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If CH can be neither proved nor disproved, and to assume it true or false can yield different results, isn't it an axiom?

iDontKnowBetter
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The Continuum Hypothesis (CH) can be used as an axiom.

To elaborate on what you've stated in your question. I'll take Zermelo-Frankel set theory + the Axiom of Choice (ZFC) as the axiom scheme of standard mathematics. If standard mathematics (ZFC) is consistent, then so is ZFC+CH. Likewise, if ZFC is consistent, then so is ZFC+$\neg$CH.

So attaching it or its negation to ZFC does not change consistency.

But to be fair, any statement can be taken as an axiom -- even a false statement. The problem is that using a false statement (or inconsistent statements) as an axiom leads to a world where everything is both true and false (which isn't very interesting).

A more interesting question is, "Why isn't CH taken as an axiom (for standard mathematics)?"

One answer might be "Because we don't need it to do what we find interesting."

Another answer is "Because CH makes some cardinal arithmetic 'too easy'. $\neg$CH leads to more interesting mathematics."

Bill Cook
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    Perhaps it could be added that CH is believed to be false by a number of people in the field. – André Nicolas Jan 10 '12 at 02:11
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    Good point. I would imagine if set theorists had a panel to vote on which statements should be axioms (like the astrophysicists vote on which rocks are planets), I bet CH would get voted down. It trivializes interesting stuff like Martin's Axiom and generally make life less interesting. – Bill Cook Jan 10 '12 at 02:14
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    @AndréNicolas What does it mean to believe CH is false? – Austin Mohr Jan 10 '12 at 03:00
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    @Austin Mohr: It means to believe, like Gödel did, that the set of countable ordinals is in fact not equinumerous with the reals. – André Nicolas Jan 10 '12 at 03:58
  • Thanks guys this is all very interesting! – iDontKnowBetter Jan 10 '12 at 04:15
  • @AndréNicolas any comments as to the existence or otherwise of an axiomatic system in which CH would be provable? – nb1 Jan 10 '12 at 07:12
  • @Nikhil Bellarykar: Well, there is the famous Axiom of Constructibility, usually called $V=L$, which in fact implies GCH. There are a number of others. There are also not unreasonable (and more interesting) axioms that imply the negation of CH. – André Nicolas Jan 10 '12 at 07:59
  • Interesting. The axiom of constructiblity does appear rather restrictive, though. Can you please provide some of other interesting axioms you mentioned? thanks in advance. – nb1 Jan 10 '12 at 08:25
  • @BillCook: I know that GCH makes "cardinal arithmetic too easy" (or rather, cardinal arithmetic is very interesting in the absence of GCH); I am not aware that CH by itself does, however. – Arturo Magidin Jan 10 '12 at 15:14
  • @ArturoMagidin Yes. GCH makes (essentially) all cardinal arithmetic "too easy". CH would just say "$2^{\omega_0}=\omega_1$" (the first step) is "too easy". That's why I said "some cardinal arithmetic..." – Bill Cook Jan 10 '12 at 15:39
  • But then, couldn't you also say that AC makes cardinals "too easy"? – celtschk Jun 28 '13 at 07:22
  • @celtschk One could say that, but I wouldn't. Getting rid of AC makes things more difficult in ways I don't find "interesting". For my taste, chucking out AC makes life unnecessarily difficult -- but that's just me. Many people thrive in such a world. :) – Bill Cook Jun 30 '13 at 23:36