A subset of a metric space is open iff it is a union of open neighborhoods.

Question

May someone please verify if this proof is correct? Let E be the union of open neighborhoods, since the union of a collection of open sets is open, it follows that E is an open set. Assume E is an open set therefore, $\forall$ p$\in$ E $\exists$ $r>0$ : $N_r(p)$ $\subset$ E. So for $p_1$ , $\exists$ $N_{r_1}(p_1)$ $\subset$ E and the same thing holds for every p. This means that every element in E has a neighborhood around it, therefore, E is contained within the union of the neighborhoods. Furthermore, may someone please tell me of possible ways to improve this proof?

Answer 1

You asked if it is possible to improve your proof that an open set is the union of open sets.

Notation. Let $(M,\rho)$ be a metric space. Fix $x_0\in M$. Fix $r>0$. "$B(x_0,r)$" is notation for "$\{x\in M:\rho(x,x_0)<r\}$."

Terminology. Let $(M,\rho)$ be a metric space. Fix $S\subseteq M$. Say $S$ is open if for each $x\in S$, there exists $r>0$ such that $B(x,r)\subseteq S$.

Proposition. Let $(M,\rho)$ be a metric space. Fix $S\subseteq M$. The following are equivalent.

• (i) $S$ is open.
• (ii) $S=\bigcup_{\alpha\in A}U_\alpha$, where $\{U_\alpha\}_{\alpha\in A}$ is a family of open sets.

Proof.

($\Rightarrow$) Assume $S$ is open. Then $S=\bigcup_{x\in S}B(x,r_x)$, where $r_x>0$ and $B(x,r_x)\subseteq S$ for every $x\in S$. Because $B(x,r_x)$ is open for every $x\in S$, we are done.

($\Leftarrow$) Assume $S=\bigcup_{\alpha\in A}U_\alpha$, where $\{U_\alpha\}_{\alpha\in A}$ is a family of open sets. To prove $S$ is open, fix $x_0\in S$. Then $x_0\in U_{\alpha_0}$ for some $\alpha_0\in A$. Because $U_{\alpha_0}$ is open, there exists $r_0>0$ such that $B(x_0,r_0)\subseteq U_{\alpha_0}$. Because $S=\bigcup_{\alpha\in A}U_\alpha$, it follows that $B(x_0,r_0)\subseteq S$. As a result, $S$ is open.