Characterization of Open 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 $W \subseteq S$.

Then $W$ is open in $T$ if and only if:

$\forall U \in \UU: W \cap U$ is open in $\struct{U, \tau_U}$

Proof

Necessary Condition

This follows immediately from the definition of subspace topology.

$\Box$

Sufficient Condition

Let:

$W \cap U$ be open in $\struct{U, \tau_U}$ for each $U \in \UU$

From Open Set in Open Subspace:

$\forall U \in \UU : W \cap U$ is open in $T$

We have:

\(\ds W\)

\(=\)

\(\ds W \cap S\)

Intersection with Subset is Subset

\(\ds \)

\(=\)

\(\ds W \cap \paren{\bigcup \UU}\)

Definition of Open Cover

\(\ds \)

\(=\)

\(\ds \bigcup \set{W \cap U : U \in\UU}\)

Intersection Distributes over Union

By Open Set Axiom $\paren {\text O 1 }$: Union of Open Sets:

$W$ is open in $T$

$\blacksquare$