Can we construct a function that has uncountable many jump discontinuities?

I know that Dirichlet function has uncountable many discontinuities. I think they are removable, because the discontinuities can be removed by redefining the function values of the rational numbers as 0.

So Dirichlet function is a function that has uncountable many removable discontinuities, then my question is can we construct a function with uncountable many jump discontinuities? If not, how do we prove it is impossible? Thank you.

An odd but similar question is can we have a function that has uncountable many infinite discontinuities?

Solution 1:

$f$ has a jump discontinuity at $x=a$ if $f(a-) = \lim_{x \to a-} f(x)$ and $f(a+) = \lim_{x \to a+} f(x)$ exist, but $f(a-) \ne f(a+)$. Let $S(a) = |f(a+) - f(a-)$ be the size of the jump discontinuity at $a$. Let $J(n)$ be the set of jump discontinuities $a$ with $S(a) > 1/n$. Then I claim $J(n)$ is a discrete set. Since a discrete subset of $\mathbb R$ is countable, and the union of countably many countable sets is countable, this implies that the set of jump discontinuities of $f$ is countable.

To prove the claim, suppose $a \in J(n)$, and take $\epsilon < 1/(2n)$. There is $\delta > 0$ so that for $a -\delta < x < a$, $|f(x) - f(a-)| < \epsilon$ and for $a < x < a + \delta$, $|f(x) - f(a+)| < \epsilon$. Thus if $x$ and $y$ are both in $(a-\delta, a)$ or both in $(a, a+\delta)$, $|f(x) - f(y)| < 2 \epsilon < 1/n$. This implies that if $x$ is a jump discontinuity in one of these intervals, $S(x) \le 2 \epsilon < 1/n$.