It's well known that, if GCH holds, then every cardinal can be well-ordered. However, I'm sure we don't need full power of GCH to prove it for specific cardinal, e.g. continuum. I have been wondering, what is the minimum amount cardinal numbers $A$ for which non-existence of cardinal $B$ with $A<B<2^A$ guarantees that $\frak{c}$ can be well ordered (or, simplier, GCH for which cardinals already implies existence of well-ordering of $\frak{c}$)? By looking through the proof I think it's enough for it to hold for $A\in\{\mathbb{R},2^\mathbb{R},2^{2^\mathbb{R}}\}$.

Thanks in advance for help!

up vote 26 down vote accepted

Yes, you are right. This is a theorem of Specker. If there are no intermediate cardinals between $A,\mathcal P(A)$ and $\mathcal{P(P(}A))$, then $A$ can be well-ordered.

You can find nice details in:

Akihiro Kanamori, David Pincus, "Does GCH imply AC locally?, in "Paul Erdős and his mathematics, II (Budapest, 1999)", Bolyai Soc. Math. Stud., 11, János Bolyai Math. Soc., Budapest, (2002), 413–426.

Link to the paper on Kanamori's website.

And on this relevant thread.

I had discussed this answer with Yair Hayut earlier this evening, and then I noticed the following peculiar thing.

Suppose that $\sf CH$ holds for $\Bbb R$ and $\aleph_1$, then $2^{\aleph_0}=\aleph_1$ or $2^{\aleph_0}=\aleph_2$. In either case, it is well-orderable.

  1. First we observe that since there is a definable surjection from $\Bbb R$ onto $\omega_1$, it is necessarily the case that $\aleph_1$ and $2^{\aleph_1}$ are both $\leq2^{2^{\aleph_0}}$, with $\aleph_1$ being strictly smaller. Therefore $\aleph_1\leq2^{\aleph_0}\leq2^{\aleph_1}\leq2^{2^{\aleph_0}}$, but our assumptions tell us it is impossible that all three are strict inequalities. Either the middle one is equality, or both the left and right ones are equality (but not both, of course).

  2. If $2^{\aleph_0}<2^{\aleph_1}$, then we have that $\aleph_1\leq2^{\aleph_0}<2^{\aleph_1}$, and therefore $2^{\aleph_0}=\aleph_1$.

  3. If $2^{\aleph_1}=2^{\aleph_0}$ then $\aleph_2<2^{2^{\aleph_0}}$ (since there is a definable surjection from $2^{\omega_1}$ onto $\omega_2$, so by this assumption there is a surjection from $\Bbb R$ onto $\omega_2$). If $2^{\aleph_0}$ and $\aleph_2$ are incomparable then we get $2^{\aleph_0}<2^{\aleph_0}+\aleph_2<2^{2^{\aleph_0}}$ which is a contradiction (this is the same argument by $\aleph_1\leq2^{\aleph_0}$). Therefore $\aleph_2\leq 2^{\aleph_0}$. If they are equal, wonderful.

    Otherwise, $\aleph_1<\aleph_2<2^{\aleph_0}=2^{\aleph_1}$ which is a contradiction to our assumption that $\sf CH(\aleph_1)$ holds.

So while it doesn't decrease the number of sets which required here, it does give a nice twist on the things. I'm not sure if we can get rid of the assumption of $\sf CH(\aleph_1)$, maybe at the cost of requiring $\sf CH(\aleph_2)$, and to remove that assumption we would need to go on and on. I'm not sure what happens at the limit, though.

Therefore, this raises a nice question:

Suppose that $\aleph(\Bbb R)=\aleph^*(\Bbb R)=\aleph_\omega$, does this imply that $\sf CH(\Bbb R)$ fails, or is it consistent that $\sf CH(\Bbb R)$ holds? What if we replace $\omega$ by an arbitrary limit ordinal?

  • I thought you will also need no cardinal between P(P(A)) and P(P(P(A))) because, I believe, latter is the upper bound on Hartogs' number of A. Also, thanks for link to the paper! – Wojowu Jul 2 '14 at 8:36
  • 2
    As you can read in the paper, this was the original result by Sierpinski, but Specker improved it. – Asaf Karagila Jul 2 '14 at 8:37
  • I got lost in point 2. $2^{2^{\aleph_0}}=2^{2^{\aleph_1}}>2^{\aleph_1}$, right? – Emil Jeřábek Jul 2 '14 at 19:59
  • What is $\aleph^*(\mathbb{R})$? – Wojowu Jul 2 '14 at 21:07
  • 1
    @Wojowu: The least ordinal such that there is no surjection from the real numbers onto it. (What I like to call the Lindenbaum number of a set.) – Asaf Karagila Jul 3 '14 at 1:15

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