Opens in a manifold I am studying for my exam of manifolds and I dno't understand why the following is true.
Let $(U,\phi)$ be a local map at $p$ in the manifold $M$, and let $\phi(p)=0$. Then there exists an open set $W$ such that $p \in W \subset \bar W \subset U$, where $\bar W$ is compact.
Why can we find such $W$? I can't find an argument for this.
Thanks!
 A: Remember that, by definition $\phi$ is a homeomorphism. Hence $\phi(U)$ is an open subset of $R^n.$ From that there exist a radius $r$ such that open ball $K(0,r)\subset\phi(U).$ We put ball at $0$ because of assumption that $\phi(p)=0.$ Now consider a closed ball $B_1=\bar{K}(0,r/2)$ which is compact. Consider now open ball $B_2=K(0,r/2)$ and set $W=\phi^{-1}(B_2).$ Again by homeomorphity (I assume it is such word) of $\phi$ we get that $W$ is open and again by homeomorphity of $\phi$ we can get the closure in and out of preimage in the following:
$$W=\phi^{-1}(B_2)\subset\phi^{-1}(B_1)=\phi^{-1}(\overline{B_2})=\overline{\phi^{-1}(B_2)}=\overline{W}=\phi^{-1}(\overline{B_2})\subset\phi^{-1}(K(0,r))\subset\phi^{-1}(\phi(U))=U.$$
It is left to show that $p\in W,$ but it is straightforward because $p=\phi^{-1}(0)$ and $0\in B_2.$
A: As Nick noted, $\Bbb R^n$ is locally compact. This means that every point in a manifold has a compact neighborhood $K$, so we can take $W$ to be (for instance) the interior of $K$.
