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Let $X$ be any set.
> [!proposition]+ Proposition. ([[basis-nestling characterization of comparing topologies]])
> Let $\mathscr{B}$ and $\mathscr{B'}$ be bases for [[topological space|topologies]] $\tau$ and $\tau'$, respectively, on $X$. Then the following are equivalent:
> 1. $\tau'$ is [[comparable topologies|finer]] than $\tau$;
> 2. $\mathscr{B}'$ [[nestles in]] $\mathscr{B}$, i.e., For each $x \in X$, and each [[basis for a topology|basis element]] $B \in \mathscr{B}$ containing $x$, there is a basis element $B' \in \mathscr{B'}$ s.t. $x \in B' \subset B$.
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> [!proof]+ Proof. ([[basis-nestling characterization of comparing topologies]])
> $(2) \implies (1)$. Suppose $(2)$ and let $U \in \tau$; we want to show $U \in \tau'$. Let $x \in U$. Since $\mathscr{B}$ [[topology generated by a basis|generates]] $\tau$, there exists $B \in \mathscr{B}$ such that $x \in B \subset U$. By $(2)$, there exists $B' \in \mathscr{B'}$ s.t. $x \in B' \subset B$. Then $x \in B' \subset U$, meaning that $U \in \tau'$ by [[topology generated by a basis|definition]].
>
> $(1) \implies (2)$. Suppose $(1)$ and let $x \in X$ with $x \in B \in \mathscr{B}$. Since $\tau' \supset \tau$, $B \in \tau'$. Since $\tau'$ is generated by $\mathscr{B'}$, there is an element $B' \in \mathscr{B'}$ s.t. $x \in B' \subset B$.
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####
#reformatReviseBBatch5
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#### References
> [!backlink]
> ```dataview
TABLE rows.file.link as "Further Reading"
FROM [[]]
FLATTEN file.tags
GROUP BY file.tags as Tag
> [!frontlink]
> ```dataview
TABLE rows.file.link as "Further Reading"
FROM outgoing([[]])
FLATTEN file.tags
GROUP BY file.tags as Tag