# I Do you believe that continuum is Aleph-2, not Aleph-1?

1. Sep 15, 2016

### tzimie

This negation of CH is based on Woodin's work: https://en.wikipedia.org/wiki/Ω-logic

Of course, you can only believe his result because you need to believe his axioms first. But for me it is really convincing for multiple reasons:

1. While it is, of course, a negation of CH, it does not really break everything because the sequence of Aleph numbers is preserved, just the names assigned to different alephs change
2. His conclusion is based on the quantification over possible forcings, and it looks really powerful - as forcing is used to prove independence of so many large cardinal axioms, so quantification over forcings must be extremely powerful. So it is like (in physics) expanding universe of sets into the multiverse!
3. $\omega_1$ is now less than continuum. And $\omega_1$, at least for me, looks intuitively "almost" countable, as the sequence of ordinals is explicitly well ordered. Of course, any set can be well ordered if we assume AC, but often no constructive example of such ordering can be provided.
4. Interestingly enough, Goedel himself had suspected that continuum = $\aleph_2$

I am Platonist, so for me it sounds more like a discovery. Not like a formal game (with this axiom we can do this, and with another we can do that). Do you feel the same?

2. Sep 15, 2016

### micromass

Staff Emeritus
Many set theorists believed that the continuum does not have cardinality $\omega_1$. This includes Gödel and Cohen. For example, in Herrlich's book on the axiom of choice, he calls the GCH something that is widely seen as something false.

You know, I don't believe in a single set theory. To me there are multiple ones, some which satisfy CH and some which don't. You can transfer between these set theories very easily using topos theory, which I think is the future of this kind of math.

3. Sep 15, 2016

### zinq

[In what follows, we identify each cardinal with the least ordinal having that cardinality. Thus we may use cardinals as subscripts of alephs.]

Hugh Woodin told me about five years ago that he no longer believed his earlier work, that the continuum c satisfies

c = aleph2;​

I don't know what he believes now.

Also, about 20 years ago I spoke with the late Paul Cohen, who said it was possible that c satisfies

c = alephc

(!!!). In fact, he pointed out that the ZFC axioms are consistent with the hypothetical axiom that

c = alephk

for any ordinal k whose cofinality cf(k) is not equal to aleph0.

(Note: The cofinality cf(k) of any ordinal is the least ordinal having the order type of a cofinal subset of k. A cofinal subset X of an ordinal k is a subset such that for every element y ∈ k, there is an element x ∈ X such that y ≤ x.)

This observation, derived from Koenig's Theorem, implies for instance that it must be the case that

c ≠ aleph(aleph0).

4. Sep 15, 2016

### pwsnafu

Yay, someone who actually likes topos theory. Everyone I talk to doesn't know it exists. Makes me happy.

5. Sep 16, 2016

### tzimie

I've seen links in Wiki to topos theory, but Wiki claims that this theory is strongly inaccessible to the idiots like me:
https://en.wikipedia.org/wiki/History_of_topos_theory
But is there some kind of simple explanation? )))
Is it some kind of Multivese?

6. Sep 16, 2016

### tzimie

What a pity...
If there are so many cardinalities in between $\aleph_0$ and continuum (not just $\omega_1$), then these cardinalities create a fuzzy set like in Banach-Tarsky paradox, and no example of set of these cardinalities can be provided constructively.

7. Oct 3, 2016

### Stephen Tashi

I'm curious whether "the continuum" refers to a structure that satisfies a specific set of axioms - or whether it is a term of common speech (at least among set theorists) - similar to terminology like "the universe", which refers to a common notion, but not one that is defined by a unique set of axiom.

8. Oct 3, 2016

### pwsnafu

See linear continuum.

9. Oct 4, 2016

### Stephen Tashi

I see.

That is an axiomatic definition of "a" linear continuum. Is there a theorem that any two linear continuua have the same cardinality? That would justify speaking of "the" continuum - at least as far as the property of cardinality goes.

10. Oct 4, 2016

### zinq

"See linear continuum."

This is not correct. A "linear continuum" can be any of various totally ordered sets that do not all have the same cardinality. For example, the "long line" L in that Wikipedia article and the set R of real numbers satisfy

card(L) > card(R).​

But in the sense it is used above, "the continuum" — often denoted in math just by the letter c (but for obvious reasons this is a bad idea in physics) — refers to a specific cardinality. This is usually described as either the cardinality of the real numbers, or equivalently as the cardinality of the set of all subsets of the integers.

In terms of other cardinalities, the continuum is usually expressed as the cardinal power

2aleph0,​

where, as usual, 2 denotes the cardinality of the set {0, 1} and aleph0 denotes the cardinality of the integers.

