# An open subset of a manifold is a manifold

Let $M$ be an $n$-manifold. I would like to show that any open subset $A$ of $M$ is an $n$ manifold. Let $x\in A$. since $M$ is a manifold, there exists a neighborhood $U_x\subseteq M$ of $x$ such that $U_x$ is homeomorphic to an open subset $V$ of $\mathbb R^n$, let $h:U_x\to V$ be such homeomorphism. Consider the restriction of $h$ to $A\cap U_x$. The set $A\cap U_x$ contains the open set $A$ that contains $x$ so $A\cap U_x$ is a neighborhood of $x$. The proof is finished if i could say that $h(A\cap U_x)$ is an open subset of $\mathbb R^n$. this is obvious if $U_x$ were open because in this case i would say that $A\cap U_x$ is open in $M$ and since $h$ is homeomorphism then it is an open map hence $h(A\cap U_x)$ is an open subset of $\mathbb R^n$. But the Problem is that $U_x$ need not be open..

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Charts are usually required to have open domains $U_x$. Where are you getting this definition? –  Dylan Moreland Mar 19 '12 at 15:13
I suppose it depends on what your definition of *neighborhood... of x" is, as applied to $U_x$. Many authors would say this means an open set containing $x$. –  hardmath Mar 19 '12 at 15:14
All references i know are talking about neighborhoods not of open domains, and a neighborhood need not be open it has to contain an open set containing $x$ without being open itself. am i wrong? –  palio Mar 19 '12 at 15:16

Notice that you can assume without loss of generality that $U_x$ is open. After all, the definition of neighborhood is that $U_x$ contains an open subset, say $U_x'$, which contains the point $x$. Then $U_x'$ is an open neighborhood of $x$ which is homeomorphic to an open subset of $\mathbb R^n$, namely $h(U_x')$.