Let F:R→RF : \mathbb{R}\to \mathbb{R} be a monotone function. Then

FF has no discontinuities.

FF has only finitely many discontinuities.

FF can have at most countably many discontinuities.

FF can have uncountably many discontinuities.

Which is the correct answer and why ?

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1 Answer

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The answers is 3.

Wihout loss of any generality, suppose ff is increasing. Let DD be the set of discontinuities of ff.

If d∈Dd\in D, thenlimx→d−f(x)\lim\limits_{x\rightarrow d-}f(x) and \lim\limits_{x\rightarrow d+}f(x)\lim\limits_{x\rightarrow d+}f(x) exists. Thus, for every point of discontinuity dd we asign an interval I_d=(\lim\limits_{x\rightarrow d-}f(x),\lim\limits_{x\rightarrow d+}f(x))I_d=(\lim\limits_{x\rightarrow d-}f(x),\lim\limits_{x\rightarrow d+}f(x)). Notice that all this intervals are disjoint. Now, let g:D \rightarrow \mathbb{Q}g:D \rightarrow \mathbb{Q} such that g(d)=q_dg(d)=q_d for some q_d\in\mathbb{Q}q_d\in\mathbb{Q}. Therefore gg is injective because all the intervalls are disjoint, hence \left\vert{D}\right\vert \leq \left\vert\mathbb{Q}\right\vert=\aleph_0\left\vert{D}\right\vert \leq \left\vert\mathbb{Q}\right\vert=\aleph_0.

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In order to show that the answer is (3), rather than (1) or (2), the OP should also construct a specific monotone function with a countably infinite set of discontinuities. (For example, a step function will work.)

– Mitchell Spector

2 days ago