11. Oct 4, 2016

### tzimie

I guess if is a consequence of AC.

12. Oct 4, 2016

### tzimie

But intuitively, do you interpret $\omega_1$ (as well-ordered sequence of all countable ordinals) as continuum ?
For me it is weaker than continuum...

(Note: this question does not make sense to a formalist, but I am Platonist)

13. Oct 19, 2016

### Demystifier

14. Oct 19, 2016

### tzimie

Thank you.

But how is it different from a collection of axiomatic systems? How is it different from saying "take any axiomatic system and do what you want"?

15. Oct 19, 2016

### tzimie

P.S.
Demystifier, and am surprised and happy to meet you here, in math, not in physics subforum, so I can't resist asking you as physicist:
Do you believe that stuff (CH, Large cardinal axioms etc) has any (potential) relation to physics?

P.P.S
What interpretation of mathematics do you prefer - formalism or platonism?

16. Oct 19, 2016

### Demystifier

At the moment I don't see any relevance for physics, but one day, who knows.

When I want an intuitive understanding of abstract math concepts, I am a platonist. When I need to compute something or formally prove a theorem, I am a formalist. When I think philosophically about mathematical ontology, I am often a constructivist and finitist.

17. Oct 19, 2016

### Demystifier

Category theory is not a replacement for logic and axiomatic systems. As a foundation for mathematics, category theory is a kind of replacement for set theory.

18. Oct 19, 2016

### micromass

Staff Emeritus
It can be.

19. Oct 19, 2016

### stevendaryl

Staff Emeritus
My favorite intuitive argument against the continuum hypothesis is Freiling's "Throwing Darts at the Number Line". It's a simple enough argument that I think I can reproduce it here.

Imagine that there is some process for randomly selecting a real number with a flat probability distribution in $[0,1]$. You can't actually select a random real, because a real number requires an infinite amount of precision, but for the sake of argument, suppose that you can. Freiling describes it as "throwing a dart at the number line", and wherever the dart sticks, that's your random real. Now, elementary measure theory tells you that for any countable set $X$ you pick ahead of time, the chance that your randomly chosen real, $x$ will be in $X$ is zero. The measure of any countable set is zero.

Now, suppose that you have a function $F(x)$ which takes a real number in $[0,1]$ and returns a countable subset of $[0,1]$. Then two players, Alice and Bob, can use this function to play the following game:
1. Alice randomly picks a real $a$ in $[0,1]$.
2. She computes a countable set $F(a)$
3. Bob then randomly picks a different real, $b$.
4. He also computes a countable set $F(b)$
5. If Bob picks a real from Alice's set, he wins. If Alice picks a real from Bob's set, she wins.
The question is: What are the odds that Bob will win? After Alice has picked her real, she can reason as follows: "There are only countably many reals in $F(a)$. So the odds that Bob will pick a real from that set is zero. So Bob's chance of winning is zero."

We can also ask what are the odds of Alice winning. If Bob went first, then he could have used the same argument as Alice to argue that Alice has a zero chance of winning. Intuitively, it shouldn't matter who went first, so the conclusion should be that with probability 1, neither Alice nor Bob is going to win. This means that it is very likely (probability 1) that $a \notin F(b)$ and $b \notin F(a)$. This should be true, no matter what the function $F$ is (as long as it always returns a countable set).

This argument motivates the following conjecture:
(The argument actually suggests that most such pairs of numbers have this property, but for what follows, it's enough that at least one pair has this property.)

But Freiling gives a simple proof that the above Symmetry Axiom contradicts the Continuum Hypothesis.

Proof: Assume the continuum hypothesis. Then that means that it is possible to arrange the reals in $[0,1]$ in a well-ordering of type $\omega_1$, which means that we can map each real to an ordinal less than $\omega_1$. Since $\omega_1$ is the first uncountable ordinal, that means that we can associate each real $x$ with a countable ordinal $ord(x)$. So assume we have such a mapping, then we define a function $F(x)$ as follows: $F(x)$ is the set of all reals $y$ in $[0,1]$ such that $ord(x) > ord(y)$. Since every countable ordinal has only countably many smaller ordinals, that means that for every $x$, $F(x)$ is countable. But clearly, for any two reals $a$ and $b$, either $ord(a) < ord(b)$, or $ord(b) < ord(a)$. So for any two reals $a$ and $b$, either $a \in F(b)$ or $b \in F(a)$. This contradicts the Symmetry Axiom above.

Assuming that you find that plausible, Freiling goes on to argue, along similiar lines, that the continuum can't be $\omega_2, \omega_3, ...$

20. Oct 20, 2016

### Demystifier

Can you elaborate?