Is the following function continuous on the irrationals in $[0,1] \times [0,1]$? $$f(x,y) = \begin{cases} 0 & x\text{ irrational}\\ 0 & x\text{ rational, }y\text{ irrational}\\ \frac{1}{q} & x\text{ rational, }y = \frac{p}{q}\text{, lowest terms} \end{cases} $$
I believe that it is, but I'm having a hard time showing why. For any epsilon ball, I think that even $\delta = \epsilon$ probably works, but how do I show that any rational, $\frac{p}{q}$, in $(y - \delta, y+\delta)$ is such that $y-\frac{p}{q} > 0-\frac{1}{q}$, which is how I think I should show continuity.
Yes, it is continuous on the points where both are irrational. Choosing a specific $\delta$ is hard, but it is easy to claim it exists. Given irrational $y$ there are only finitely many rationals in the square with denominators less than $\frac 1\epsilon $ Take $\delta$ less than the distance from $y$ to the closest one and you are home.