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Let $(f_n)$ a sequence of functions that converges uniformly to $f$, and each $f_n$ have a finite number of discontinuities.

How I can show that the limit function $f$ at most have a countable number of discontinuities? What kind of formal argument I can write?

The unique proofs that I saw and involve some cardinality appear in the context of set theory but, how I can do the same in the context of the limit of some sequence?

This question surely is elementary but I dont know very much about ZFC to set up the concept of limit in a pure set-theoretic context, and surely this is not necessary.

Maybe it is enough to invoke the theorem that says that $$|A_n|=|\Bbb N|,\forall n\in\Bbb N \implies|\bigcup_{n\in\Bbb N}A_n|=|\Bbb N|$$ ?? But anyway I dont know eactly how to fit it formally in the context of the question.

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    Here is a little hint: can you possibly _guess_ what the set of possible discontinuities might be? Or a set which the set of discontinuities is contained? – 2017-02-08
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    @Wojowu sometimes these kind of guess can be hard to guess... I dont follow exactly, sorry. – 2017-02-08
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    Unfortunately, this is so for very many problems one tries to solve; to even begin, one has to have some idea. Let me try to give you one more (a bit less vague, I hope) hint: do you know any results which relate uniform convergence and continuity? – 2017-02-08
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    @Wojowu I know some basic stuff but I never had to deal with discontinuities and uniform convergence. I know how to prove the statement through the kind of discontinuities of the $f_n$ but this is not directly related to the cardinality of the set of discontinuities of each $f_n$. Maybe my question is wrong, Im assuming that exists a simple formal way (something like induction or similar) to these kind of tasks. – 2017-02-08

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