Open Sets in Manifolds

# Open Sets in Manifolds

Definition: Let $M$ be an $m$-dimensional manifold with maximal atlas $\mathcal A = \{ (U_{\alpha}, \varphi_{\alpha} ) : \alpha \in \Gamma \}$. A set $U \subseteq M$ is said to be Open in $M$ if for all $\alpha \in \Gamma$ we have that $\varphi_{\alpha} (U \cap U_{\alpha})$ is open in $\mathbb{R}^m$. |

*By "open in $\mathbb{R}^m$" we mean in a topological sense. We say that a set $A$ is open in $\mathbb{R}^m$ with the usual Euclidean topology if $A = \mathrm{int} (A)$, i.e., for all $x \in A$ there exists an $r > 0$ such that $B(x, r) \subseteq A$.*

There are a few important things to notice. First, if $U$ is an open set in a manifold $M$ then $U$ is itself a manifold for which we can define a maximal atlas on $U$ by:

(1)\begin{align} \quad \mathcal A^* = \{ (U \cap U_{\alpha}, \varphi_{\alpha} \mid_{U \cap U_{\alpha}}) : \alpha \in \Gamma \} \end{align}

Secondly, the notion of open sets in a manifold naturally gives rise to that manifold having a topology determined by such open sets as we prove in the following proposition.

Proposition 1: Let $M$ be an $m$-dimensional manifold with maximal atlas $\mathcal A = \{ (U_{\alpha}, \varphi_{\alpha}) : \alpha \in \Gamma \}$. Then:a) $M$ and $\emptyset$ are open in $M$.b) If $\{ U_i : i \in I \}$ is an arbitrary collection of open sets in $M$ then $\displaystyle{\bigcup_{i \in I} U_i}$ is open in $M$.c) If $\{ U_1, U_2, ..., U_n \}$ is a finite collection of open sets in $M$ then $\displaystyle{\bigcap_{i=1}^{n} U_i}$ is open in $M$. |

**Proof of a)**Notice that for all $\alpha \in \Gamma$ we have that $M \cap U_{\alpha} = U_{\alpha}$ (since $\mathcal A$ covers $M$). So for all $\alpha \in \Gamma$, $\varphi_{\alpha} (M \cap U_{\alpha}) = \varphi_{\alpha} (U_{\alpha})$ which is open in $\mathbb{R}^m$ by definition of $(U_{\alpha}, \varphi_{\alpha})$ being a chart on $M$. Therefore, $M$ is open in $M$.

- Furthermore, notice that for all $\alpha \in \Gamma$ we have that $\emptyset \cap U_{\alpha} = \emptyset$. So for all $\alpha \in \Gamma$, $\varphi_{\alpha} (\emptyset \cap U_{\alpha}) = \varphi_{\alpha} (\emptyset) = \emptyset$ which is trivially open in $\mathbb{R}^m$. Therefore, $\emptyset$ is open in $M$.

**Proof of b)**Let $\{ U_i : i \in I \}$ be an arbitrary collection of open sets in $M$. Then for all $\alpha \in \Gamma$:

\begin{align} \quad \varphi_{\alpha} \left ( \left ( \bigcup_{i \in I} U_i \right ) \cap U_{\alpha} \right ) = \varphi_{\alpha} \left ( \bigcup_{i \in I} (U_i \cap U_{\alpha}) \right ) = \bigcup_{i \in I} \varphi_{\alpha]} (U_i \cap U_{\alpha}) \end{align}

- Since for each $i \in I$, $U_i$ is open in $M$ we have that $\varphi_{\alpha} (U_i \cap U_{\alpha})$ is open. So from above we see that $\displaystyle{\varphi_{\alpha} \left ( \left ( \bigcup_{i \in I} U_i \right ) \cap U_{\alpha} \right )}$ is an arbitrary union of open sets in $\mathbb{R}^m$ which is open in $\mathbb{R}^m$ for all $\alpha \in \Gamma$. So $\displaystyle{\bigcup_{i \in I} U_i}$ is open in $M$.

**Proof of c)**Let $\{ U_1, U_2, ..., U_n \}$ be a finite collection of open sets in $M$. Then for all $\alpha \in \Gamma$:

\begin{align} \quad \varphi_{\alpha} \left ( \left ( \bigcap_{i=1}^{n} U_i \right ) \cap U_{\alpha} \right ) = \varphi_{\alpha} \left ( \bigcap_{i=1}^{n} U_i \cap U_{\alpha} \right ) = \bigcap_{i=1}^{n} \varphi_{\alpha} (U_i \cap U_{\alpha}) \end{align}

- Since for each $i \in I$, $U_i $$ is open in $M$ we have that $\varphi_{\alpha} (U_i \cap U_{\alpha})$ is open. So from above we see that $\displaystyle{\varphi_{\alpha} \left ( \left ( \bigcap_{i=1}^{n} U_i \right ) \cap U_{\alpha} \right )}$ is a finite intersection of open sets in $\mathbb{R}^m$ which is open in $\mathbb{R}^m$ for all $\alpha \in \Gamma$. So $\displaystyle{\bigcap_{i=1}^{n} U_i}$ is open in $M$. $\blacksquare$