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Many thanks to B.S Thompson for such a great explanation of Froda's theorem. I am a bit troubled by the assumption of a countable number of points i.e. $J_n$, between the limits of the interval (which is I realise part of the official proof), because it seems on that basis you could prove there is a countable number of real values between e.g. $0$ and $1$. There seems to be a circular argument here. I'm probably being dense.

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Edwin Lightstone
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$J_n$ is the set of points in $[a,b]$ where the monotone non-decreasing function $f$ has a jump of size greater than $\frac1n$. For each $x\in J_n$ let

$$I_x=\big(f(x-0),f(x+0)\big)\;;$$

this is an open interval of length greater than $\frac1n$. If $x<y\in J_n$, then $f(x+0)\le f(y-0)$, so $I_x\cap I_y=\varnothing$. Thus, $\{I_x:x\in J_n\}$ is a pairwise disjoint family of open intervals of length greater than $\frac1n$. Moreover, all of these intervals are subsets of the closed interval $[f(a),f(b)]$. Suppose that $A\subseteq J_n$, and $|A|=m$. Then on the one hand the total length of the pairwise disjoint intervals $I_x$ with $x\in A$ is greater than $\frac{m}n$, but on the other hand the total length is at most $f(b)-f(a)$. Thus,

$$\frac{m}n<\sum_{x\in A}\operatorname{length}(I_x)\le f(b)-f(a)\;,$$

and it follows that $m<\big(f(b)-f(a)\big)n$. In particular, this shows that we must have

$$|J_n|<\big(f(b)-f(a)\big)n\;,$$

so $J_n$ must in fact be finite.

Brian M. Scott
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