Additionally, is there any intuitive way to visualize the cardinalities that result?
$\begingroup$
$\endgroup$
2
-
1$\begingroup$ I don't understand what you mean by "look like". The first uncountable ordinal will have this property, but I'm not sure if that's what you're looking for. $\endgroup$– Richard DoreCommented Oct 16, 2009 at 16:17
-
$\begingroup$ To expand on Richard's comment: if CH is false, then the set of all countable ordinal numbers will be a relatively "natural" set with cardinality between aleph-0 and the continuum. Though we could debate all day about whether or not anybody really knows what this set "looks like"... $\endgroup$– John GoodrickCommented Oct 16, 2009 at 20:07
Add a comment
|
10 Answers
$\begingroup$
$\endgroup$
2
I think your reading is wrong. Set theorists have studied all sorts of additional axioms, some implying CH, some being strictly weaker than CH, and many contradicting CH. My understanding is that most set theorists today, if they have any opinion on the matter, prefer to think that CH is false. In particular, Woodin has recently advanced a philosophical argument for a strong additional axiom of set theory that implies that the continuum is $\aleph_2$.
However, the fact of the matter is that CH has very little effect on "ordinary mathematics". You can come up with a few down-to-earth seeming combinatorial statements that are equivalent to CH, but it really just never comes up if you never deal with objects that are either very infinite or are infinite and don't have a lot of structure attached to them. For reference, it is consistent for the cardinality of the real numbers to be almost any uncountable cardinality (the only requirement is that it have uncountable cofinality.
I don't have time now, but if you want more references on this I can try and add some later.
answered Oct 16, 2009 at 16:02
-
2$\begingroup$ You're right that CH seems to have very little effect on, say, modern algebraic geometry. But CH (or rather the Generalized Continuum Hypothesis) does have some very nice consequences for the study of so-called Abstract Elementary Classes ("AECs"), where the goal is to generalize Morley's Theorem in first-order model theory to the setting of infinitary logics (more or less). Shelah and others have recently proved some interesting results on this topic. $\endgroup$ Commented Oct 16, 2009 at 20:23
-
3$\begingroup$ Well, but you're talking about consequences in Model Theory here, and I think Eric is sticking to the convention (customary among logicians) that "ordinary mathematics" means "all fields of mathematics except logic". $\endgroup$ Commented Dec 8, 2013 at 18:52
$\begingroup$
$\endgroup$
1
Reading the comments, I just realized that part of your question asks about what sets of intermediate cardinality would look like. As Richard and John Goodrick pointed out, there are straightforward definitions of ordinals of intermediate cardinality. But I assume the question is more about whether there are any sets of real numbers with intermediate cardinality, and what such sets of reals would look like.
It's not too hard to use the ordinals to cook up some sets of real numbers. (For instance, there's probably a definable way to represent each countable ordinal by a real number, and then the set of representations will itself be of intermediate cardinality.) But as far as I know, no one has come up with any direct characterization of a set of real numbers that could have intermediate cardinality.
There are a lot of negative results though. The first is the Cantor-Bendixson theorem, which states that no closed set can have intermediate cardinality. (The theorem shows that every closed set is either countable or has a http://en.wikipedia.org/wiki/Perfect_set_property>perfect subset. Since perfect sets have cardinality of the continuum this means they can't be intermediate.)
In fact, some difficult work of Martin and others in the '70s showed (using just ZF) that Borel determinacy is true, which among other things entails that every Borel set is either countable or has a perfect subset. If we further assume projective determinacy (which set theorists tend to believe) then the same is true for projective sets.
Thus, any set of intermediate cardinality has to be pretty weird. It can't be closed, it can't even be Borel, and (if set theorists are right about projective determinacy) it can't even be projective. Thus, it must be pretty crazy, just as we know about non-measurable sets. (Though there's no guarantee that intermediate cardinality goes along with non-measurability - Martin's Axiom guarantees that in fact every set of intermediate cardinality is measurable and has measure 0.)
answered Oct 19, 2009 at 7:50
-
$\begingroup$ This is what I meant by the question, thank you. $\endgroup$ Commented Oct 19, 2009 at 23:41
$\begingroup$
$\endgroup$
2
Under ZFC, the negation of CH is actually equivalent to Freiling's "Axiom of Symmetry". Wikipedia says this equivalence was proven by Sierpinski.
Freiling's axiom is fairly intuitive, in that it basically says that (countable) discrete invariants on the reals disagree at points, for "probabilistic" reasons. I think it makes a pretty cool argument against CH :)
answered Oct 16, 2009 at 16:04
-
$\begingroup$ As is discussed on the Wikipedia page, the "intuitive" argument here is based on the fallacy that all sets should be measurable. $\endgroup$ Commented Oct 16, 2009 at 18:16
-
$\begingroup$ Yes, I've seen that fallacious argument as support for the axiom. $\endgroup$ Commented Oct 16, 2009 at 19:42
$\begingroup$
$\endgroup$
Let {fj} be a collection of analytic functions, with j ranging over some index set J. Suppose that for every fixed complex number z, the set {fj(z)} is countable. Does it follow that J itself is countable? In other words, can you have uncountably many different analytic functions such that they only obtain countably many values at each point?
