Continuity on a neighbourhood of a point If a function $f$ is continuous at $x$, then there exists a neighbourhood of $x$, on which $f$ is continuous. Is it true of false?

False. Let $f:\Bbb R\to \Bbb R$ be the function $$f(x)=\begin{cases}
x &\text{, if } x\in\Bbb Q\\
0 &\text{, else}\\
\end{cases}$$
It is continuous at $0$, but not continuous anywhere else (which is easy to prove). Now, if $U$ is a neighborhood of $0$ such that $f|_U:U\to\Bbb R$ is continuous, then $f:\Bbb R\to\Bbb R$ would be continuous at every $y\in\text{int}(U).$
Note that the phrasing $f$ is continuous on the neighborhood $U$ can mean "$f$ is continuous at each $y\in U$" but it can also mean "the restriction $f|_U$ is continuous". However, if there is a nbh satisfying the first, then there is also a nbh satisfying the second interpretation, and vice versa.
