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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?

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  • $\begingroup$ 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. $\endgroup$ – David K Mar 1 '15 at 22:45
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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.

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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.

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