Why is $\mathbb{R}^2\setminus \mathbb{Q}^2$ not a topological manifold?

Question

I would like to understand why $\mathbb{R}^2\setminus \mathbb{Q}^2$ endowed with the subspace topology is not a topological manifold. It seems to me it is Hausdorff and second countable. So I am wondering. Why is it not locally Euclidian?

Answer 1

Any neighbourhood of any point in $\Bbb R^2\setminus\Bbb Q^2$ is not simply connected, so no it cannot be homeomorphic to $\Bbb R^n$.

Answer 2

Because it has lots of holes!

One way to make this precise is to show that $\mathbb{R}^2\setminus \mathbb{Q}^2$ is not locally compact. In fact, no point has a compact neighborhood.

Let $p\in \mathbb{R}^2\setminus\mathbb{Q}^2$ and suppose for contradiction that $K$ is a compact neighborhood of $p$, so we have $p\in U\subseteq K$ for some open $U$. Then $U = U'\cap (\mathbb{R}^2\setminus\mathbb{Q}^2)$ for some open set $U'\subseteq \mathbb{R}^2$, so $U'$ contains a point $q\in \mathbb{Q}^2$. Pick some sequence $(q_i)_{i\in \mathbb{N}}$ in $U$ converging to $q$. Then no subsequence of $(q_i)_{i\in \mathbb{N}}$ converges in $\mathbb{R}^2\setminus\mathbb{Q}^2$, contradicting compactness of $K$. (Since $\mathbb{R}^2\setminus\mathbb{Q}^2$ is a metric space, compactness is equivalent to sequential compactness.)