> [!definition] > > Let $(X, \topo)$ be a [[Topological Space|topological space]]. A [[Set|set]] $C \subseteq X$ is **closed** if its complement is an [[Open Set|open set]]. > [!theorem] > > A set is closed if and only if it contains all of its [[Boundary|boundary]] (it's its own [[Topological Closure|topological closure]]) > $ > (X - A) \in \topo \Leftrightarrow X = \overline{X} > $ > [!theorem] > > Let $(X, \topo)$ be a topological place. > 1. Let $U$ be an open set and $E \subset X$, then $U \cap E \ne \emptyset$ if and only if $U \cap \ol{E} \ne \emptyset$. > 2. Let $\seqf{E}$ be a family of sets, then $\ol{\bigcup_{i = 1}^nE_i} = \bigcup_{i =1}^n\ol{E_i}$. > 3. Let $\seqi{E}$ be a family of sets, then it is locally finite if and only if $\seqi{\ol{E}}$ is locally finite. > 4. If $\seqi{E}$ is locally finite, then $\ol{\bigcup_{i \in I}E_i} = \bigcup_{i \in I}\ol{E_i}$. > > *Proof*. Suppose that $U \cap \ol{E} \ne \emptyset$. Let $x \in U \cap \ol{E}$, then $x$ is an [[Topological Closure|adherent point]] to $E$. Therefore $V \cap E \ne \emptyset$ for all $V \in \cn^o(x)$, and as $U \in \cn^o(x)$, $U \cap E \ne \emptyset$. > > Since the closure is the smallest closed set containing the union, $\ol{\bigcup_{i = 1}^nE_i} \subset \bigcup_{i =1}^n\ol{E_i}$. As $E_i \subset \bigcup_{j = 1}^nE_j$ for all $i$, $\ol{\bigcup_{i = 1}^nE_i} \supset \bigcup_{i =1}^n\ol{E_i}$. > > Suppose that $\seqi{E}$ is locally finite and let $x \in X$, then there exists $U \in \cn^o(x)$ such that only finitely many sets in the family intersect $U$. Since $E_i$ intersects $U$ if and only if $\ol E_i$ intersects $U$, $U$ only intersects with finitely many $\ol{E_i}$s. > > Let $x \in \ol{\bigcup_{i \in I}E_i}$, and let $U \in \cn^o(x)$ that only intersect with finitely many $E_i$s, say $\seqf{E_i}$. Then any open subset of $U$ would intersect $\bigcup_{i = 1}^nE_i$, and $x \in \ol{\bigcup_{i = 1}^nE_i} = \bigcup_{i = 1}^n\ol{E_i}$, so $x \in \bigcup_{i \in I}\ol{E_i}$.