Why do non-Decreasing Functions have countable discontinuities [duplicate]

I was reading some notes and one of the results in it implicitly used a result which fell along the lines of "non-decreasing functions have countable discontinuities". I don't completely understand why.

The notes were essentially describing the properties of a CDF of a "weird" random variable. They made the above statement and then concluded (after some more proofs) that the CDF is "cadlag".

marked as duplicate by Rahul, mrf, Did, Stefan Hansen, Old JohnFeb 8 '13 at 8:57

• Short answer: Each discontinuity jumps across at least one rational, and rationals are countable. Long answer: Surely this is a duplicate of a previous question; that question will have a long answer. – Rahul Feb 8 '13 at 6:55
• @ℝⁿ Yes, I agree this is duplicate. It didn't turn up somehow. Thanks! – Inquest Feb 8 '13 at 7:13

• @JonathanGallagher We're assuming the function is defined in an interval (which could be whole $\mathbb{R}$). Hence, you cannot have a horizontal line with holes poked in it. – EA304GT Oct 4 '16 at 23:00
If you look at how many jumps are larger than $\frac{1}{k}$ in the interval $[n,n+1)$ for any integer $n$ and positive integer $k$, this must be finite as otherwise the function becomes infinite in that interval.
So the number of jumps in the interval $[n,n+1)$ must be finite or countable as the limit of a finite sequence. So the total number of jumps on the real line must be finite or countable as a countable sum.