I want to show that a map $F: \mathbf{R} \to \mathbf{R}$ has at most countable discountinuities, if $F(x) \leq F(y)$ whenever $x \leq y$.

Here's the idea. Let's use standard notation $F(x^+), F(x^-)$ for upper and lower limits of $F$ around $x$. And let $D(F) = \{x : F(x^+) \neq F(x^-)\}$, which is the set of discountinuities for a monotone function. Now suppose that $D(F)$ is uncountable. Suppose $x, y \in D(F)$ are distinct. Then $(F(x^-), F(x^+)) \cap (F(y^-), F(y^+)) = \emptyset$. This is because with $r = d(x, y)$, and assuming $x < y$, if $F(x^+) > F(y^-)$, this means that $\inf_{x \leq s < x + r/2} F(x) > \sup_{y - r/2 < t \leq y} F(t)$, which happens if and only if for some particular $x \leq s < x + r/2$ and $y - r/2 < t \leq y$, we have $F(s)> F(t)$. But this is a contradiction since as stated, $s \leq t$. Hence we have $F(x^+) \leq F(y^-)$, which means the intervals are disjoint as required.

We have shown that $D(F)$ uncountable implies the existence of uncountably many disjoint intervals $(F(x^-), F(x^+))$, where $x$ ranges over $D(F)$. But this can't happen because each open interval contains a distinct rational, which is a countable set.

  • $\begingroup$ Your proof looks good. :) $\endgroup$ – Tki Deneb Dec 8 '18 at 20:59
  • $\begingroup$ Yes this is exactly the proof I learned. It might bear mentioning why $F(x^+)$ and $F(x^-)$ exist (which is easy) but this is right. $\endgroup$ – user25959 Dec 8 '18 at 20:59

I assume your question is, is this proof valid? The answer is yes!

To be clearer, you should probably point out that $x + r/2 = y - r/2$, which is how you get $s < x + r/2 = y - r/2 < t$ and therefore $s \le t$ (you say this is true "as stated", but I don't see you stating it).

You might also want to say more about why a discontinuity in a monotone function must have $F(x^+) \ne F(x^-)$.


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