# Basis for a topology

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## Definition

Let [ilmath](X,\mathcal{ J })[/ilmath] be a topological space and let [ilmath]\mathcal{B}\in\mathcal{P}(\mathcal{P}(X))[/ilmath] be any collection of subsets of [ilmath]X[/ilmath][Note 1]. We say [ilmath]\mathcal{B} [/ilmath] is a basis for the topology [ilmath]\mathcal{J} [/ilmath] if both of the following are satisfied:

1. [ilmath]\forall B\in\mathcal{B}[B\in\mathcal{J}][/ilmath] - every element of [ilmath]\mathcal{B} [/ilmath] is an open set of [ilmath](X,\mathcal{ J })[/ilmath]
2. [ilmath]\forall U\in\mathcal{J}\exists\{B_\alpha\}_{\alpha\in I}\subseteq\mathcal{B}[\bigcup_{\alpha\in I}B_\alpha=U][/ilmath] - every open set in [ilmath](X,\mathcal{ J })[/ilmath] is the union of some arbitrary family of basis elements[Note 2]

The elements of [ilmath]\mathcal{B} [/ilmath] are called basis elements.

## Basis criterion

Let [ilmath](X,\mathcal{ J })[/ilmath] be a topological space and let [ilmath]\mathcal{B}\in\mathcal{P}(\mathcal{P}(X))[/ilmath] be a topological basis for [ilmath](X,\mathcal{ J })[/ilmath]. Then:

• [ilmath]\forall U\in\mathcal{P}(X)\big[U\in\mathcal{J}\iff\underbrace{\forall p\in U\exists B\in\mathcal{B}[p\in B\subseteq U]}_{\text{basis criterion} }\big][/ilmath][Note 3]

If a subset [ilmath]U[/ilmath] of [ilmath]X[/ilmath] satisfies[Note 4] [ilmath]\forall p\in U\exists B\in\mathcal{B}[p\in B\subseteq U][/ilmath] we say it satisfies the basis criterion with respect to [ilmath]\mathcal{B} [/ilmath]

## Topology generated by a basis

Let [ilmath]X[/ilmath] be a set and let [ilmath]\mathcal{B}\in\mathcal{P}(\mathcal{P}(X))[/ilmath] be any collection of subsets of [ilmath]X[/ilmath], then:

• [ilmath](X,\{\bigcup\mathcal{A}\ \vert\ \mathcal{A}\in\mathcal{P}(\mathcal{B})\})[/ilmath] is a topological space with [ilmath]\mathcal{B} [/ilmath] being a basis for the topology [ilmath]\{\bigcup\mathcal{A}\ \vert\ \mathcal{A}\in\mathcal{P}(\mathcal{B})\}[/ilmath]
• we have both of the following conditions:
1. [ilmath]\bigcup\mathcal{B}=X[/ilmath] (or equivalently: [ilmath]\forall x\in X\exists B\in\mathcal{B}[x\in B][/ilmath][Note 5]) and
2. [ilmath]\forall U,V\in\mathcal{B}\big[U\cap V\neq\emptyset\implies \forall x\in U\cap V\exists B\in\mathcal{B}[x\in W\wedge W\subseteq U\cap V]\big][/ilmath][Note 6]
• Caveat:[ilmath]\forall U,V\in\mathcal{B}\ \forall x\in U\cap V\ \exists W\in\mathcal{B}[x\in W\subseteq U\cap V][/ilmath] is commonly said or written; however it is wrong, this is slightly beyond just abuse of notation.[Note 7]