If $f$ is continuous at $a$, is it continuous in some open interval around $a$?

Solution 1:

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?).

Solution 2:

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 $0\mapsto 0$.