Continuity of $f(x) = xy$ for $xy > 0$, $0$ otherwise

Question

The function is continuous on the non negative plane, and along the edges since setting $y=ax$ the $f(x,y) \rightarrow 0$ as $x \rightarrow 0$. So it is continuous there, but then is it continuous everywhere? This seems incredibly simple and like I am missing something obvious.

Answer 1

In your case, you can define the open set $A=\{(x,y)\in\mathbb R^2~:~x,y>0\}$. Since $f\mid_A(x,y)=xy$ you get $f$ is continuous on $A$. And since $f\mid_{\mathbb R^2\setminus \overline A}(x,y)=0$ you get $f$ is continuous on $\mathbb R^2\setminus \overline{A}$.

You just need to check the continuity on $$ \partial A=\{(x,y)~:~x=0,y\geq 0\vee x\geq 0,y=0\}. $$ Choose $(0,y)\in\partial A$ and an arbitrary sequence $(x_n,y_n)_n$ such that $(x_n,y_n)\to (0,y)$ and show that $f(x_n,y_n)\to f(0,y)=0$.

Since $y_n\to y$, you now, that $y_n$ is bounded by $C>0$. You get either $0\leq f(x_n,y_n)\leq Cx_n$. This holds even is $f(x_n,y_n)=0$ for some $n$. Therefore you can conclude $f(x_n,y_n)\to 0=f(0,y)$.

You argue the same way for $(x,0)\in\partial A$ and your are done.