Closure of Subset of Ordered square

Question

Let $A=[(x,x^2):x\in[0,1]\setminus\mathbb{Q}]$ be a subset of the ordered square $I_0^2$. How would you find the closure of this set? My first guess would be a parabola, but I'm not entirely comfortable with finding closures of the ordered square.

Answer 1

The closure of $A$ is $A\cup ((0,1]\times \{0\})\cup ([1,0)\times \{1\}$. Check that open intervals around these extra “endpoints” all contain points of $A$. All points with second coordinate in $(0,1)$ that don’t already lie in $A$ are not in the closure.