Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Are there any fundamental/interesting results that are a consequence of assuming the continuum hypothesis as an additional axiom?

I'm sorry if this question was already asked. I'm also sorry if there is no rigour at all in the way I asked it.

Thanks!

share|improve this question
2  
A related thread: math.stackexchange.com/q/79346/5363 –  t.b. Aug 2 '12 at 15:33
    
There was a complex analysis theorem I saw, I think in "Proofs from the Book," that was true if and only if CH was true. I'd have to look it up, but the proof was fairly simple. –  Thomas Andrews Aug 2 '12 at 15:47
2  
See some of the answers to this MO question. –  Zhen Lin Aug 2 '12 at 15:50
4  
Ah, here is the question I was looking for. Let $\{f_\alpha\}$ be a family of analytic functions such that for each $z\in\mathbb C$, $\{f_\alpha(z)\}$ is countable. Does it follow that $\{f_\alpha\}$ is countable? If CH is false, the answer is "yes," if true, the answer is "no." renyi.hu/~p_erdos/1964-04.pdf –  Thomas Andrews Aug 2 '12 at 16:07
1  
Sierpinski's 1934 book Hypothèse du Continu (written in French) is devoted to equivalences and consequences of the continuum hypothesis. I don't know if there's a copy freely available on the internet, but the Bulletin of the AMS review of Sierpinski's book is freely available. I also don't know if there's an English translation, but here is someone who has apparently translated some of it. –  Dave L. Renfro Aug 2 '12 at 20:35

1 Answer 1

up vote 2 down vote accepted

There are many cardinals which have aleph values undetermined in ZFC, but for which we can say that they are either infinite and less than continuum, or uncountable and no greater than continuum, both of which would immediately resolve their (aleph) value if we had CH.

Many of those can be resolved by the somewhat weaker Martin's axiom. It is strictly weaker than CH (if ZFC is consistent, so is ZFC with the axiom and the statement that $\mathfrak c=\omega_2$, but CH implies it by Rasiowa-Sikorski lemma).

For the consequences of Martin's axiom, a good source is David Fremlin's book aptly named "Consequences of Martin's Axiom". Quite many of the results in this book are somewhat trivial if you assume CH, though, if I recall correctly...

share|improve this answer
1  
Many of the consequences of MA apply to uncountable cardinals in the interval $(\omega,\frak c)$. Assuming CH simply makes this interval become empty... conclusion: CH makes a lot of things boring. (On the other hand, it makes combinatorics easier) –  Asaf Karagila Aug 2 '12 at 20:13

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.