If $f$ is continuous at $a$, is it continuous in some open interval around $a$? If $f: \mathbb{R} \to \mathbb{R}$ is continuous at $a$, is it continuous in some open interval around $a$?
 A: Fix an enumeration of the rationals, $q_n$ for $n\in\Bbb N$. We define the following function:$$\large f(x)=\sum_{q_n<x}\frac1{2^n}$$
That is, we sum $\frac1{2^n}$ for all the $n$ such that $q_n$ is strictly smaller than $x$. This function is continuous at every irrational number, but discontinuous at every rational number. Therefore it cannot be continuous on any interval.
A: No, that's not true. Take the function $f : \mathbb R \to \mathbb R$ defined by $f(x) = 0$ if $x \in \mathbb Q$ and $f(x) = x$ if $x \not\in \mathbb Q$. Then $f$ is discontinuous everywhere except at $0$ (can you prove this?).
A: No. Take $$x\mapsto \left\lfloor\frac 1 x\right\rfloor^{-1}$$
It is continuous at $x=0$; but has discontinuities on every $x_n=\dfrac 1 n$
ADD On a side note, it is not hard to show this function is continuous at $0$. Use the squeeze theorem with $x$ and $2x$ for $x\to 0^+$ and $x$ and $x/2$ for $x\to 0^{-}$. Of course, we need to set $0\mapsto 0$.
A: No. Consider the function 
$$f(x)=\begin{cases} x &\text{if } x\in \mathbb Q\\
-x &\text{otherwise}
\end{cases}$$
which is continuous at $0$, but not at any other real number.
A: Not necessarily:
$$f(x)=\begin{cases}
x,&\text{if }x\in\Bbb Q\\
-x&\text{if }x\in\Bbb R\setminus\Bbb Q
\end{cases}$$
is continuous only at $0$.
A: Here is a function that is continuous at every irrational point, and discontinuous at every rational one.  So the answer is no.
