Basis and subbasis for a topology, and the topology generated by a family of sets

Let $(X,\mathcal{T})$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

A family $\mathcal{B}\subseteq \mathcal{T}$ is a basis for $\mathcal{T}$ if every open set is a union of members of $\mathcal{B}$: for every $U\in \mathcal{T}$ there is ${\mathcal{B}}_U\subseteq \mathcal{B}$ with $U=\bigcup {\mathcal{B}}_U$. Equivalently, and this is the form used in proofs,

$\text{for every }U\in \mathcal{T}\text{ and every }x\in U\text{ there is }B\in \mathcal{B}\text{ with }x\in B\subseteq U\mathrm{.}$

The two forms say the same thing. If every open $U$ is such a union and $x\in U$, then $x$ lies in one of the sets united, which is a member of $\mathcal{B}$ inside $U$. Conversely, if the displayed condition holds then $U=\bigcup \{B\in \mathcal{B}:B\subseteq U\}$, since each such $B$ is contained in $U$ and each $x\in U$ lies in one of them. Note that $\bigcup \mathrm{\varnothing }=\mathrm{\varnothing }$, so the empty open set is covered by the empty subfamily and needs no member of $\mathcal{B}$. The members of a basis are called basic open sets.

The topology generated by a family. Let $\mathcal{S}\subseteq \mathcal{P}(X)$ be any family of subsets of $X$. Then

$⟨\mathcal{S}⟩:=\bigcap \{{\mathcal{T}}^{\mathrm{\prime }}:{\mathcal{T}}^{\mathrm{\prime }}\text{ is a topology on }X\text{ with }\mathcal{S}\subseteq {\mathcal{T}}^{\mathrm{\prime }}\}$

is a topology on $X$, it contains $\mathcal{S}$, and it is contained in every topology on $X$ that contains $\mathcal{S}$. It is called the topology generated by $\mathcal{S}$, and $\mathcal{S}$ is a subbasis for a topology $\mathcal{T}$ when $\mathcal{T}=⟨\mathcal{S}⟩$.

This is well posed, and the obligation is discharged here. The collection being intersected is nonempty, because $\mathcal{P}(X)$ is a topology on $X$ containing $\mathcal{S}$; so the intersection is an intersection of a nonempty family of subsets of $\mathcal{P}(X)$ and is a set. It is a topology: $\mathrm{\varnothing }$ and $X$ lie in every topology on $X$, hence in the intersection, which is (T1); if ${\mathcal{S}}^{\mathrm{\prime }}\subseteq ⟨\mathcal{S}⟩$ then ${\mathcal{S}}^{\mathrm{\prime }}$ is a subfamily of each ${\mathcal{T}}^{\mathrm{\prime }}$ in the collection, so $\bigcup {\mathcal{S}}^{\mathrm{\prime }}$ lies in each ${\mathcal{T}}^{\mathrm{\prime }}$ and hence in the intersection, which is (T2); and the same argument with $U\cap V$ gives (T3). It contains $\mathcal{S}$ because every ${\mathcal{T}}^{\mathrm{\prime }}$ in the collection does, and it is contained in each such ${\mathcal{T}}^{\mathrm{\prime }}$ because an intersection is contained in each of its members. So $⟨\mathcal{S}⟩$ is the coarsest topology on $X$ containing $\mathcal{S}$ (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison), and in particular it is uniquely determined by $\mathcal{S}$.

Neither notion is intrinsic to the family alone. "$\mathcal{B}$ is a basis for $\mathcal{T}$" and "$\mathcal{S}$ is a subbasis for $\mathcal{T}$" are relations between a family and a topology, not properties of the family. The question of which families are a basis for some topology, and how the topology generated by a subbasis is computed from it, is settled by the next item.

• Every topology is a basis for itself, so a basis always exists; the point of a basis is to be smaller and more explicit than $\mathcal{T}$, and the point of a subbasis is to be smaller still at the cost of one round of finite intersections.

• Two extremes of the generated topology. $⟨\mathrm{\varnothing }⟩$ is the indiscrete topology, since $\{\mathrm{\varnothing },X\}$ is a topology containing the empty family and is contained in every topology. At the other end, $⟨\mathcal{P}(X)⟩=\mathcal{P}(X)$ is the discrete topology.

• Generation is monotone and idempotent. If ${\mathcal{S}}_1\subseteq {\mathcal{S}}_2$ then every topology containing ${\mathcal{S}}_2$ contains ${\mathcal{S}}_1$, so $⟨{\mathcal{S}}_1⟩\subseteq ⟨{\mathcal{S}}_2⟩$; and $⟨⟨\mathcal{S}⟩⟩=⟨\mathcal{S}⟩$ because $⟨\mathcal{S}⟩$ is itself a topology containing $\mathcal{S}$. Both are used silently below.