Set of Discontinuities for a function $f$ Take $f$ to be a function over the reals. I want to show that
a set of discontinuities of the first kind for $f$ are
countable. This is the discontinuity type at point $P \in \mathbb{R}$
where $lim_{x \rightarrow P^{-}} f(x)$ and
$\lim_{x \rightarrow P^{+}} f(x)$ both exist but either do not equal
each other or do not equal $f(P).$ The suggestion given to me is
that if there is a discontinuity but the right and left limits
exist, I can slip a rational number between them. Hence, I am
considering showing this countability by developing a bijection
between the set of discontinuities and the rationals, although I
am having difficulty formalizing this in the language of functions.
Any assistance will be appreciated.
 A: This is a minor simplification of the proof of theorem 7.7 from here which contains a slightly more general result.
For each positive integer $k$ let $A_k = \{ a \in \mathbb{R} : f(a+) \text{ and } f(a-) \text{ exist and } |f(a+) - f(a-)| > \frac{1}{k}|\}.$
Since the union of $A_k$'s consists of the set of all jump discontinuities, it is sufficient to show each $A_k$ is countable. 
If $A_k$ is not empty for each $t \in A_k$ we can choose a $\delta(t) >0$ such that for any $x \in (t,t+\delta(t))$ we have $|f(t+) - f(x)| < \dfrac{1}{8k},$ note that for any $u,v \in (t,t+\delta(t))$ we have $|f(u) - f(v)| \leq |f(u) - f(t+)| + |f(t+) - f(v)| < \dfrac{1}{4k}.$
If we can show that for distinct $a,b \in A_k$ the intervals $(a,a+\delta(a))$ and $(b,b+\delta(b))$ are disjoint we are done. For then the mapping $a \to (a,a+\delta(a))$  will be 1-1 from $A_k$ to a disjoint collection of open intervals which is countable.
To see this assume $a,b$  with $a < b$ are distinct points in $A_k$. If the intervals $(a,a+\delta(a))$ and $(b,b+\delta(b))$ are not disjoint we would have $ a < b < a + \delta(a) $.  
Since there is a jump discontinuity of size $\frac{1}{k}$ at $b$ we can find $c,d$ with $ a < c < b < d < a + \delta(a)$ such that $$|f(c) - f(d)| > \frac{1}{2k}$$ this follows from the fact that 
$$
\begin{align}
|f(c) - f(d)| & \geq  |f(b+) - f(b-)| - |(f(b+) -f(d)| - |f(b-) - f(c)| \\
& \geq \frac{1}{k} - |(f(b+) -f(d)| - |f(b-) - f(c)|
\end{align}
$$
and we can choose $c$ and $d$ to make $|f(b-) - f(c)|$ and $|(f(b+) -f(d)|$ as small as we wish. 
However from the definition of $\delta(a)$ we should also have $|f(c) - f(d)| \leq \frac{1}{4k}.$ 
This contradiction proves $(a,a+\delta(a))$ and $(b,b+\delta(b))$ are disjoint and we are done.
