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Or alternative you need to assume CH is false.

I know several proofs that use axiom of choice. Heine Borel theorem is the best example I can think off. Zorns lemma is heavily used in the non commutative algebra course I'm studying.

However, I have yet to see CH be used to prove stuff.

Anyone give a theorem that comes up in undergrad degree like last year of it that uses CH or not CH is used?

As personally, I can't think where I've seen some sort of argument that assumes say CH or looks like you are assuming it. The same can't be said about the axiom of choice, which in about 50% of the proofs.

CH scares me abit. As I can't see how it can be independent.

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Cohen constructed a model of ZFC in which CH fails, and Gödel constructed a model of ZFC in which CH is true. Perhaps you should study those two constructions? –  Zhen Lin Nov 5 '11 at 22:57
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The statement "If a set X is smaller in cardinality than another set Y, then X has fewer subsets than Y." follows from the generalized continuum hypothesis but can fail without it (even in relatively harmlessly-seeming situations). See Joel David Hamkins's answer here –  t.b. Nov 5 '11 at 22:58
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I think, Zorn's Lemma is more useful than CH. –  Erno Nemecsek Nov 5 '11 at 23:04
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@Karatug: Depends on the context though.However in undergrad mathematics, I agree. –  Asaf Karagila Nov 5 '11 at 23:26
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Medial limits (which exist in models of CH) can simplify some functional analytic arguments. I was going to write up more, but nothing I write can compare to t.b.'s answer to this question: math.stackexchange.com/questions/54554 –  user83827 Nov 6 '11 at 17:43
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up vote 17 down vote accepted

I cannot recall a single proof that I have seen during my undergrad which assumes the continuum hypothesis.

Some proofs are "nicer" assuming it, but this assumption is usually not necessary. One could get away with just $\frak c$ instead of $\aleph_1$. However as you progress in set theory you run into things which depend on the continuum hypothesis.

For example, Freiling's axiom of symmetry holds if and only if the continuum hypothesis is false.

Jensen defined the combinatorial principle $\lozenge$ (Diamond) which implies the continuum hypothesis. This means that by assuming the diamond principle, a very useful principle in infinitary combinatorics, one also assumes $\frak c=\aleph_1$.

Saharon Shelah proves in Diamonds [Sh:922] a generalization of this principle [1]. Generalizations of $\lozenge$ to higher cardinals are implied by continuum hypothesis-like principles.

All this comes to show that the continuum hypothesis, as well "localized" versions of it are important in combinatorial proofs. Since $V=L$ (Goedel's axiom of constructibility) implies the generalized continuum hypothesis it means that a lot of these arguments hold in $L$, however the above proofs show that if one wants to use the arguments in a different model he still has to make such and such assumptions.

In other places "slightly" outside set theory, one can see uses of the continuum hypothesis in topology, functional analysis and other topics which are close to the fine line of the continuum.

One example by Todd Eisworth is this paper and the immediate references mentioned in the introduction (and further as well) which come to show that the continuum hypothesis has some effect on topological properties.

[1] I was corrected that it may not be a result of Shelah, and if it is his result then it was known prior to the publication of the paper I linked.


As for the independence of the continuum hypothesis, it is slightly harder to understand how this can be independent since the natural numbers have a "fixed" number of subsets. However this is apparently not true, and one can think about it as the claim that $2$ has a square root in every field.

It is boldly false in $\mathbb Q$ and in $\mathbb Q[\sqrt 3]$, however it is true in $\mathbb Q[\sqrt 2]$. In this analogy, $\mathbb C$ is a bit like $V=L$. It is a very "constrictive" model (since the theory of algebraically closed fields is complete up to the characteristics of the field), so it decides everything.

In that manner, $V=L$ while not yet complete, is still pretty decisive on many axioms. So it implies $\frak c=\aleph_1$, but without this axiom it might be not very clear.

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Nasty! (more characters) –  Jonas Teuwen Nov 5 '11 at 23:14
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