Is $\mathbb{Q}\times \mathbb{Q}$ closed in $\mathbb{R^2}$?

Question

I reviewed some of the general topology materials to enhance skills in real analysis. The following problem struck me to the core:

Problem: Is there a set that is not bounded, not closed, and not open in $\mathbb{R^2}$

My answer: Take $\mathbb{Q}\times \mathbb{Q}$.

I wonder if this is a true example. Are there few other examples? Thanks. WY.

Answer 1

A closed set is a set with open complement. This is equivalent to saying it contains its limit points (I leave this as an exercise). Consider the sequence $\{(p_n,0):p_n\in\mathbb{Q}\}$. By choosing our $p_n$, we can make it converge to an irrational value, thus, $\mathbb{Q}^2$ does not contain all its limit points, and is not closed.

It is also not open because its complement is not closed (look at sequences of irrational numbers converging to $0$).