
  so there must be intervals to the left and right of a simple continuity on which no other simple discontinuity can exist.


That is incorrect. Consider an enumeration $r_n$ of the rationals, and let

$$f(x) = \sum_{\substack{n \in \mathbb{N}\\r_n &lt; x}} 2^{-n}.$$

Then $f$ is a strictly monotonic function that has a jump discontinuity in every rational number.

However, as the points of discontinuity approach any fixed $y \in \mathbb{R}$, the jumps tend to $0$, in fact, the sum of all jumps in the discontinuities $\neq y$ contained in a neighbourhood of $y$ tends to $0$ when the neighbourhood shrinks to a point.

Prove that any real function defined on a real open interval has at most countably many simple discontinuities.
On Careers, when filling out my CV, it is listed as being 145% complete.  Could someone tell me why this is?
