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Thomae's function,
$$ f(x)= \begin{cases} 0 & \text{$x$ is irrational}\\ \frac{1}{p} & \text{$x = \frac{p}{q}$ where $\gcd(p,q) = 1$} \end{cases}, $$
is an example of a function from $(0,1)\to [0,1]$ which is continuous at the irrationals and discontinuous at the rationals.
I was wondering: is there a two-dimensional analogue of this? Namely, can we construct a function from $[0,1]^2\to [0,1]^2$ which is discontinuous at points with rational coordinates and continuous elsewhere?
If your "rational coordinates" means $x,y$ both being rational, namely, that $(x,y)\in\Bbb Q^2$, then @kingW3's example won't satisfy your "continuous elsewhere".
Take the point $(\sqrt2/2,1/2)$ as an example, it is not in $\Bbb Q^2$, and the $f(x,y)$ so defined in @kingW3's comment is not continuous at this point. To see the discontinuity, you may just fix $y=1/2$, and let $x\to\sqrt2/2$, then you will of course find that, however small an $x-$neighborhood $(-\delta+\sqrt2/2,\delta+\sqrt2/2)$ you confine your $x$ to, there are both infinitely many rational $x$s and infinitely many irrational $x$s in it. When you hit a rational $x=q/p$, $f(x,y)=(1/p,1/2)$ (assuming $\delta$ to be small enough, $1/p$ will be extremely small); and when you hit an irrational one, $f(x,y)=(0,0)$, hence the discontinuity.
To answer your question, consider this modified version $$ f(x,y)= \begin{cases} (0,0) &x,y\text{ both irrational}\\ (\frac1p,0) &x=\frac{p}{q}\, \text{in the lowest term},\,y \,\text{irrational}\\ (0,\frac1r) &x\,\text{irrational},\,y=\frac{s}{r}\,\text{in the lowest term}\\ (\frac1p,\frac1r) &x=\frac{q}{p},y=\frac{s}{r}\,\text{in the lowest term} \end{cases} $$
The reasoning is simple and similar to that of the Riemann-Thomaes function you mentioned in your post.
