Countability of removable discontinuities I saw a problem of proving that number of removable discontinuities in a function is countable. I was not able to prove it and tried many things. Can anyone do it? 
Thanks in advance.
 A: For $n\in \{1,2,\dots,\}$, let $E_n$ be the set of removable discontinuities $x_0$ where the difference between $f(x_0)$ and $\lim_{x\to x_0}f(x)$ is at least $1/n$. If we can prove $E_n$ is countable for all $n$, we will be done, since the set of removable discontinuities is the union of the sets $E_n$.
Given $x_0\in E_n$, let $\lim_{x\to x_0}=L$. There is a neighborhood around $x_0$ where points in the neighborhood are all with $\frac1{4n}$ of $L$. Therefore, they are all within $\frac1{2n}$ of each other. This means there cannot be any removable discontinuities with a gap of $\frac1n$ in that neighborhood, so this neighborhood is disjoint from the rest of $E_n$. Therefore, the points in $E_n$ are all isolated from each other, and therefore countable.
A: Let $x_{0}$ be removable discontinuity, then $$\lim_{x\to x_{0}^{-}}=\lim_{x\to x_{0}^{+}}=L$$then for $\epsilon=0.5$ $\exists \delta >0$ s.t $\forall x\in [x_{0}-\delta,x_{0}+\delta] \neq x_{0}$ we have:$$L-0.5 \leq f(x) \leq L+0.5$$ now from the density of the rational number there exists a rational number $q\in [x_{0}-\delta,x_{0}+\delta]$ and we can show that $x$ is the only discontinuity in this interval, thus we assign a unique rational number for every removable discontinuity, but since the rationals are countable so are the removable discontinuities
