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My question is a little more specific than the title may lead to believe. In the article The set-theoretic multiverse (J.D. Hamkins), the author writes the following:

[...] the continuum is something like a light-switch, which can be turned on and off by moving to ever larger forcing extensions. In each case, the forcing is relatively mild, with the new universes, for example, having all the same large cardinals as the original universe. After decades of experience and study, set-theorists now have a profound understanding of how to achieve the continuum hypothesis or its negation in diverse models of set theory [...]

In a model where $\mathsf{CH}$ is false, would $\aleph_1$ become any more of a concrete object? Would we be able to construct sets with $\aleph_0<\text{card}\,S<\mathfrak{c}$, and if so, how? Or would these models only imply the existence of such sets?

Edit: to make my own question more concrete: if $\neg \mathsf{CH}$, can we construct $S\subset \mathbb{R}$ such that $\text{card}\,S=\aleph_1$, for example?

I realise this is quite a naive question, and that these things are probably not the 'point' of these models, but hopefully my question is not complete nonsense.

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  • $\begingroup$ It's not clear what you mean by "construct." Or what you mean by "any more concrete." Any more than what? $\endgroup$ – Thomas Andrews Jan 4 '14 at 16:38
  • $\begingroup$ @ThomasAndrews Well we can explicitly construct sets of cardinality $\aleph_0,\;2^{\aleph_0},\;2^{2^{\aleph_0}}$. Could similar things be achievable for $\aleph_1$ (for example) if we assumed $\neg \mathsf{CH}?$ $\endgroup$ – user118224 Jan 4 '14 at 16:42
  • $\begingroup$ It's not clear what is more explicit about constructing a set of size $2^{\aleph_0}$ (as the power set of $\omega$) than constructing a set of size $\aleph_1$ (as the set of classes of well-orders of $\omega$). $\endgroup$ – Miha Habič Jan 4 '14 at 16:45
  • $\begingroup$ But a model isn't an axiom system. If you take as your added axiom $\lnot CH$ then does that axiom let you "construct" $\aleph_1$? $\endgroup$ – Thomas Andrews Jan 4 '14 at 16:47
  • $\begingroup$ @MihaHabič Maybe this is not a good question. If $\neg \mathsf{CH}$, could we construct some subset $S$ of the reals such that $|S|=\aleph_1$ for example? $\endgroup$ – user118224 Jan 4 '14 at 16:50
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If we assume the axiom of choice then we can construct a set of every possible cardinality. This is easy because cardinalities are really just particular sizes of ordinals, so we just need to go to the particular ordinal of interest and we have a set of the wanted cardinality.

What is, however, much harder is to point out a set of size $\aleph_1$ of real numbers. This sort of trickery requires the axiom of choice, and in fact, once one is versed enough in this sort of idea, it becomes quite easy. We can construct a surjection from $\Bbb R$ onto an ordinal of size $\aleph_1$, and using the axiom of choice we can construct an injective inverse whose range is such a set.

But, comes the million dollar question, can we write down an injection from the real numbers into an ordinal of size $\aleph_1$? And this, in fact, is the continuum hypothesis.

Assuming the continuum hypothesis holds, or fails, doesn't change our construction of the real numbers, nor it changes our construction of the ordinals. It only changes the answer to the above question. Assuming the continuum hypothesis the answer is positive, and assuming it fails the answer is negative.

It should also pointed out that the negation of equality is merely inequality. So if the continuum hypothesis fails, we did not yet decide the exactly cardinality of the continuum. It might be $\aleph_2$ or it might be $\aleph_{\omega_1}$. If we do know what is the cardinality of the continuum, then by definition we know that there is a bijection between the real numbers and a particular ordinal, restrict it to meet only certain (smaller) ordinals would result in sets of real numbers whose cardinality is between the natural numbers and the real numbers.

But if we don't know the cardinality of the continuum, just that it is not $\aleph_1$, then we can't say much more than "There is a set of real numbers of size $\aleph_1$, and the real numbers themselves are not a set of size $\aleph_1$". However, their construction from Dedekind cut would be exactly the same.

As tomasz suggests in the comments, we can point at a few values which are impossible for the continuum. The first cardinal which has infinitely many infinite cardinals smaller than itself, also known as $\aleph_\omega$, cannot be the cardinality of the continuum. This is due to Koenig's theorem about cardinal arithmetics. The other point of interest is that Cohen proved that the continuum is consistently $\aleph_2$, but this result was extended by Solovay and it was shown that pretty much any value which is not forbidden by Koenig's theorem is consistently the cardinality of the continuum.

I should say a word about the axiom of choice, that without it it might be the case that the real numbers might not be equipotent with an ordinal to begin with, and it might be that there is no set of size $\aleph_1$ of real numbers; it still might be the case that there are intermediate cardinalities between the real numbers and the natural numbers, and it is possible that there are aren't any. But without the axiom of choice a lot of things become harder to work out anyway. Not the construction of the real numbers, though. That's just as simple as with the axiom of choice.

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  • $\begingroup$ It can't be $\aleph_\omega$, though. It might a tad out of scope, but I think mentioning Easton's theorem at least in passing wouldn't be a bad idea. :) $\endgroup$ – tomasz Jan 4 '14 at 17:12
  • $\begingroup$ Perhaps you meant Solovay's theorem. :-) $\endgroup$ – Asaf Karagila Jan 4 '14 at 17:22
  • $\begingroup$ Well, that works, too. ;) $\endgroup$ – tomasz Jan 4 '14 at 22:35
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It would look like a world where Freiling's axiom of symmetry (AX) is true. This axiom says the following: Let A be the set of functions mapping real numbers in the unit interval [0,1] to countable subsets of the same interval. The axiom AX states: For every f in A, there exist x and y such that x is not in f(y) and y is not in f(x). Sierpinski proved that this plausible axiom is equivalent to the negation of CH. The plausibility of AX stems from probabilistic dart-throwing experiments.

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