You asked "What if the continuum hypothesis were false". Well, if it were, then it would follow that J must be countable. If, on the other hand, CH is true, then you can build an uncountable family of such functions. It's a peculiar instance where a somewhat "pathological" construction requires CH to be true, while if it is false then no such pathology can occur.
This is just one random example - as Eric mentioned, a lot of people have looked into what CH and not-CH imply and are implied by. Saharon Shelah, in particular, published extensively on the question of what Aleph could the cardinality of the continuum possibly be.
answered Oct 16, 2009 at 17:02
$\begingroup$
$\endgroup$
Usually it is difficult to prove something from just not CH, much like if the parallel postulate fails, you want to distinguish between hyperbolic and elliptic geometries. There are lots of interesting models of ZFC where CH is false. One common example is a model where Martin's Axiom holds. Not CH + Martin's Axiom does lots of interesting things like provide a counterexample to the Whitehead Problem, or prove Kaplansky's Conjecture. Woodin's Pmax forcing produces a model in which CH is false in an "effective" way. The model produced also has absoluteness properties similar to Godel's constructible universe L (the canonical model where CH is true).
Also, I would say most people (at least set theorists) who think CH has an answer (whatever that means) think that it is probably false.
answered Oct 16, 2009 at 16:07
$\begingroup$
$\endgroup$
4
Alon: "Saharon Shelah, in particular, published extensively on the question of what Aleph could the cardinality of the continuum possibly be."
Actually this question (and much more) is answered by a theorem of W.B. Easton from 1970. It turns out that the continuum could be any infinite cardinality of uncountable cofinality! (At least if we assume that the axioms ZFC are consistent, which most people are happy to do.)
See Easton's Theorem, which is proved by a variation of the forcing method originally used by Paul Cohen to get that ZFC + not-CH is equiconsistent with ZFC:
http://en.wikipedia.org/wiki/Easton%27s_theorem
So, e.g. the continuum could be aleph_1, or aleph_17, or even the 23rd Woodin cardinal (if you believe in such things), but it could NOT be aleph_omega (this would violate Konig's Lemma).
Shelah has made many contributions to "cardinal arithmetic" and what kinds of bounds can be proved for cardinal exponents just in ZFC. This gets pretty technical pretty fast, but you can get the flavor of it here:
-
1$\begingroup$ Actually, Easton's Theorem is much stronger than just specifying the value of the continuum; it specifies 2^\kappa for all regular \kappa simultaneously. I'm certain the result for just the continuum was known before that; if it wasn't in Cohen's original work it was discovered shortly afterward. $\endgroup$ Commented Oct 16, 2009 at 20:18
-
$\begingroup$ You're right, thanks; and I have no idea who first specified just the possible values of the continuum, but certainly it was a bit before Shelah's time. I've edited my answer accordingly. $\endgroup$ Commented Oct 16, 2009 at 20:42
-
5$\begingroup$ There's a little mistake in what you said: the continuum cannot be the 23rd Woodin cardinal. It in fact cannot be any strongly inaccessible cardinal, because the very definition of strongly inaccessible prevents it (if $\kappa$ is strongly inaccessible then $\mathfrak c=2^{\aleph_0}<\kappa$ because $\aleph_0<\kappa$). $\endgroup$ Commented Dec 8, 2013 at 18:58
-
2$\begingroup$ (continued) What you can do, following Cohen's work, is start with (say) the 23rd Woodin cardinal $\kappa$ and add $\kappa$ many Cohen reals. After doing that, we live in a new model of set theory where $2^{\aleph_0}=\kappa$, but also $\kappa$ is no longer a Woodin cardinal and, in fact, not even strongly strongly inaccessible. $\endgroup$ Commented Dec 8, 2013 at 18:59
$\begingroup$
$\endgroup$
People have done a lot on CH. I don't know how to answer the comments but at a recent conference in Bedlewo, Hugh Woodin actually claimed that CH is true in the true model of ZFC. A new picture of V has been emerging in which CH is true along with GCH. There are of course many set theorists who don't believe in true V, but Woodin's work, if everything works out, will be a phenomenal piece of set theory which will identify unarguable the true core model of ZFC. Think of the situation like this. If 0^# doesn't exist that there are few who will argue that L isn't the core model of ZFC, ``core" here just means the largest canonical. In that sense, he basically identified the model without putting any large cardinal restrictions on the universe. Whether the model is itself the true universe of set theory is an entirely different matter.
There is an alternative approach to CH via forcing axioms and etc. Justin Moore has obtained some really deep recent results on this.
