Characterization of Closed Set by Open Cover

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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 $T$ if and only if:

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

Proof

Necessary Condition

This follows immediately from Closed Set in Topological Subspace.

$\Box$

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:

\(\ds \forall U \in \UU: \, \)

\(\ds U \setminus \paren{F \cap U}\)

\(=\)

\(\ds U \setminus F\)

Set Difference with Intersection

\(\ds \)

\(=\)

\(\ds \paren{U \cap S} \setminus F\)

Intersection with Subset is Subset

\(\ds \)

\(=\)

\(\ds U \cap \paren {S \setminus F}\)

Intersection with Set Difference is Set Difference with Intersection

Hence:

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

From Characterization of Open Set by Open Cover:

$S \setminus F$ is open in $T$

By definition of closed set:

$F$ is closed in $T$

$\blacksquare$