# What does the continuum hypothesis imply?

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!

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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
See some of the answers to this MO question. – Zhen Lin Aug 2 '12 at 15:50
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
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

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).
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