If CH can be neither proved nor disproved, and to assume it true or false can yield different results, isn't it an axiom?
Why is the Continuum Hypothesis not the Continuum Axiom?
-
7The name reflects history. – 2012-01-10
-
1For $46$ pages of interesting history see Juris Steprans, [History of the Continuum in the 20th Century.](http://www.math.yorku.ca/~steprans/Research/PDFSOfArticles/hoc2INDEXED.pdf) – 2012-01-10
-
0@BillDubuque Thanks Bill, I'll check it out. – 2012-01-10
1 Answers
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."
-
1Perhaps it could be added that CH is believed to be false by a number of people in the field. – 2012-01-10
-
1Good 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](http://en.wikipedia.org/wiki/Martin%27s_axiom) and generally make life less interesting. – 2012-01-10
-
1@AndréNicolas What does it mean to believe CH is false? – 2012-01-10
-
1@Austin Mohr: It means to believe, like Gödel did, that the set of countable ordinals is in fact not equinumerous with the reals. – 2012-01-10
-
0Thanks guys this is all very interesting! – 2012-01-10
-
0@AndréNicolas any comments as to the existence or otherwise of an axiomatic system in which CH would be provable? – 2012-01-10
-
0@Nikhil Bellarykar: Well, there is the famous [Axiom of Constructibility,](http://en.wikipedia.org/wiki/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. – 2012-01-10
-
0Interesting. The axiom of constructiblity does appear rather restrictive, though. Can you please provide some of other interesting axioms you mentioned? thanks in advance. – 2012-01-10
-
0@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. – 2012-01-10
-
0@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..." – 2012-01-10
-
0But then, couldn't you also say that AC makes cardinals "too easy"? – 2013-06-28
-
0@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. :) – 2013-06-30