# In set theories where Continuum Hypothesis is false, what are the new sets?

So, say we are working with non-CH mathematics. This means, AFAIK, that there is at least one set $S$ in our non-CH mathematics, whose cardinality is intermediate between $|\mathbb{N}|$ (card. of naturals) and $|\mathbb{R}|=2^\mathbb{N}$, the continuum.

Question: what kind of objects would we find in this set $S$?

Also: is this mathematics radically different from the one where CH holds?

Specifically, are there results that are used in everyday math , at a relatively introductory level, which do not hold on our non-CH math.?. What results that we find in everyday math would not hold in our new math? Would there, e.g., still exist non-measurable sets? Maybe more specifically: what results depend on the CH?

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The question of what happens when CH fails is, of course, intensely studied in set theory. There are entire research areas, such as the area of cardinal characteristics of the continuum, which are devoted to studying what happens with sets of reals when the Continuum Hypothesis fails.

The lesson of much of this analysis is that many of the most natural open questions turn out to be themselvesd independent of ZFC, even when one wants ¬CH. For example, the question of whether all sets that are intermediate in size between the natural numbers and the continuum should be Lebesgue measure 0, is independent of ZFC+¬CH. The question of whether only the countable sets have continuum many subsets is independent of ZFC+¬CH. There are a number of cardinal characteristics that I mention here, whose true nature becomes apparant only when CH fails. For example, must every unbounded family of functions from ω to ω have size continuum? It is independent of ZFC+¬CH. Must every dominating family of such functions have size continuum? It is independent of ZFC+¬CH. Those question are relatively simple to state and could easily be considered part of "ordinary" mathematics.

However, much of the rest of what you might think of as ordinary mathematics is simply not affected by CH or not CH. In particular, the existence of non-measurable sets that you mentioned is provable in ZFC, whether or not CH holds. (This proof requires the use of the Axiom of Choice, however, unless large cardinals are inconsistent, a result proved by Solovay and Shelah.)

Nevertheless, there is a growing body of research on some sophisticated axioms in set theory called forcing axioms, which have powerful consequences, and many of these new axioms imply the failure of CH. This topic began with Martin's Axiom MAω1, and has continued with the Proper Forcing Axiom, Martin's Maximum and now many other variations.

Lastly, in your title you asked what are the new sets like. The consistency of the failure of the Continuum Hypothesis was proved by Paul Cohen with the method of forcing. This highly sophisticated and versatile method is now used pervasively in set theory, and is best thought of as a fundamental method of constructing models of set theory, sharing many affinities with construction methods in algebra, such as the construction of algebraic or transcendental field extensions. Cohen built a model of ZFC+ ¬CH by starting with a model V of ZFC+CH, and then using the method of forcing to add ω2 many new real numbers to construct the forcing extension V[G]. Since V and V[G] have the same cardinals (by a detailed combinatorial argument), it follows that the set of reals in V[G] has size at least (in fact, exactly) ω2. In particular, the old set of reals from V, which had size ω1, is now one of the sets of reals of intermediate size. Thus, these intermediate sets are not so mysterious after all!

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You might not get new sets -- you might get fewer bijections between the sets you already have! Sometimes getting more cardinals means having fewer sets.

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Indeed, the method of forcing shows that every model of ZFC has a forcing extension where CH holds. That is, one can force CH to hold by adding new functions, just as Adam suggests. In fact, one can turn CH on and off again by moving to ever larger forcing extensions. Nevertheless, every model of ZFC where CH fails contains a model of ZFC+CH, namely, the constructible universe. In this sense, if CH fails it must be that there are new reals and sets of reals over a model of ZFC+CH. –  Joel David Hamkins Dec 31 '09 at 14:08
I love the expression "one can turn CH on and off"! ;) –  Mariano Suárez-Alvarez Dec 31 '09 at 22:55
@Mariano: The technical term for a statement like this, that can be made true and then false again by moving to larger forcing extensions, is a "switch". In contrast, a "button" is a statement that can be made true, in such a way that it cannot become false again in a larger universe. So you can turn a switch on and off as you like, but once you push a button, you cannot unpush it. –  Joel David Hamkins Jan 3 '10 at 2:39
A sequence of statements $\phi_n$ forms a "ratchet", if each is a button, each necessarily implies the previous, and in any model of set theory in which $\phi_n$ is not true, you can force to make $\phi_n$ true without making $\phi_{n+1}$ true. Thus, you can turn the sound louder and louder, but never quieter. –  Joel David Hamkins Dec 20 '14 at 12:53

CH is a very subtle question. To complement the other answers, I'll briefly explain why it is so difficult to see the effects of CH in everyday mathematics.

