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Let $X$ be a set; let $\mathscr{B}$ be a [[topology generated by a basis|basis for a topology]] $\tau$ on $X$.
> [!proposition] Proposition. ([[open sets are unions of basis elements]])
> Then $\tau$ equals the collection of all unions of elements of $\mathscr{B}$.
> [!proof]- Proof. ([[open sets are unions of basis elements]])
> We'll show a two-way inclusion.
> 1. Since $\mathscr{B} \subset \tau$, any union of elements of $\mathscr{B}$ is in $\tau$ by definition of [[topological space]].
> 2. Consider $U \in \tau$. For each $x \in U$ we can find $B_{x} \in \mathscr{B}$ such that $x \in B_{x} \subset U$. Then $U=\bigcup_{x \in U} B_{x}$ and so we've written each element $U$ of $\tau$ as a union of basis elements. $\textcolor{Apricot}{\text{Warning:}}$ This expression for $U$ is not unique!
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#### References
> [!backlink]
> ```dataview
> TABLE rows.file.link as "Further Reading"
> FROM [[]]
> FLATTEN file.tags as Tag
> WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom"
> GROUP BY Tag
> ```
> [!frontlink]
> ```dataview
> TABLE rows.file.link as "Further Reading"
> FROM outgoing([[]])
> FLATTEN file.tags as Tag
> WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom"
> GROUP BY Tag
> ```