Bases in Locally Euclidean spaces

**Drexel28** · March 30th 2010, 05:59 PM

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

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

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

Comments?

**Tinyboss** · March 30th 2010, 07:36 PM

Seems right to me, and I don't see anything that's redundant.

**Drexel28** · March 30th 2010, 07:40 PM

Originally Posted by Tinyboss

Seems right to me, and I don't see anything that's redundant.

Thanks friend!

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