I was thinking about the continuity discussion that took place in the spanish comunity and I thought if we have:
$$f: A \longrightarrow \mathbb{R},\quad A \subset \mathbb{R}.$$
Such that $Graph(f)$ is connected but it is not path connected so $f$ is not continuos in his domain, For example: $$ h(x)=\left\{\begin{array}{cl} \sin \left(\frac{1}{x}\right) & x \in(0, 1) \\ 0 & x=0 \end{array}\right. $$
$Graph(h)$ is connected because it can’t be expressed as the disjoint union of two non empty open sets. But it is not path connected because it includes the point $(0,0)$ but there is no way to link the function to the origin so as to make a path.
In conclusion if $graph(f)$ is connected but it is not path connected so it is not continuos and if it is path connected it is continuos?.
Thanks in advance.
