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

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?

• No proof is needed, because the very definition of an open set in a metric space is a set which is a union of open balls. Jan 16, 2019 at 2:17
• But is what I said, equivalent to the definition? Jan 16, 2019 at 2:19
• Furthermore, is the union of neighborhoods necessarily a neighborhood? Jan 16, 2019 at 2:20
• The definition of an open set, similar to any mathematical definition, allows you to automatically translate back and forth between the two statements "$E$ is open" and "$E$ is a union of open balls", without any proof. That's what definitions do. Jan 16, 2019 at 2:26
• @LeeMosher that's not the definition Rudin gives. And it's pretty clear from the question that mathsssislife isn't using your definition. Jan 16, 2019 at 3:35

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)."

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.