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Suppose $f:\mathbb{R}\rightarrow \mathbb{R}$ is a function (not necessarily continuous) from the real line to the real line,

how to prove that the set $V\triangleq \{ a\in \mathbb{R}| \overline{\lim}_{x\mapsto a+} f(x)\neq \overline{\lim}_{x\mapsto a-}f(x) \}$ is countable?

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    Related: http://math.stackexchange.com/questions/84870/how-to-show-that-a-set-of-discontinuous-points-of-an-increasing-function-is-at-most-countable2011-11-25
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    could you give some explanation, how to define a monotonic function from $f(x)$ to reduce this question to the link you mentioned?2011-11-25
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    I didn't say it was reducible to there, just that it was similar.2011-11-25
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    What is $\overline{\textrm{lim}}$?2011-11-25
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    @user7530, it's an alternative notation for [limsup](http://en.wikipedia.org/wiki/Limit_superior).2011-11-25
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    Another hint. For each pair of rationals $u$a$ can have $$\overline{\lim_{x\mapsto a+}}f(x) < u < v < \overline{\lim_{x\mapsto a-}}f(x)$$. – 2011-11-25

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Following GEdgar's hint:

For each pair of rationals $uv>u>\limsup_{n\rightarrow a^-} f(x) \}. $$

Now, fix an $A_{u,v}$, and suppose $a$ is a limit point of $A_{u,v}$. Then, we may (and do) choose a sequence $\{a_n\}$ from $A_{u,v}$ that converges monotonically to $a$. Assume $a_n\nearrow a$.

Now, for each $n$, there is a sequence $\{b^n_m\}_m$ with $b^n_m\downarrow a_n$ and $f( b^n_m)>v$ for each $m$. This implies the existence of a sequence $\{ c_n\}$ with $c_n\nearrow a$ and $f(c_n)>v$. But then, this implies $\limsup\limits_{x\rightarrow a^-} f(x)\ge v$. And, thus, $a\notin A_{u,v}$.

A similar argument shows that $a\notin A_{u,v}$ if $a_n\searrow a$.

So, $A_{u,v}$ is a set that either has no limit points or does not contain any of its limit points. $A_{u,v}$ must therefore be countable (in fact, all but countably many points of an uncountable set A is a limit point of A).

In a similar mannar, one can show that the sets $$ B_{u,v} =\{ x: \limsup_{x\rightarrow a^-}f(x) >v>u>\limsup_{n\rightarrow a^+} f(x) \}, $$ where $u

It folows that $V=\bigcup A_{u,v} \cup \bigcup B_{u,v} $ is countable.