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The title pretty much says it all. What is the cardinality of $G$, the group of all functions $f: \mathbb{R} \to \mathbb{R}$ such that $\forall x,y\in \mathbb{R} \left( x>y\Rightarrow f(x)>f(y)\right)$? Obviously $G$ is a subgroup of the symmetric group on $\mathbb{R}$, which has cardinality $\beth_2$, and $G$ is also obviously uncountable, so the real question here is whether the cardinality is $\beth_1$ or $\beth_2$.

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This is a simple exercise. Hint: Consider the images and preimages of sets of the form $(-\infty,x)$. –  Douglas Zare Nov 12 '12 at 2:34
The alleged group isn't actually a subgroup (or even a subset) of the symmetric group on $\mathbb R$, because it contains functions that are not surjective. –  Andreas Blass Nov 12 '12 at 3:49
A tangential remark: your question seems to suggest that the only uncountable cardinal less than $\beth_2$ is $\beth_1$, but this is not a consequence of ZFC. –  Trevor Wilson Nov 12 '12 at 4:30
Ahh, you're correct, Andreas - I simply said "function" when I should have said "bijection." I'm also familiar with the fact that the existence of cardinals between $\beth_1$ and $\beth_2$ is independent of ZFC, but if my question has an answer in ZFC, then, logically, it cannot be an answer that does not even necessarily exist, which led me to the conclusion that, the possible existence of intermediate cardinals notwithstanding, the answer must be either the one or the other. –  Kyle Ferendo Dec 2 '12 at 8:25

1 Answer 1

up vote 10 down vote accepted

Such a function can have only countably many points of discontinuity, since any discontinuity will be a jump discontinuity and hence the range will skip over an interval unique to that point, and so we can associate each point of discontinuity with a distinct rational number, meaning there are only countably many.

So the function is determined by specifying its values at the members of this countable set, plus specifying values on a countable dense set. This is altogether a countable amount of information, and so the number of such functions is $2^{\aleph_0}=\beth_1=\frak{c}$.

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Ahh, yes, I see now that the crux of the question is that such a function is uniquely determined by its effect on the rationals, since the irrationals can be determined from the images of certain Cauchy sequences. I also appreciate that you addressed discontinuities, although in my original question, I meant (but forgot) to specify that the functions should be bijections. –  Kyle Ferendo Dec 2 '12 at 8:43

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