Show that $\left\{(x, y) \in \mathbb{R}^{2} \mid x y=0\right\}$ it is not a locally Euclidean topological space.

Asked Mar 23 '24 at 23:55

Active Mar 24 '24 at 17:23

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I have tried to reach a contradiction by working with the definition:

$M$ is locally Euclidean of dimension $\boldsymbol{n}$ : each point of $M$ has a neighborhood that is homeomorphic to an open subset of $\mathbb{R}^n$.

Consider the following: I assumed that there exists an open set around $(0,0)$, let’s call it $U$. Then I defined $O = U - \{(0,0)\}$. Now, $O$ is not connected; in fact, it has 4 connected components. Does this somehow lead to an contradiction? And how can I determine that there is no homeomorphism to an open set in $\mathbb{R}^{2}$ ?

edited Mar 24 '24 at 15:32

asked Mar 23 '24 at 23:55

Mainred

Hint. Every open set in the plane is a union of open disks. Think about an open disk containing the origin. – Ethan Bolker Mar 24 '24 at 00:01
Does this answer your question? Show that $xy=0$ is a variety but not a manifold or this? Why is this set not a manifold? – Anne Bauval Mar 24 '24 at 17:29
Or this? https://math.stackexchange.com/questions/1280770/explain-why-this-set-is-not-a-differentiable-manifold/1285489 – Anne Bauval Mar 24 '24 at 17:34
yes! thank you so much – Mainred Mar 24 '24 at 18:18

Show that $\left\{(x, y) \in \mathbb{R}^{2} \mid x y=0\right\}$ it is not a locally Euclidean topological space.

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