Graph connected does not imply $f$ is continuous

Question

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 $\R \times \R \times \R$).

Answer 1

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 $(0,0)$.

Answer 2

Let:

$$f(x,y)=\sin\left(\frac{1}{x^2+y^2}\right)$$

for $(x,y) \neq (0,0)$

and $f(0,0)=0$

Then graph of this function is:

$$\{(x,y,\sin\left(\frac{1}{x^2+y^2}\right))|(x,y) \neq (0,0)\} \cup \{(0,0,0)\}$$

Closure of this graph is:

$$\{(x,y,\sin\left(\frac{1}{x^2+y^2}\right))|(x,y) \neq (0,0)\} \cup \{(0,0,x): x \in [-1,1]\}$$

There is a theorem that if closure of set is connected then set is connected. You can check by definition that closure is connected.

Answer 3

Define $$ f(x, y) = \begin{cases} \sqrt{x^{2} + y^{2}} & \text{if $x \neq 0$ and $y/x$ is rational,} \\ 0 & \text{otherwise.} \end{cases} $$

Theorem: The function $f$ has (path-)connected graph, but is continuous at precisely one point.

Proof: The graph of $f$ is the union of:

• Lines (through the origin) of irrational slope in the $(x, y)$-plane;

• The $y$-axis;

• Absolute value graphs over lines (through the origin) of rational slope in the $(x, y)$-plane.

Since each of these sets is (path-)connected and contains the origin, their union is (path-)connected.

Since lines (through the origin) of irrational slope are dense in the plane, $f$ is continuous at $(x, y)$ if and only if $\sqrt{x^{2} + y^{2}} = 0$, if and only if $(x, y) = (0, 0)$.