$\begingroup$
$\endgroup$
1
I have an objection to the way in which the question and some of the answers are phrased. To me it seems meaningless to ask whether the Continuum Hypothesis is "true" or "false". It is of course meaningful to ask whether CH is true or false in a given model. But unless you can reveal to me some objective world of sets, I'll have no idea what you mean by CH being "true" or "false" in an absolute sense.
It's like asking whether the parallel postulate is "true" or "false". There are geometrical systems in which it's true, and geometrical systems in which it's false; end of story. Similarly, we know that from any model of ZFC, one can build (i) another model of ZFC in which CH is true, and (ii) another model of ZFC in which CH is false.
I know what the original question is asking, but I'm surprised that so many people continue to use this language. The same goes for the Axiom of Choice; it amazes me when people argue about whether it's "true" or "false".
answered Oct 27, 2009 at 22:18
-
$\begingroup$ I wanted to simplify the text, so I reworded the question while I was at it. To be fair, I think everyone was speaking of true or false in the sense you speak, I've only heard (of late) philosophers make absolute types of arguments. $\endgroup$ Commented Oct 28, 2009 at 12:55
$\begingroup$
$\endgroup$
2
A more interesting question is what cardinalities would be like if the Axiom of Choice were false. AC was proven independent of ZF by Cohen at the same time as CH was (the positive consistency results were both proven by Godel in the 1930s with the "constructible hierarchy" L, while Cohen's method of forcing produced two different models giving the negative consistency results). AC is of course equivalent to the well-ordering theorem, which states that every set can be well-ordered. Thus, with AC, we can always represent every cardinality by the least ordinal of that size. Among other things, this ensures that cardinalities are linearly ordered
Without AC we no longer get this constraint. Cantor's theorem tells us that the power set of a given set is larger than that set. There is also a function, called "Hartog's aleph" (I don't know if the name has one or two "g"s - google has several hits with each spelling, but not many, and the first hit on one seems to be some lecture notes I wrote a while ago while the first hit on the other is the book I first learned about this from) that has similar behavior. We know that there is no set whose cardinality is greater than that of every ordinal (because then there would be a set of all ordinals, which gives rise to the http://en.wikipedia.org/wiki/Burali-Forti_paradox>Burali-Forti paradox). Since the ordinals are well-ordered, for any given set S we can define \aleph(S) to be the least ordinal whose cardinality is not less than that of S. If P denotes the power set function, then it's not hard to show that P(P(P(P(S)))) > \aleph(S), and I think with a little work you can get that down to 3 or maybe even 2 iterations of P. However, in general P(S) and \aleph(S) will be incomparable, even if S itself was well-orderable. (This contrasts with the situation with choice, where we know that P(S) \geq \aleph(S).)
If we further assume the Axiom of Determinacy as a replacement for choice, I think there is also strange behavior, like \aleph_1 turns out to be a measurable cardinal, and there are something like only five cardinals strictly smaller than the continuum.
But there's probably lots of interesting stuff in this area.
answered Oct 18, 2009 at 8:06
-
1$\begingroup$ One can definitely prove aleph(P^3(S)) > aleph(S). The proof is a good hard exercise in basic set theory (it's in chapter 1 of Kunen). It doesn't require knowing anything sophisticated, just running the diagonalization argument carefully. I don't know offhand what better bounds can be achieved than that. $\endgroup$ Commented Oct 18, 2009 at 17:53
-
1$\begingroup$ Start with L and add omega2 many Cohen reals. In the extension, the reals of L is a definable set of reals with intermediated cardinality. $\endgroup$ Commented Oct 29, 2009 at 0:08
$\begingroup$
$\endgroup$
1
You must start slightly further back. The Axiom of Choice (AC) and the Axiom of Determinacy (AD) are both very desirable, but contradict one another. We usually feel AC has more claim to truth since AC is equivalent to "a cartesian project of non-empty sets is non-empty". So set theorists found weaker versions of AD that don't contradict AC, one being the Axiom of Projective Determinacy.
These determinacy axioms basically say that infinite sets behave slightly more like finite sets than one might otherwise imagine. In particular, determinacy axioms have always been found to be equi-consistent with various large cardinal axioms, i.e. axioms that create some cardinal(s) far larger than any that came before. A good example of a large cardinal is just the omega, set of all integers, i.e. the axiom of infinity. You cannot prove that large cardinal axioms are consistent outright, yet they have extremely clear motivation and are unlikely to be inconsistent.
CH seems independent of these large cardinal axioms themselves, but Woodin showed that CH contradicts one compelling determinacy axiom.
Btw, Freiling's axiom of symmetry (AX) is apparently equivalent to not CH.
-
2$\begingroup$ The situation is better than equiconsistency -- if your universe has sufficient large cardinals, then AD is outrightly true in L(R). $\endgroup$ Commented Oct 18, 2009 at 17:57