I've never seen a text that includes GCH in the ZFC axioms. I presume this means that GCH has not achieved widespread acceptance. This seems surprising to me, given that:

  1. The cardinal numbers encountered in "ordinary" mathematics are the beth numbers $\beth_\alpha$, and
  2. ZFC can't prove much about the beth numbers without at least GCH or something stronger, and
  3. ZFC+GCH has been shown equiconsistent with ZFC.

So my question is, why hasn't GCH achieved more widespread acceptance (or has it)?

  • $\begingroup$ Few people have any intuition either way, and to many of us it seems implausibly neat. (And $\omega_1$, as distinct from $2^\omega$, does occur fairly naturally in mathematics.) $\endgroup$ – Brian M. Scott Mar 31 '13 at 4:47
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    $\begingroup$ Because set theorists are interested in investigating aspect of set theory where $GCH$ fails. It is the same reason why the foundation axioms $\textit{is}$ part of ZFC. Foundation is not particularly useful for ordinary mathematics, but quite helpful in studying models of set theory. $\endgroup$ – William Mar 31 '13 at 5:34

Unfortunately, I do not feel the arguments you mention are strong enough to make a convincing case. Before addressing $\mathsf{GCH}$ proper, note that in fact:

  1. $\omega_1$ and other cardinals do appear in mathematics. For example, transfinite iterations of length $\omega_1$ are not uncommon in analysis (a simple instance of this is that the Borel sets can be presented as the union of an increasing chain of length $\omega_1$, and this representation is useful beyond simply saying that they are the $\sigma$-algebra generated by the open sets). The cardinal $\omega_1$ appears naturally in descriptive set theory (tied up to the boldface $\Pi^1_1$ sets) and other cardinals (not necessarily beth numbers) appear also in this context.
  2. $\mathsf{ZFC}$ can prove a great deal about beth numbers without any restrictions in the cardinal arithmetic. For example, as pointed out in this answer, the beth numbers appear naturally when studying the spectrum problem in model theory. That answer also mentions one of Shelah's best known results in cardinal arithmetic, that $\aleph_\omega^{\aleph_0}$ is either the continuum, or less than $\aleph_{\omega_4}$. These results are invisible under strong cardinal arithmetic restrictions. Cardinal arithmetic in fact has a lot to say when we do not "trivialize" it by assuming $\mathsf{GCH}$ or the like. For example, there are results that are known to hold from $\beth_\omega$ on but may fail for an initial segment of the cardinals. The fact that we do not have as extensive a list of results without strong cardinal arithmetic restrictions is simply because such restrictions were typically adopted in classical results, and only relatively recently we have been exploring the landscape without them. There is also the obvious fact that the study of cardinal invariants of the continuum disappears if we adopt $\mathsf{CH}$; there is significant structure here, provably in $\mathsf{ZFC}$, but $\mathsf{CH}$ hides it.
  3. Equiconsistency does not seem a good argument for the adoption of an axiom. Replacement makes $\mathsf{ZFC}$ strictly stronger than $\mathsf{ZC}$, and it is this strength that makes it a desirable axiom to adopt. Large cardinal axioms go beyond $\mathsf{ZFC}$, and are routinely taken for granted when studying the projective hierarchy or determinacy. In fact, the ladder of consistency strength hierarchy is one of the most remarkable mathematical objects we have uncovered. One could argue instead that extensions of $\mathsf{ZFC}$ should be as strong as possible to accommodate all mathematical discourse, that $\mathsf{ZFC}$ is in strength too weak to do this properly.

With that out the way, we need to consider what it is that we expect to accomplish with extensions of $\mathsf{ZFC}$. For example, we may want a rich mathematical universe, with as many examples as possible, even at the level of the reals. $\mathsf{CH}$ certainly gives us that: Under this assumption, there are many non-isomorphic linear orders of size $\omega_1$, there are discontinuous homomorphisms between Banach algebras, etc. Analysts tend to adopt $\mathsf{CH}$ for this reason. When we want instead a very orderly universe, with strong classification results, alternative axioms are adopted, usually in the form of forcing axioms. I elaborate this to some extent, and provide some references, here.

