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Definition: A topological space $X$ is locally Euclidean if there exists $n \in \mathbb{N}$ so that for all $x \in X$ there is a neighborhood $U$ of $x$ which is homeomorphic to an open subset $V \subset \mathbb{R} ^n$.

I want to show that $$X = \{ (x,y) \in \mathbb{R^2} : x^2+y^2 = 1 \} \cup \{ (x,0) : -1 \leq x \leq 1 \}$$ is not locally Euclidean.

I proved that $X$ is not homeomorphic to the circle $S^1 \subset \mathbb{R^2}$, is this enough? If not, how could we proceed/conclude? I am using Introduction to Topology by Munkres.

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  • $\begingroup$ You should focus on finding a point where it does not look Euclidean. I would suggest to take a look at the point $(1,0)$ for example. What do you see? Is that homeomorphic to $\mathbb{R}^n$ for any $n$? $\endgroup$ – Maxime Scott Feb 22 '18 at 14:39
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Points $(-1,0)$ and $(1,0)$ do not have any euclidean neighborhood. If there was an homeomorphism from an open set of $\mathbb{R}^n$ to one of these neighborhoods $U$, what would the counterimage of $U-\{(1,0)\}$ be?

Certainly an open set of $\mathbb{R}^n$ consisting in not more than two connected components, while $U-\{(1,0)\}$ will have three distinct connected components (unless of course you choose $U$ to cointain $(-1,0)$, but you don't have to).

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Consider the point $x=(1,0)$ or $x=(-1,0)$, then any neighborhood of this points contains an arc of circle together a line segment. Note that any open set in any $\mathbb{R}^n$ can not be homeomorphic to such set.

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