The cardinals $\aleph_1$ and $2^{\aleph_0}$ are remarkably different even if they are consistently equal. In fact, there is no natural way to compare them. The mere existence of an injection from $\aleph_1$ into $\mathbb{R}$ implies the existence of a non-measurable set. So in the Solovay model (where all subsets of $\mathbb{R}$ are measurable) the two cardinals $\aleph_1$ and $2^{\aleph_0}$ are incomparable. However, CH is still true in the Solovay model in the sense that there are no sets with intermediate cardinality between $\aleph_0$ and $2^{\aleph_0}$.

Although CH is seemingly irrelevant to everyday mathematics, it has found many uses and it is a common tool in some branches of mathematics. Especially in certain areas of functional analysis. It is most often used to construct examples or counterexamples when attempting to generalize a result from the separable case to the non-separable case. Such constructions are usually diagonalisation arguments. As Joel pointed out, these constructions can often be generalized to the non-CH case by the use of forcing axioms such as MA or PFA, or more fine-tuned arguments involving cardinal invariants of the continuum.

The fact remains that CH is a seemingly invisible question for most of mathematics. The deeper reason behind this rests on the fact that the majority of relevant objects of mathematics are either countable or separable. Classes of such objects and questions about them can often be formulated using only first- or second-order quantifiers (i.e., using quantification over natural numbers, real numbers, sets of natural numbers, but not quantification over sets of real numbers or higher type objects) even if the natural formulation is not of this kind. Assuming some large cardinal axioms (which is often not necessary) sets obtained in this way can never be of intermediate cardinality between $\aleph_0$ and $2^{\aleph_0}$. Thus, morally speaking, CH is virtually true for a majority the objects and questions of everyday mathematics.

Of course, this is not the entire story, there are natural examples of objects which have size $\aleph_1$. For example, Ulm invariants in Abelian group theory. However, since the methods that lead to objects of size $\aleph_1$ and $2^{\aleph_0}$ are radically different, the difference between such objects is usually striking and the comparison question does not even arise.

So why is CH a question at all? Well, in all areas of mathematics, when there are two completely different methods of constructing two objects of similar type, it is a completely natural question to ask just how far this similarity goes. In this case, we have two different sets $\aleph_1$ and $2^{\aleph_0}$ which are both uncountable. Comparing them is completely natural. What makes the question both difficult and interesting is that sets have essentially no structure, so there are no obvious ways of distinguishing $\aleph_1$ and $2^{\aleph_0}$.

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Generally, people don't assume CH or not CH, because it's independent. However, a few of the answers in this question might be relevant here. But as a rule, results in ordinary math are going to hold in ordinary math + CH and ordinary math + not CH, so you won't find much along the lines of what you're looking for.

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I think people generally don't assume CH or not CH, not because it is independent of ZFC (after all, the axiom of choice is independent of ZF, and yet it has frequent and widespread use), but because it is not very relevant for most mathematics (unlike AC), as you say. –  Reid Barton Dec 31 '09 at 4:34
That's a good point, though I think that part of it is that it's not relevant because it's independent, and unlike choice, people didn't just think it was true uncritically for awhile. If instead of CH, AoC was a Hilbert Problem of the form "is this true?" I think we'd have less choice-based mathematics, for purely sociological reasons, and if people had just said "Oh, CH is clearly true, here's a proof" and used something that was equivalent but hadn't been proven, we might have more CH-based math. –  Charles Siegel Dec 31 '09 at 4:47

If AC is true, then the reals can be well-ordered. Now look at the set of all reals that have only countably many predecessors in that well-ordering. That set has cardinality $\aleph_1$. If $\aleph_1 < 2^{\aleph_0}$, then there's the sought example.

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A very natural question for algebraists is the Whitehead problem: is there an abelian group A which is not free but for which Ext^1(A, Z) = 0? Shelah showed that this natural problem was independent of ZFC. If you assume CH is false and Martin's axiom is true, then such a group exists. On the other hand, V=L, which implies CH, also implies such a group does not exist.

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Assuming CH is not enough to prove that every Whitehead group of size $\aleph_1$ is free. –  Asaf Karagila Oct 5 '12 at 22:37
Reference for my previous comment [Sh:97] Shelah, Whitehead groups may not be free, even assuming CH. II -- Israel J Math 35 (1980) 257-285. –  Asaf Karagila Oct 5 '12 at 23:58