Open Set is Union of Elements of Basis

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Theorem

Let $T = \left({S, \tau}\right)$ be a topological space.

Let $B$ be a basis of $T$.

Let $V$ be an open subset of $S$.


Then $V = \bigcup \left\{ {G \in B: G \subseteq V}\right\}$


Proof

Let $x$ be arbitrary.

We will prove that

$x \in V \implies \exists Y \in \left\{ {G \in B: G \subseteq V}\right\}: x \in Y$

Assume that

$x \in V$

By definition of basis:

$\exists F \subseteq B: V = \bigcup F$

By definition of union:

$\exists Y \in F: x \in Y$

By Set is Subset of Union/General Result:

$Y \subseteq V$

Thus by definition of subset:

$Y \in \left\{ {G \in B: G \subseteq V}\right\}$

Thus $x \in Y$

$\Box$

Assume that

$\exists Y \in \left\{ {G \in B: G \subseteq V}\right\}: x \in Y$

By assumption:

$Y \subseteq V$

Thus by definition of subset:

$x \in V$

Hence by definition of union:

$V = \bigcup \left\{ {G \in B: G \subseteq V}\right\}$

$\blacksquare$


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