Continuity in each variable Consider the function
I want to show that the function
$F(x,y)=\frac{xy}{x^2+y^2}$ defined to be 0 at (0,0)
Is continuous in each variable separately and then to show that the function is not continuous at 0
How do I show this topologically (inverse image of open sets)? If the denominator is not zero, then holding one variable fixed, we see that we have a rational function is continuous.
Why don’t we have continuity at 0?
 A: You have
$$  F(x,y) = \begin{cases}
\frac{x y}{x^2 + y^2}  ,& (x,y) \neq (0,0)  \\
0  ,&  (x,y) = (0,0)
\end{cases}  \text{.}  $$
"...at (0,0) Is continuous in each variable separately" means we must show $F(x,0) = 0$ and $F(0,y) = 0$ are continuous, respectively, in $x$ and $y$.  They're constant, so the preimage of any open set omitting $0$ is the empty set, so is open, and the preimage of any open set containing $0$ is $\Bbb{R}^2$, so is open.  Therefore, $F$ is continuous in each variable separately at $(0,0)$.
"Show the function is not continuous at $(0,0)$" (fixed that last word for you) means we must find an open set whose preimage is not open.  (As a hint: This function's global maximum is $1/2$, attained on (a subset of) the line $y = x$.  Its global minimum is $-1/2$, attained on (a subset of) the line $y = -x$.  Those two lines intersect...)  Consider the open set $(-1/2, 1/2)$.  The preimage of this open set is the union of five simple pieces,  \begin{align*}
\{(0,0)\} &\cup \{(x,y) \mid 0 < x, -x < y < x\}  \\
    &\cup \{(x,y) \mid x < 0,  x < y < -x\}  \\
    &\cup \{(x,y) \mid 0 < y, -y < x <  y\}  \\
    &\cup \{(x,y) \mid y < 0,  y < x < -y\}  \text{,}
\end{align*}
which union is not open -- $(0,0)$ is in the boundary of this set.  (There is a shorter way to write the set : the plane minus the two diagonals plus the origin.  Perhaps this makes it clearer that the origin is the obstruction to openness.)
A: If $f$ is continuous at $(0,0)$ then for any sequence $(a_n, b_n) \to (0,0)$ in $\Bbb R^2$ we have that $f(a_n, b_n) \to f(0,0)=0$. This holds in any topological space.
But then note that $(\frac1n, \frac1n) \to (0,0)$ and $f(\frac1n, \frac1n)= \frac{\frac{1}{n^2}}{\frac{1}{n^2} + \frac{1}{n^2}} = \frac{1}{2}$ for all $n$ so that the sequence $f(\frac1n, \frac1n) \not\to 0$. So $f$ is not continuous at $(0,0)$.
A: If $F$ were continuous then the inverse $F^{-1}(-1/3,1/3)$ of the open real interval would be open in $\Bbb R^2$.
And $(0,0)\in F^{-1}(-1/3,1/3)$.
So there would exist $r>0$ such that $\{(x,y)\in \Bbb R^2: x^2+y^2<2r^2\}\subseteq F^{-1}(-1/3,1/3)$.
Which would imply $(r,r)\in F^{-1}(-1/3,1/3)$.
Which is absurd because $F(r,r)=1/2$ when $r>0.$