It is true that quite a few set theorists see untamed cardinal arithmetic as a pathology rather than a virtue. (Shelah is a notorious exception.) Even so, $\mathsf{GCH}$ seems unforgivably restrictive, typically because it is fairly easy to violate it with mild forcing. For this reason, $\mathsf{SCH}$, the singular cardinal hypothesis, is considered a more reasonable statement to desire. (I am not doing this here, but) one could even make the case that a version of this statement should be part of whatever extension of $\mathsf{ZFC}$ we end up agreeing upon in the future. Unlike $\mathsf{GCH}$, we have that $\mathsf{SCH}$ is typically preserved by forcing. It is in fact seen as "cofinally true" in the universe, since it holds beyond strongly compact cardinals. It is implied by a variety of reflection principles and forcing axioms, so it is somewhat expected to hold under the heuristic that the universe of sets should be both as tall and as wide as possible.

On the other hand, one might expect that the universe of sets should be close to a fine-structural model. Models possessing a fine structure satisfy strong versions of $\mathsf{GCH}$, so this is not by any means an unreasonable axiom to aspire to. In fact, Woodin's recent "ultimate $L$" program aims at developing a theory of the universe that would be compatible with all large cardinals and would possess fine-structural features. If this program succeeds, we would have a template from which we could prove the consistency of any reasonable statement by means of forcing, so the universe could be seen as a forcing extension of a very well-behaved fine structural core. Any other competing "completion" is expected to be mutually interpretable with this alternative.

Now, the mention of mutual interpretability brings us to an interesting point. The argument can be made that there may not be a "distinguished" completion of $\mathsf{ZFC}$, that any two theories are equally desirable if they are mutually interpretable. (A multiverse view, if you wish.) This recent paper by John Steel elaborated on this view. Under this approach, the question of whether $\mathsf{GCH}$ should be adopted is moot.

Even if we do not follow that route, I do not think we have strong evidence to advocate fully for $\mathsf{GCH}$. However, one can make the case that $\mathsf{CH}$ is a desirable axiom to have. Woodin has proved a maximality result, showing that $\mathsf{CH}$ "decides" all $\Sigma^2_1$ statements. (Somewhat more formally, under appropriate large cardinals, any such statement that can be forced actually holds in the presence of $\mathsf{CH}$. Note that $\mathsf{CH}$ itself is a $\Sigma^2_1$ statement.)

On the other hand, a few years ago, Woodin presented a program that seemed to conclude on fairly general grounds that $\lnot\mathsf{CH}$ was the reasonable option. Oversimplifying, the argument was: From large cardinals, we have a "complete" theory of $H(\omega_1)$ and, in particular, its theory cannot be modified by set forcing under large cardinal assumptions. Similarly, we can have a "complete" theory of $H(\omega_2)$; for example, Woodin introduced a principle $(*)$ that implies this, and implies the negation of $\mathsf{CH}$ (again, under large cardinals). In contrast, it was claimed that no such "complete" theory exists compatible with $\mathsf{CH}$. However, a flaw was found in the proof of this claim, so the argument for $\lnot\mathsf{CH}$ on these grounds has lost strength.

Nowadays, advocates of the negation of $\lnot\mathsf{CH}$ are willing to give up the "conditional maximality" that $\mathsf{CH}$ provides. But typically one does not argue for $\lnot\mathsf{CH}$ directly, but rather for principles that (usually for technical reasons) end up implying $\lnot\mathsf{CH}$.

My take on $\mathsf{GCH}$, such as it is, is that first of all, we are far from having enough mathematical evidence to make a decision. And second, whatever we end up adopting should not be directly this axiom (or its negation), but whether it holds or not should be a consequence of deeper structural principles, and the precise nature of those principles is not yet developed to the point where we can venture a guess. I confess I am partial to large cardinals, and to strong reflection principles, so I am not opposed to $\mathsf{SCH}$. (But note that I am not advocating for large cardinals or forcing axioms or the like here, I am just pointing out that arguments in their favor have been advanced, and I find them more compelling and encompassing than the arguments I am aware of for their negation.) I find the structure of the continuum provided by forcing axioms somewhat more appealing than the wild untamed riches brought about by $\mathsf{CH}$.

But, again, as I said, whether strong cardinal arithmetic restrictions end up being part of the accepted axiomatization of set theory, I expect that this will be a consequence of other principles rather than just adopted as an axiom on its own.


I heard about one of the famous set theorists from the 1960's (but I don't recall the exact name) who was almost certain that $V=L$ is going to be accepted as an axiom "soon enough", because it seemed so natural at the time.

But then came Cohen and showed that it's very easy to destroy $V=L$ by adding new sets, and forcing was developed.

