Let $F:\mathbb R\times \mathbb R\to \mathbb R$ be defined by the equation
$$F(x,y)=\begin{cases} \frac{xy}{x^2+y^2} & \text{if }(x,y)\neq (0,0)\\ 0 & \text{if } (x, y) = (0,0).\end{cases}$$
Show that $F$ is not continuous.
I got a solution. But I am not convinced.
It says "$F(x,y)=1/2$ implies $(x,y)\neq (0,0)$ and $(x−y)^2=0$ . Therefore, $$F^{−1}(\{1/2\})=\{(x,x)|x\neq 0\}$$ is not closed."
I can not understand how $F^{−1}(\{1/2\})=\{(x,x)|x\neq 0\}$ is not closed. Of course this is not closed in $\mathbb R^2$ in standard topology. But in ' Dictionary Order Topology' It is closed.
The set $\{(x,x):x\neq 0\}$ is indeed closed in the dictionary order topology, and not closed in the standard topology. Obviously, this problem is talking about the standard topology, not the dictionary order topology. (It's called "standard" for a reason--it's the default topology that is assumed unless stated otherwise.)