# Graph connected does not imply $f$ is continuous

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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 path-connected, but $f$ is not continuous at $(0,0)$.

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### user137957

Updated on August 01, 2022

• 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 years
Any 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 years
See below for my example. It is a real analysis problem.
• Andrew D. Hwang over 8 years
Thanks. :) If you add your example to your question I'll vote to re-open, and will undelete my example, which is has a path-connected 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 years
Your 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 years
I 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 path-connected (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 y-axis.)
• Ben Grossmann over 8 years
It'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 years
Actually, that works.