Let $(\mathbb{R},\mathcal{B}(\mathbb{R}),\mathbb{P})$ be a probability space, where $\mathcal{B}(\mathbb{R})$ is the Borel $\sigma$-algebra and $F:\mathbb{R}\to[0,1]$ be the distribution function associated to $\mathbb{P}$ defined by \begin{align*} F(x) = \mathbb{P}((-\infty,x]) \end{align*} where $x\in\mathbb{R}$. Prove that $F$ has at most a countable number of discontinuities.
Suggestion: how many points $x\in\mathbb{R}$ can exist such that $\mathbb{P}(\{x\}) > 1/n$ for each $n\in\mathbb{N}_{>0}$?
This is what I have tried:
For $n = 1$, there is no $x\in\mathbb{R}$ such that $\mathbb{P}(\{x\}) > 1$.
For $n = 2$, there is at most one point $x\in\mathbb{R}$ such that $\mathbb{P}(\{x\}) > 1/2$
For $n = 3$, there are at most two points $x\in\mathbb{R}$ such that $\mathbb{P}(\{x\}) > 1/3$.
In general, for $n = k$, there are at most $k - 1$ points $x\in\mathbb{R}$ such that $\mathbb{P}(\{x\}) > 1/k$.
But then I get stuck.
Can anyone help me on this?
EDIT
As far as I have understood, $n$ counts the number of discontinuities. For each $n$, there are $n - 1$ ways to rearrange the size of each discontinuity: there may be just $1$ value above $1/n$, two values above $1/n$, and so on. At the end, there are at most $n - 1$ values above $1/n$ which exhausts all the possibilities.
Can someone double check or improve my argument?