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from $\mathbb{N}$ to $\mathbb{N}$ there are $2^{\aleph_0}$ as you can define $2^{\aleph_0}$ functions as following: For every subset of $\mathbb{N}$ keep the numbers in the subset the same, and add 1 to the rest.

What about the cardinality of all montonic functions from $\mathbb{R}$ to $\mathbb{R}$?

Is it ${\mathfrak c}$ or $2^{\mathfrak c}$, ${\aleph_1}$ or ${\aleph_2}$

My intuition is to say that its cardinality is the same as the cardinality of the set of real continuous functions which is ${\mathfrak c}$, as monotonic functions are continuous except possibly at a countable number of points, but I couldn't come up with a concrete proof.

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    $\begingroup$ The cardinality of the set of real continuous functions is ${\aleph_1}$? I'd like to see a proof of that. $\endgroup$ – Professor Vector Jun 7 '17 at 9:30
  • $\begingroup$ Since $\mathfrak c=\aleph_3$ is consistent with ZFC, we can safely say that neither $\aleph_1$ nor $\aleph_2$ are answers. $\endgroup$ – user228113 Jun 7 '17 at 9:50
  • $\begingroup$ @ProfessorVector math.stackexchange.com/questions/477/… $\endgroup$ – user844541 Jun 7 '17 at 9:55
  • $\begingroup$ So you claim $\mathfrak c=\aleph_1$? $\endgroup$ – Professor Vector Jun 7 '17 at 9:59
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    $\begingroup$ $\aleph_1$ is the least uncountable cardinal. $2^{\aleph_0}$ is the cardinal of the set of all subsets of a countably infinite set. Their equality is not provable nor disprovable , from the axioms of Set Theory, unless those axioms are inconsistent with each other. $\endgroup$ – DanielWainfleet Jun 7 '17 at 16:40
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Every monotonic function from $\mathbb{R}$ to $\mathbb{R}$ is continuous except possibly at countably many points where it has jump discontinuity. So, there are $|\mathbb{R}|^{\aleph_0} =|\mathbb{R}|$ number of possible choices of such discontinous points. After the discontinuities are specified, we then only need to fill in the "gap" between the discontinuities. But a continuous function is uniquely determined if its value is already determined at densely many points. This means that $$\text{(the number of ways to fill in the gap)} \leq |\mathbb{R}|^{|\mathbb{Q}|} = |\mathbb{R}|.$$

To sum up, there are at most $|\mathbb{R}|\cdot|\mathbb{R}| = |\mathbb{R}|$ number of monotonic functions from $\mathbb{R}$ to $\mathbb{R}$, but it is also easy to show that there are at least $|\mathbb{R}|$ number of monotonic functions (for instance, $x+c$ is monotonic for each $c\in \mathbb{R}$). Therefore, the number of such functions is $|\mathbb{R}|$.

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  • $\begingroup$ Somewhat similarly, the cardinal of the set of continuous real functions is $|\mathbb R|$ because each such function is completely determined by its values on $\mathbb Q,$ and $|\mathbb R^{\mathbb Q}|=|\mathbb R|. $ $\endgroup$ – DanielWainfleet Jun 7 '17 at 16:44

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