A monotone function has at most countably many discontinuities

Question

Suppose that $f: S \rightarrow T$ is increasing, therefore $f(x-)$ and $f(x+)$ exist when $x \in S$. The fact to show is that $f$ has at most countably many jump points. A proof is given as follows on Page 4 in Chung's A Course in Probability Theory: $x_1$ is a jump point such that $f(x_1-) < f(x_1+)$. Let $I_{x_1} = (f(x_1-), f(x_1+))$. Let $x_2$ denote another jump point and $x_2 > x_1$, then by monotonicity, we have $f(x_1+) \leq f(x_2-)$. Let $I_{x_2} = (f(x_2-), f(x_2+))$ Therefore, $I_{x_1}$ and $I_{x_2}$ are disjoint. For each jump point, such an interval exists and all of them are disjoint. Then, for any two jump point $x_1$ and $x_2$, Chung says that one can find two rationals, $r(x_1) \in I_{x_1}$ and $r(x_2) \in I_{x_2}$ such that $r(x_1) < r(x_2)$. Therefore the sets of these jump points are in a one-to-one correspondence with the rationals.

My question is: Since all of these intervals are nonempty and disjoint, we can also find two irrationals $ir(x_1)\in I_{x_1}$ and $ir(x_2)\in I_{x_2}$ such that $ir(x_1) < ir(x_2)$. Therefore $ir$ is a one-to-one correspondence from the set of jump points to the set of irrationals. Could anyone help me: What mistake did I make?

Answer 1

Perhaps this is what you mean:

Since there are uncountable irrational numbers, and one irrational number can be chosen from each interval, there is a one-to-one correspondence between such intervals and irrational points.

You are right that you can choose an irrational number from such an open interval, but how can you conclude that each irrational number is associated with an open interval? This is not true. Thus, in this case, you cannot conclude anything.

We can choose a rational point from each open interval, and since there are countably many rational numbers, there are $\textbf{at most}$ countably many jump points.