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Suppose that a real-valued function function $f$ defined on an open interval $I$ has the property that at every point $a \in I$ that either the limit of $f$ as $x$ approaches $a$ from the left exists or the limit of $f$ as $x$ approaches $a$ from the right exists (or both). In this context, I am trying to understand why the following statement must be true:

For every $\epsilon > 0$ there exists either an open interval $J_L = (L, a) \subset I$ such that

$ |f(s) - f(t)| < \epsilon \; \; \forall s,t \in J_L $

or an open interval $J_R = (a, R) \subset I$ such that $ |f(s) - f(t)| < \epsilon \; \; \forall s,t \in J_R $

How can I see that this is a true statement?

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    @3Sphere You can post your solution as an answer, so that this question will appear as answered in the future.2011-09-26

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Per Srivatsan's suggestion, I am posting my "answer" as an Answer:

Following the hint given by robjohn, suppose that the left limit exists. By definition of limit, for every $\epsilon/2 > 0$ we have $|f(s) - a| < \epsilon/2$ for all $s$ in some open interval of the form $(u,a)$ and $|f(t) - a| < \epsilon/2$ for all $t$ in some open interval of the form $(v,a)$ Let $J_L = (u, a) \cap (v, a)$ Then,

$ |f(s) - a| + |f(t) - a| < \epsilon \implies |f(s) - f(t)| < \epsilon \;\; \forall s, t \in J_L $

where the last step follows from the triangle inequality. The case where the limit from the right exists proceeds similarly.