About countable set

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Let $f$ be a function in [a,b] with a countable break points (note by $S$). Is it true that the closure of S is countable?

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No. The rationals are dense so all you need to do is take the rationals in the interval. The closure is $[a,b]$, which is uncountable.

The answer to your new question is no. One can construct a function that is discontinuous only at the rationals in the interval.

Let ${c_n}$ be a sequence of positive numbers such that $\sum c_n$ converges. Put the rationals into sequence, ${x_n}$. Then define $$f(x)=\sum_{x_n\lt x}c_n$$, where $f(x)$ is zero if there are no $x_n$ to the left of $x$.

$f$ has the required property.