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?

  • $\begingroup$ 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. $\endgroup$
    – Lee Mosher
    Jan 16, 2019 at 2:17
  • $\begingroup$ But is what I said, equivalent to the definition? $\endgroup$ Jan 16, 2019 at 2:19
  • $\begingroup$ Furthermore, is the union of neighborhoods necessarily a neighborhood? $\endgroup$ Jan 16, 2019 at 2:20
  • $\begingroup$ 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. $\endgroup$
    – Lee Mosher
    Jan 16, 2019 at 2:26
  • $\begingroup$ @LeeMosher that's not the definition Rudin gives. And it's pretty clear from the question that mathsssislife isn't using your definition. $\endgroup$ Jan 16, 2019 at 3:35

1 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.


($\Rightarrow$) 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.


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