Characterization of Closed Set by Open Cover

Theorem
Let $T = \struct{S, \tau}$ be a topological space.

Let $\UU$ be an open cover of $T$.

For each $U \in \UU$, let $\tau_U$ denote the subspace topology on $U$.

Let $F \subseteq S$.

Then $F$ is closed in $T$ :
 * $\forall U \in \UU: F \cap U$ is closed in $\struct{U, \tau_U}$

Necessary Condition
This follows immediately from Closed Set in Topological Subspace.

Sufficient Condition
Let:
 * $F \cap U$ be closed in $\struct{U, \tau_U}$ for each $U \in \UU$

By definition of closed set:
 * $U \setminus \paren{F \cap U}$ is open in $\struct{U, \tau_U}$ for each $U \in \UU$

We have:

Hence:
 * $\forall U \in \UU : U \cap \paren {S \setminus F}$ is open in $\struct{U, \tau_U}$ for each $U \in \UU$

From User:Leigh.Samphier/Topology/Characterization of Open Set by Open Cover:
 * $S \setminus F$ is open in $T$

By definition of closed set:
 * $F$ is closed in $T$