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Axiom of choice is discussed very often, because it should be a lot of paradoxes (Banach-Tarski paradox, for example) and in general it is considered by many non-obvious (for uncountable case, of course). But continuum hypothesis remains aside; it doesn't seem so strange. How many people — so many opinions. But so many theorems only from calculus use AC; it seems that it is easier to accept than reject AC.

And what about CH? There was a question on SE, but I would like to know what the (interesting) theorems using the CH; I don't know any such theorem. Math with AC and math without AC seems to me very different; but math with and without CH... Who cares?)

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    $\begingroup$ Related: math.stackexchange.com/questions/79346/… $\endgroup$ – Asaf Karagila Jul 14 '15 at 6:08
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    $\begingroup$ Also, remember that everything that can be done naively in set theory (which, from a "working mathematics" point of view, is most things) can be done without using CH, which is not "intuitive" as AC (or weak forms which suffice for analysis). Since many mathematicians shy away from independence, there's an obvious selection bias against theorem which depend on the truth value of CH. So who cares? Apparently some people do, but not sufficiently many of them so that CH became a common everyday assumption. $\endgroup$ – Asaf Karagila Jul 14 '15 at 6:10
  • $\begingroup$ @AsafKaragila, many thanks. What about another theorems (more simple theorems, actually)? :) $\endgroup$ – Michael Galuza Jul 14 '15 at 6:12
  • $\begingroup$ Like what? I don't know what you have in mind. $\endgroup$ – Asaf Karagila Jul 14 '15 at 6:29
  • $\begingroup$ @AsafKaragila, like theorems of calculus: equivavalency of Cauchy and Heine definitions of limit, for example $\endgroup$ – Michael Galuza Jul 14 '15 at 6:31
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Kaplansky's Conjecture for Banach Algebras is an example of an interesting statement with no obvious set theoretic content whose truth depends on CH.

The conjecture is that for every algebraic homomorphism from C(X) (where X is some Compact Haussdorff space) to a Banach Algebra is continuous.

Under CH, the conjecture is false for any infinite choice of X. Without CH, the conjecture may be true.

In general there's a character to CH which makes pathological uncountable things more easily "constructable", because you can always build them out of countable pieces through a process of transfinite induction. I no longer remember (if I ever knew) how the construction of the discontinuous homorphism works, but I'd expect it's something along these lines.

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I posted this previously, but am unable to find it:

There is a theorem by Erdös, that may be relevant to you:

The following are equivalent

  • $CH$
  • There is an uncountable family $\mathcal F$ of entire functions, such that for all $z \in \mathbb C$ the set $\{ f(z) \mid f \in \mathcal F \} $ is countable.
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    $\begingroup$ You're looking for: P. Erdos. An interpolation problem associated with the continuum hypothesis. Michigan Mathematical Journal, 11, 1964. (which is available here as open access: projecteuclid.org/euclid.mmj/1028999028 ) $\endgroup$ – DRMacIver Jul 20 '15 at 13:15

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