Nonexamples:: *[[Nonexamples]]*
Constructions:: *[[Constructions|Used in the construction of...]]*
Specializations:: *[[Specializations]]*
Generalizations:: *[[Generalizations]]*
Justifications and Intuition:: *[[Justifications and Intuition]]*
Properties:: [[condition for obtaining a basis from a topology]]
Sufficiencies:: *[[Sufficiencies]]*
Equivalences:: *[[Equivalences]]*
Examples:: [[singletons are a basis for the discrete topology]]
---
Let $X$ be any set.
> [!definition]+ Definition. ([[basis for a topology]])
> A **basis** for a [[topological space|topology]] on $X$ is a collection $\mathscr{B}$ of subsets of $X$, called *basis elements*, such that
> 1. (Covering) For each $x \in X$, at least one *basis element* contains $x$;
> 2. (Nestling) $\mathscr{B}$ [[nestles in]] its intersections[^1]: if $x \in B_{1} \cap B_{2}$ with $B_{1}, B_{2} \in \mathscr{B}$, then $\ex B_3 \in \mathscr{B}$ for which $x \in B_{3} \subset B_{1} \cap B_{2}$.
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> [!basicexample]+
> Let $X$ be a finite set at $\tau \subset \mathcal{P}(X)$ a [[topological space|topology]] on $X$.
>
> **1.** Prove that for any $x \in X$ there is an [[topological space|open]] [[neighborhood]] $B_x \subset X$ of $x$ that is contained in every other open [[neighborhood]] of $x$.
>
> **Proof of 1.** Let $x \in X$. Indexing the open [[neighborhood]]s $U_{\alpha}$ of $x$ with $\alpha \in J$, $J$ an index set, we define $B_{x} := \bigcap_{\alpha}^{}U_{\alpha}$. By construction, $B_{x} \subset U_{\alpha}$ for any open [[neighborhood]] $U_{\alpha}$, so we just have to show that $B_{x}$ is indeed an open neighborhood of $x$. To begin, $\mathcal{P}(X)$ is finite and so WLOG $J$ is finite and hence $B_{x}$ is indeed an open set in $X$ as an intersection thereof. Also $x \in B_{x}$ because $x \in U_{\alpha}$ for all $\alpha$. Every open set containing $x$ is also an open neighborhood of $x$, of course. So, the proposition is shown.
>
> **2.** Prove that $\mathscr{B}:=\{ B_{x}: x \in X \}$ is a [[basis for a topology|basis for the topology]] $\tau$.
>
> **Proof of 2.** Certainly $\mathscr{B}$ is a [[basis for a topology]]... that $\mathscr{B}$ covers $X$ is immediate by construction: to each $x$ there corresponds $B_{x} \in \mathscr{B}$ since $B_{x}$ is an open [[neighborhood]] of $x$. And letting $B_{1}, B_{2} \in \mathscr{B}$ with $x \in B_{1} \cap B_{2}$, we have $B_{3}:=B_{x} \ni x$ s.t. $B_{x} \subset B_{1} \cap B_{2}$, since by definition $B_{x} = \bigcap_{\alpha}^{}U_{\alpha} \subset B_{1} \cap B_{2}$.
>
> Denote by $\tau'$ the [[topology generated by a basis|topology generated by]] $\mathscr{B}$. We want to show $\tau'=\tau$.
>
> For one inclusion, let $U \in \tau'$. So, for all $x \in U$ there exists $B_{x} \in \mathscr{B}$ s.t. $x \in B_{x} \subset U. Since $B_{x} \in \tau$, the union $\bigcup_{x \in X}^{}B_{x}$ is contained in $\tau$. Moreover, union equals $U$, so we have $U \in \tau$. Hence $\tau \supset \tau'$.
>
> For the other inclusion, let $O \in \tau$. The "open neighborhoods of $x \in X
quot; are the open sets in $X$ containing $x". So, taking $x \in O$ and obtaining $B_{x}$ as in **1**, we have $x \in B_{x} \subset O$. Thus $O \in \tau'$ and hence $\tau \subset \tau'$. This completes the proof.
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[^1]: That is, $\mathscr{B}$ [[nestles in]] the set $\mathcal{I}_{\mathscr{B}}$ of all pairwise intersections of elements in $\mathscr{B}$.
---
#### 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
#reformatrevisebbatch2