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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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    Consider the function $f(0)=1$ and $f(x)=0$ for $x\neq 0$ on the interval $(-1,1)$. This function admits the property you've described - moreover, both limits exists in any point. On the other hand, clearly the statement does not hold at $a = 0$. Or maybe $a\notin N_\varepsilon(a)$?2011-09-26
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    So, the "local" uniform continuity argument doesn't work; I'm still trying to see why the statement holds.2011-09-26
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    Gortaur's point is not that a particular argument doesn't work, it is that the statement you're trying to prove is false (in which case there'd better be _no_ argument that works). Unless an "one-sided neighborbood" of $a$ can be something that doesn't include $a$.2011-09-26
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    In view you your comments, I have reworded the question to better reflect what I am trying to ask.2011-09-26
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    **Hint:** apply the definition of a limit and the triangle inequality.2011-09-26
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    @robjohn I have updated the question to provide my proposed answer that makes use of your hint; thanks.2011-09-26
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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.