# Only removable/jump discontinuities

Definitions

Consider R with the usual metric and let $f:A \to R$ , $A \subset R$ be a function.

We say that $f$ has a jump discontinuity at a point $p \in A$ iff
$\lim_{x \to p^-}f(x)$ and $\lim_{x \to p^+}f(x)$ exists but have different values.

We say $f$ has a removable discontinuity at a point $p \in A$ iff
$\lim_{x \to p}f(x)$ exists but $\lim_{x \to p}f(x) \neq f(p)$ .

Question

I'm trying to find two functions with $R$ as co-domain that are everywhere discontinuous ( discontinuous at every point of the domain ) such that one only has jump disconinuities and the other only has removable discontinuities.
Is there any pair of examples showing this is possible,or can we prove that this is not possible ? $A$ is an arbitrary subset of $R$.

• Do you mean discontinuous at every point of the domain? Oct 14, 2014 at 16:48
• Yes, that's what i mean by everywhere discontinuous. Oct 14, 2014 at 17:26
• What is $A$? An arbitrary subset of $\mathbb R$? An interval? An arbitrary topological space? A subset of $\mathbb R$ to be chosen carefully so that the problem can be solved? Oct 14, 2014 at 17:28
• For suitable $A$, it is possible. Can $A$ be chosen? Oct 14, 2014 at 17:39
• It should be a subset of R to be chosen so that the problem can be solved. I'm interested in what functions ( each function should specify its domain, and hence A ) have removable or jump discontinuities everywhere. Oct 14, 2014 at 17:49

Choose $A = (0,1) \cap \mathbb{Q}$ (or $\mathbb{Q}$ if you want $A$ dense in all of $\mathbb{R}$).

Let

$$f(x) = \frac{1}{n}$$

if $x = \frac{m}{n}$ with $\gcd(m,n) = 1$. That is the restriction of Thomae's function to the rationals in $(0,1)$, which may (if one has already dealt with Thomae's function) help seeing that $f$ is discontinuous at every point of $A$, but has only removable discontinuities.

Let

$$g(x) = \sum_{\substack{y\in A\\y < x}} f(y)^3.$$

Then $g$ is a strictly increasing funtion with a jump discontinuity at every $x\in A$.