Topological manifolds are defined to be locally Euclidean (e.g. John Lee). That is, any point is in an open set that is homeomorphic to either $\mathbb{R}^n$, an open ball in $\mathbb{R}^n$ or an open subset of $\mathbb{R}^n$.

I understand why "$\mathbb{R}^n$" and "open ball in $\mathbb{R}^n$" are equivalent definitions of locally Euclidean: any open ball is homeomorphic to $\mathbb{R}^n$, and the composition is again a homeomorphism. But why is "open subset of $\mathbb{R}^n$" also equivalent? Since an open subset that is not connected is not homeomorphic to any open balls.

Attempt: suppose $p$ is a point on the manifold and $U$ is the open set containing $p$ that is homeomorphic to an open subset $V$ of $\mathbb{R}^n$. There is an open ball centered at the homeomorphic image of $p$ in $V$.

The inverse image of this open ball is an open set $W$ in the manifold. Therefore, $p$ is in an open set $W$ that is homeomorphic to an open ball (by restricting the previous homeomorphism to the inverse image). For the other direction: an open ball is an open set.

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    $\begingroup$ What you wrote in your "attempt" is perfectly fine to answer your question. $\endgroup$ – Seub Dec 18 '13 at 1:24
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    $\begingroup$ Thanks! I wasn't very sure about its validity. $\endgroup$ – Jean Valjean Dec 18 '13 at 18:04

As noted in the comments, my "attempt" is a valid answer.


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