kappa \geq \aleph_0$, for example. This does not require the use of a supercompact cardinal, but large cardinals are necessary to fully control the sizes of the powersets of singular cardinals. – François G. Dorais Mar 5 '10 at 4:10

One thing that is made very easy by assuming that $2^\alpha < 2^\beta$ whenever $\alpha <\beta$ is showing that for sets $S_1$ and $S_2$ of different cardinalities, the Banach spaces $\ell_\infty(S_1)$ and $\ell_\infty(S_2)$ are not isomorphic. This follows immediately from the fact that for an infinite set $S$, the smallest possible cardinality of a dense subset of $\ell_\infty(S)$ is $2^{|S|}$.

If one does not assume that $2^\alpha = 2^\beta$ implies $\alpha =\beta$, then one must revert to some other argument. I have not thought about this problem at length, so there may be much simpler reasons why $\ell_\infty(S_1)$ and $\ell_\infty(S_2)$ cannot be isomorphic when $S_1$ and $S_2$ have different cardinalities (without assuming $2^\alpha = 2^\beta\Rightarrow\alpha =\beta$), but one way is to instead consider the least possible cardinality of a norm dense subset of an arbitrary weakly compact subset of $\ell_\infty(S)$. In this case, a result of Haskell Rosenthal asserts that for $S$ infinite and a weakly compact set $W \subseteq \ell_\infty (S)$, $W$ admits a norm dense subset of cardinality no greater than $|S|$. On the other hand, a standard argument shows that $\ell_2 (S)$ (in fact, any Banach space of density character not exceeding $|S|$) is isomorphic to a subspace of $\ell_\infty (S)$ (generalise the standard argument that every separable Banach space embeds into $\ell_\infty$). Thus, if $\alpha$ and $\beta$ are cardinals with $\alpha <\beta$, then $\ell_2 (\beta )$ is isomorphic to a subspace of $\ell_\infty (\beta )$, but not to a subspace of $\ell_\infty (\alpha)$. Hence $\ell_\infty (\alpha )$ and $\ell_\infty (\beta)$ are not isomorphic.

The result of Rosenthal cited above is from his paper "On injective Banach spaces and the spaces $L_\infty (\mu )$ for finite measures $\mu$".

François's and Joel's answers to the general question are correct.

As for the "dimension of a free group is uniquely determined, even if it's infinite", there's a better argument. Let G be a free group on two sets of generators x and y; we wish to show |x| = |y|. First let A = Gab be the free abelian group on generators x and y; then let V = AZ F2 be the vector space over F2 with bases x and y, so that |V| = 2|x| = 2|y| (strike that, it's false for infinite cardinalities). (We could use any field here, but the argument is more intricate.)

If |V| is finite, then |x| = |y| follows immediately. So suppose |V| is infinite, so that |x| and |y| are infinite. Each element of x is the sum of finitely many elements of y, and every element of y must show up in at least one such sum (why?). This means |y| ≤ |x|•ℵ0 = |x| (cardinal arithmetic). Similarly |x| ≤ |y|, so |x| = |y|.

• Right. This has already been said in the comments following the question. – Pete L. Clark Mar 5 '10 at 5:56
• Fair enough, and duly upvoted. I'm pretty sure those comments weren't there at the time, but maybe I missed them. – Chad Groft Mar 6 '10 at 4:42