prove a set in $\mathbb{R}$ is countable 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?
 A: Following GEdgar's hint:
For each pair of  rationals $u<v$, let $A_{u,v}$ be defined
by
$$
A_{u,v} =\{ x: \limsup_{x\rightarrow a^+}f(x) >v>u>\limsup_{n\rightarrow a^-} f(x) \}.
$$
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 $u<v$ are rationals, are countable.
It folows that  $V=\bigcup A_{u,v} \cup \bigcup B_{u,v} $ is countable.
