Showing if $f_n \to f$ uniformly and each $f_n$ has at most $10$ discontinuities, then so does $f$

Question

Suppose that $f_n:[a,b] \to \Bbb R$ and $f_n$ uniformly converges to $f$ as $n$ goes to infinity. How to prove that if each $f_n$ has at most ten discontinuities (the discontinuities for each $f_n$ might be different) then $f$ also has at most ten discontinuities?

Answer 1

If for some $x_0\in [a,b]$, there is a subsequence of $\{f_n\}$ with elements functions which all continuous at $x_0$, then $f$ is continuous at $x_0$. This is due to the following fact:

If $\{g_n\}$ converges uniformly to $g$, and all the $g_n$'s are continuous at $x_0$, then $g$ is also continuous at $x_0$.

So if $f$ is discontinuous at $x_0$, then all but finitely many $f_n$'s are discontinuous at $x_0$.

Assume now that $f$ is discontinuous at the distinct points $x_1,\ldots x_k$, with $k>10$.

Then for the $x_j$, all but finitely many $f_n$'s are discontinuous at it. Say that for $n\ge n_j$, the $f_n$ is discontinuous at $x_j$. Setting $n_0=\max\{n_1,\ldots,n_k\}$, we have that all the functions $f_n$, for $n\ge n_0$, are discontinuous at $x_1,\ldots,x_k$.

This is a contradiction, since these $f_n$'s are discontinuous to more than $10$ points.