Graph connected does not imply $f$ is continuous
1,137
Hint: Choose a function such that the overall limit at $(0,0)$ does not exist (but the limit along some path does).
Example: take $$ f(x,y) = \begin{cases} \frac{x^2}{x^2 + y^2} & (x,y) \neq (0,0)\\ 0 & x = y = 0 \end{cases} $$ Verify that $\Gamma_f$ is pathconnected, but $f$ is not continuous at $(0,0)$.
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user137957
Updated on August 01, 2022Comments

user137957 over 1 year
Show an example of a function $\newcommand{\R}{\mathbb{R}} f: \R \times \R\to \R$ such that $f$ is not continuous, but its graph $$ \Gamma_f := \left\{\bigl((x, y), f(x, y)\bigr) \mid \text{$(x, y)$ is in the domain of $f$}\right\} $$ is connected (in $\R \times \R \times \R$).

Andrew D. Hwang over 8 yearsAny thoughts of your own, or indication of how rigorous an example you're seeking? (Examples suitable for multivariable calculus or for topology may look rather different.)

user137957 over 8 yearsSee below for my example. It is a real analysis problem.

Andrew D. Hwang over 8 yearsThanks. :) If you add your example to your question I'll vote to reopen, and will undelete my example, which is has a pathconnected graph but is continuous at only one point. (With a small modification based on agha's answer, you can arrange a connected graph for a function that's discontinuous everywhere.)


Andrew D. Hwang over 8 yearsYour example seems correct, but the final assertion (a set with connected closure is connected) looks suspicious, e.g., $(1, 0) \cup (0, 1)$...? (The converse is true: The closure of a connected set is connected.)

user137957 over 8 yearsI did just figure one out: f(x, y) = (x^2 + y)/(x^2 + y^2) for (x, y) nonzero and = 0 at (0, 0). The graph is pathconnected (easy to see, since f is continuous outside of the origin), but f certainly isn't continuous at the origin itself. (The limit as (x, y) goes to (0, 0) doesn't even exist along the yaxis.)

Ben Grossmann over 8 yearsIt's not clear to me that $((0,0),0)$ is path connected to any other part of the graph. In fact, I don't think it is.

Ben Grossmann over 8 yearsActually, that works.