Once forcing was developed it seems very hard to add new axioms into the theory because you want them to be preserved in every "standard model" of the theory. If forcing can destroy $\sf GCH$, then it's not a good fit for a standard axiom. On the other hand it is impossible to contradict the axiom of choice by forcing, one has to preform another additional step of generating a certain inner model.

This is even more apparent in the case of $\sf GCH$, or in fact just $\sf CH$. Once research has been put into forcing axioms, people realized that axioms like $\sf MA$ are very useful and interesting when $\lnot\sf CH$ is assumed. And in fact, strengthening $\sf MA$ would usually result in $2^{\aleph_0}=\aleph_2$. So many of the forcing axioms that we know today in fact prove that $\sf CH$ fails.

And of course there are all the results from PCF theory which will be pointless once you assume $\sf GCH$. This is like assuming that all functions are polynomials, or analytic. Sure, it makes things easier and for people who actually use the theory (e.g. engineers) it would make like much simpler - and they hardly ever care about the rest of the cases anyway. But for an analyst this would be boring.

Similarly for a set theorist, $\sf GCH$ would be boring. Sure, it can simplifies some work for non-set theorists, and when people say "But I thought $2^{\aleph_0}$ was defined as $\aleph_1$" they would be wrong, but not entirely wrong; but regardless to that, we still prefer something which generates interest rather than something that does not.

  • $\begingroup$ Gödel was the only famous theorist who had this certainty? $\endgroup$ – MphLee Mar 31 '13 at 10:20
  • $\begingroup$ What certainty? $\endgroup$ – Asaf Karagila Mar 31 '13 at 10:51
  • $\begingroup$ That $L$ was the universe of set thoery. $\endgroup$ – MphLee Mar 31 '13 at 12:40
  • $\begingroup$ Godel himself wrote on one occasion or another, that it seems unlikely that $\sf CH$ is true in a Platonic sense of truth. $\endgroup$ – Asaf Karagila Mar 31 '13 at 12:41
  • $\begingroup$ I think the position expressed in the penultimate paragraph is a little problematic. If all functions were polynomials, then the exponential function would not exist, and $\frac{d}{dx}$ would not have a fixed point, thereby breaking quantum mechanics. If all functions were analytic, then $\mathrm{max}(g,f)$ usually would not always exist, where $g$ and $f$ denote functions of type $\mathbb{R} \leftarrow \mathbb{R}$, thereby making the language of optimzation theory that much more clunky and unwieldy. Etc. $\endgroup$ – goblin GONE Nov 18 '15 at 3:53

There are strong arguments both for and against the Continuum Hypothesis. See, for example, Believing the Axioms by Maddy (J. Symb. Logic, 1988), which has a long discsusion of the arguments in both directions.

For me, the most surprising and interesting argument against CH (and GHC) is Freiling's argument. This involves an intuitive refutation of CH by "throwing darts at the number line". If you've never seen it before, it will convince you that CH is not obviously true.

  • $\begingroup$ Welcome back!${}$ $\endgroup$ – Asaf Karagila Mar 31 '13 at 8:36
  • $\begingroup$ @AsafKaragila Thanks! $\endgroup$ – Jim Belk Mar 31 '13 at 23:09

It hasn't been accepted as an axiom mainly because set-theorists are not convinced of its truth. It's relative consistency with ZFC is a weak condition that only says that the addition of it as an axiom would not create contradictions (if ZFC is consistent), but this is not a very convincing argument for considering it as an obviously true statement about the universe of sets.

Even more, I think a majority of set theorists of a Platonistic bent believe in that the Continuum Hypothesis itself is a false statement. ($\aleph_2$ generally being considered the likely real cardinality of the continuum.)

Slight Addition: The are many simpler statements than GCH that seem like they should be absolutely true but are not accepted as basic axioms of set theory. Perhaps most astounding is the following:

If $\kappa < \lambda$ are cardinals, then $2^\kappa < 2^\lambda$.

(Consider the comments to this answer by Joel David Hamkins on MathOverflow.)

