# Thread: Bases in Locally Euclidean spaces

1. ## Bases in Locally Euclidean spaces

Hey everyone, I think I am making the following WAY too difficult. Does anyone have insight into A) whether the following is even correct or B) an easier way to do it (as I am sure there is one).

Problem: Let $\displaystyle \mathfrak{M}$ be a locally Euclidean space of dimension $\displaystyle n$. Call an open subset $\displaystyle E$ of $\displaystyle \mathfrak{M}$ a Euclidean ball if $\displaystyle E\approx O$ for some open $\displaystyle O\subseteq\mathbb{R}^n$. Prove that $\displaystyle \mathfrak{M}$ has an open base of Euclidean balls.

Proof: Let $\displaystyle x\in X$ be arbitrary and let $\displaystyle E_x$ be the guaranteed Euclidean ball containing $\displaystyle x$ with $\displaystyle \varphi_x:E_x\to O_x$ the necessary homeomorphism. Let $\displaystyle \Omega_x=\left\{B_{\varepsilon}(\varphi(x))):B_{\v arepsilon}\subseteq O_x\right\}$ (each of these balls are open in $\displaystyle \mathbb{R}^n$ for obvious reasons) and $\displaystyle \Lambda_x=\left\{\varphi_x^{-1}(\omega):\omega\in\Omega_x\right\}$. Clearly $\displaystyle \Lambda_x$ contains open subset of $\displaystyle E_x$, but since $\displaystyle E_x$ is in fact open we see that each element of $\displaystyle \Lambda_x$ is open in $\displaystyle X$. So, set $\displaystyle \Lambda=\bigcup_{x\in X}\Lambda_x$. By previous comment this is a class of open subsets of $\displaystyle \mathfrak{M}$. Also, $\displaystyle \varphi_x\mid_{\varphi_x^{-1}(\omega)}:\varphi_x^{-1}(\omega)\to\omega$ is a homeomorphism and so $\displaystyle \Lambda$ is in fact a collection of Euclidean balls. It remains to show that it's a base.

So, let $\displaystyle y\in X$ be arbitrary and $\displaystyle N$ any neighborhood of $\displaystyle y$. Thus, keeping the notation of the previous paragraph we have that $\displaystyle E_y\cap N$ is an open subspace of $\displaystyle E_y$ and so $\displaystyle \varphi_y(E_y\cap N)$ is an open subset of $\displaystyle O_y$. So, there exists some $\displaystyle B_{\varepsilon}(\varphi_y(y))\subseteq \varphi_y(E_y\cap N)$. It follows that $\displaystyle \varphi_y^{-1}\left(B_{\varepsilon}(\varphi_y(y))\right)\in\La mbda$ and $\displaystyle x\in\varphi_y^{-1}\left(B_{\varepsilon}(\varphi_y(y))\right)\subse teq E_y\cap N\subseteq N$. The conclusion follows. $\displaystyle \blacksquare$.