# Open subset of an $n$-manifold is again an $n$-manifold

If $M$ is an n-manifold, by which I mean a Hausdorff space where every point has an open nhood homeomorphic to $\mathbb{R^n}$, I am tying to show that any open subset of $M$, say $U$ is also an n-manifold.

So far I can show that if $U$ is an open subset of $M$ then every point in $U$ has a nhood homeomorphic to some open subset of $\mathbb{R}^n$. See this example: An open subset of a manifold is a manifold.

However, clearly not all open subsets of $\mathbb{R}^n$ is homeomorphic to $\mathbb{R}^n$. How can I remedy this?

• When you wrote “is also an n-nhood”, perhaps that you meant “is also a manifold”. – José Carlos Santos Sep 18 '17 at 10:16
• Yes, I did, thanks. – fosho Sep 18 '17 at 10:18

Every point $p$ of $U$ contains a neighborhood $N$ homeomorphic to $\mathbb{R}^n$. Let $\varphi$ be such a homeomorphism, such that $\varphi(p)=0$. Now, consider $U\cap N$. It is an open subset of $N$ and therefore $\varphi(U\cap N)$ is an open subset of $\mathbb{R}^n$, which contains $0$. So, take $r>0$ such that $B(0,r)\subset\varphi(U\cap N)$. Then $\varphi^{-1}\bigl(B(0,r)\bigr)$ is a neighborhood of $p$ contained in $N$, which is homeomorphic to $B(0,r)$, which, in turn, is homeomorphic to $\mathbb{R}^n$.