  • $\begingroup$ Why do you say that $\aleph_2$ is the real cardinality of the continuum? $\endgroup$ – Baby Dragon Mar 31 '13 at 6:52
  • $\begingroup$ @BabyDragon: I don't say that it is. But a Platonist would believe that the continuum has an independent existence with a well-defined cardinality among the $\aleph$-numbers. The most commonly held beliefs about this cardinality is that it is either $\aleph_1$ or $\aleph_2$. $\endgroup$ – user642796 Mar 31 '13 at 7:29
  • $\begingroup$ I suppose my question should have been phrased as "why is $\aleph_2$ generally considered the likely cardinality of the continuum". Is it simply because that $\aleph_2$ is right after $\aleph_1$ and CH is widely believed to be false? $\endgroup$ – Baby Dragon Mar 31 '13 at 7:46
  • $\begingroup$ @BabyDragon: Because many forcing axioms prove the equality $2^{\aleph_0}=\aleph_2$. $\endgroup$ – Asaf Karagila Mar 31 '13 at 7:47
  • $\begingroup$ @AsafKaragila Thank you. $\endgroup$ – Baby Dragon Mar 31 '13 at 7:48

Besides the fact that GCH has not been proven, modern mathematics tends to look at hierarchical tiers of mathematics of axioms and seeing what mathematical theorems can then be proven as opposed to lumping axioms together. I highly recommend Simson's Subsystems of Second Order Arithmetic for an exposition of this way of thinking.

  • $\begingroup$ The axioms considered in Simpson's book are also not proven. For example $WKL$ can not be proved using $\text{RCA}_0$. There are models of $\text{RCA}_0$ where $\neg WKL$ holds. In the same regard, $GHC$ can not be proved in $ZFC$, but it and its negation are relatively consistent with ZFC. What do you mean by "$GCH$ has not been proved"? Is something provable if and only if it can be proved using $ZFC$? $\endgroup$ – William Mar 31 '13 at 5:29
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    $\begingroup$ Note that GCH has been proven independent of ZFC. $\endgroup$ – goblin GONE Mar 31 '13 at 5:33
  • $\begingroup$ William, my point is that because GCH is not proven in ZFC it can be seen to exist at a higher level of mathematics, which is what the question is about. $\endgroup$ – Daniel Geisler Mar 31 '13 at 5:37
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    $\begingroup$ I think you mean Simpson, not Simson. $\endgroup$ – Asaf Karagila Apr 13 '13 at 0:23

And here is a more light-weight broad discussion from Scientific American.Dispute over Infinity Divides Mathematicians

And this one is a little bit of circumstantial evidence for CH in that a statement that was expected to witness ~CH ended up disproved Mathematicians Measure Infinities and Find They’re Equal

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    $\begingroup$ I don't see how the Malliaris-Shelah result can be understood as providing circumstantial evidence for CH. There was never any possibility of $\mathfrak{p}\not=\mathfrak{t}$ being provable (from ZFC), since $\mathfrak{p}=\mathfrak{t}$ is a consequence of CH. Malliaris and Shelah showed that even in models where CH fails, we still have $\mathfrak{p}=\mathfrak{t}$, but this has absolutely no bearing on the plausibility of CH unless we think that $\mathfrak{p}$ and $\mathfrak{t}$ should be far apart if possible. But there is no real reason to think this. $\endgroup$ – Noah Schweber Jan 9 '18 at 5:27
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    $\begingroup$ That is, just because "$\mathfrak{p}\not=\mathfrak{t}$" would witness the failure of CH, doesn't mean that its impossibility should be taken as evidence for CH. $\endgroup$ – Noah Schweber Jan 9 '18 at 5:31
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    $\begingroup$ Yes, this seems completely irrelevant, and misguided. Another example is the fact that any uncountable closed set of reals has the same size as the reals, that is, "the continuum hypothesis holds for closed sets". This has no relevance for $\mathsf{CH}$ proper either. Suggesting otherwise indicates a misunderstanding more than anything else. $\endgroup$ – Andrés E. Caicedo Jan 9 '18 at 5:44
  • $\begingroup$ Guilty as charged $\endgroup$ – Mark Kortink Jan 11 '18 at 6:49
  • $\begingroup$ Here is the paragraph that is relevant to witnessing ~CH "This past July, Malliaris and Shelah were awarded the Hausdorff medal, one of the top prizes in set theory. The honor reflects the surprising, and surprisingly powerful, nature of their proof. Most mathematicians had expected that p was less than t, and that a proof of that inequality would be impossible within the framework of set theory. Malliaris and Shelah proved that the two infinities are equal. Their work also revealed that the relationship between p and t has much more depth to it than mathematicians had realized. " $\endgroup$ – Mark Kortink Jan 12 '18 at 20:16

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