# Let $f$ have a jump discontinuity at $x_0$. Show that $f(x_1), f(x_2), \ldots$ has at most two limit-points.

This is a question I understand intuitively but am having trouble proving rigorously:

Let $f$ have a jump discontinuity at $x_0$. Show that if $x_1, x_2, . . .$ is any sequence of points in the domain of $f$ converging to $x_0$, with no $x_j$ equal to $x_0$, then the sequence $f(x_1), f(x_2), . . .$ has at most two limit-points.

Since there is a jump, there is going to be a limit point from the left and from the right. I am picturing a graph where there is a break in the line, so from the left $f(x)$ will approach say $a$, and from the right $f(x)$ will approach $b$. How can I prove that there cannot be greater than two limit points?

• It depends on your definition of "limit point". The fact should be provable for any correct definition, but some definitions can make it easier than others. – David K Mar 1 '15 at 22:45

Hint: Consider the indices $j$ such that $x_j<x_0$. Then split your sequence into those $x_j$ with $x_j<x_0$ and $x_j>x_0$. These two subsequences each converge to the different limit values of $f$ at $x_0$ and the union of the two subsequences gives the entire sequence.
It all depends on how you define jump discontinuity. If it means the limit exists from the left and also from the right, then for any sequence converging to $x_0$ you will either get arbitrarily close from the right or left (or both) and then you are basically done.