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?
Asked
Active
Viewed 281 times
0
-
What neighborhood around $(\pi,\pi)$ is locally Euclidean? – Jason DeVito - on hiatus Jun 24 '20 at 23:49
-
It is a metric space whose completion is $\Bbb{R^2}$ and so is the removed part $\Bbb{Q^2}$ you won't find any open set homeomorphic to $(0,1)^2$ – reuns Jun 24 '20 at 23:52
-
I mean ${x\in \mathbb{R}^2 : x \notin \mathbb{Q}^2}$ which by this post is not a totally disconnected set – roi_saumon Jun 24 '20 at 23:53
-
@roi_saumon Connectedness (even path connectedness) is very far from local Euclideanness. – Noah Schweber Jun 24 '20 at 23:54
2 Answers
7
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.)
Alex Kruckman
- 86,811
5
Any neighbourhood of any point in $\Bbb R^2\setminus\Bbb Q^2$ is not simply connected, so no it cannot be homeomorphic to $\Bbb R^n$.
Alessandro Codenotti
- 12,895