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I always read that an $n$-manifold is a topological space that locally looks like $\mathbb R^n$. But in the definition we require that each point has a neighborhood that is homeomorphic to "an open subset of $\mathbb R^n$" and not necessarily all of $\mathbb R^n$. Knowing that an open subset of $\mathbb R^n$ need not be homeomorphic to $\mathbb R^n$ this seems to be confusing!!

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    $\begingroup$ If $x\in V$, where $V$ is an open set in $\Bbb R^n$, then there is an open $B\in\Bbb R^n$ such that $x\in B\subseteq V$ and $B$ is homeomorphic to $\Bbb R^n$, so the conditions are equivalent. $\endgroup$ – Brian M. Scott Mar 18 '12 at 16:29
  • $\begingroup$ basically you are saying that every open subset of $\mathbb R^n$ contains a copy of $\mathbb R^n$. each $x\in M$ has a neighborhood $U_x$ homeomorphic to an open subset $V$ of $\mathbb R^n$. Now $V$ contains $B\cong \mathbb R^n$, but this does not say that $U_x\cong \mathbb R^n$. $\endgroup$ – palio Mar 18 '12 at 16:50
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    $\begingroup$ No, I’m saying that every non-empty open subset of $\Bbb R^n$ contains a copy of $\Bbb R^n$ around each of its points. As I said: $x\in B\subseteq V$, and $B$ is homeomorphic to $\Bbb R^n$. $\endgroup$ – Brian M. Scott Mar 18 '12 at 16:52
  • $\begingroup$ I think $x$ is irrelevant here since the point $x$ is picked in $M$ not in $\mathbb R^n$, and for this $x\in M$ there is a neighb $U_x$ such that $U_x$ is $\cong$ to some subset $V\subset \mathbb R^n$.. am i missing something obvious? $\endgroup$ – palio Mar 18 '12 at 17:01
  • $\begingroup$ @Pete: Will do. $\endgroup$ – Brian M. Scott Mar 18 '12 at 17:01
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Suppose that a point $x\in M$ has a nbhd $U$ homeomorphic to an open subset $V$ of $\Bbb R^n$. Let $h:U\to V$ be a homeomorphism. Since $h(x)\in V$, and $V$ is open in $\Bbb R^n$, there is an $r>0$ such that the open ball $B(h(x),r)\subseteq V$. Note that $B(h(x),r)$ is homeomorphic to $\Bbb R^n$. Now let $B=h^{-1}[B(h(x),r)]$; then $B$ is homeomorphic to $B(h(x),r)$ and hence to $\Bbb R^n$, and clearly $x\in B\subseteq U$. Thus, $x$ has a nbhd homeomorphic to $\Bbb R^n$.

Thus, requiring that a point of $M$ have a nbhd homeomorphic to an open subset of $\Bbb R^n$ is in fact equivalent to requiring that the point have a nbhd homeomorphic to $\Bbb R^n$ itself. In particular, the two definitions of manifold are equivalent.

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