For many notions of structure, particularly for topological categories, one can specify a structure by a *base* or *subbase* that generate the structure. Besides being convenient ways to specify a structure, they may even be necessary when using weak foundations.

Sometimes one says ‘basis’ instead of ‘base’, but I (Toby Bartels) think that it's safest to save the former term for the generating set of a free object (or an analogous situation), especially in an algebraic category. Although a basis and a base can both generate something, they tend to do so in very different ways. (It doesn't help that ‘bases’ is the plural of both, although the pronunciation is different.)

Typically, every structure of an appropriate type is both a base and a subbase for itself, while every base is a subbase. Bases and subbases can also be characterised independently; every subbase generates a base (which tends to be *saturated* in some way), while every base (saturated or not) generates a complete structure.

Recall that a **filter** on a poset $L$ is a subset $F$ of $L$ such that:

- If $x \leq y$ and $x \in F$, then $y \in F$.
- Some $x \in F$.
- If $x, y \in F$, then for some $z \in F$, $z \leq x, y$.

More generally, any subset $F$ satisfying (2,3) is a **filter base**.

Given a filter base $F$ in a poset, we generate a filter $\overline{F}$ by closing under (1); that is, if $F$ is a filter base on a poset $L$, then

$\overline{F} \coloneqq \{ y \;|\; \exists x \in F,\; x \leq y \}$

is a filter on $L$.

If $L$ is a meet-semilattice, then we can equivalently define a filter as a subset $F$ such that:

- If $x \leq y$ and $x \in F$, then $y \in F$ (same as before).
- The top element $\top \in F$.
- If $x, y \in F$, then $x \wedge y \in F$.

Now any subset satisfying (2,3) is a **saturated filter base**, and any subset whatsoever is a **filter subbase**.

Given a filter subbase $F$ in a semilattice, we generate a base $\vec{F}$ by closing under (2,3) in the second list; that is, if $F$ is a filter subbase on a semilattice $L$, then

$\vec{F} \coloneqq \{ \bigwedge_{i=1}^n x_i \;|\; x_i \in F \}$

is a filter base on $L$, which in fact is saturated. (Note that $\top \in \vec{F}$ follows when $n = 0$.)

Given a filter subbase $F$ in a semilattice, we can generate a filter by first generating a base $\vec{F}$ and then generating a filter $\overline{\vec{F}}$. Alternatively, we can generate the same filter by closing under (1,2,3) in the first list all at once. That is, if $F$ is a filter subbase on a semilattice $L$, then

$\overline{F} \coloneqq \{ y \;|\; \exists x_1,\ldots,x_n \in F,\; \forall z \leq x_1,\ldots,x_n,\; z \leq y \}$

is a filter on $L$. Furthermore, this is the same filter as $\overline{\vec{F}}$.

The intersection of any family of filters on a semilattice $L$ is a filter; that is, being a filter is a Moore closure property on subsets of $L$. The filter generated by a filter subbase $F$ (which is an arbitrary subset of $L$, remember) is the same as the Moore closure of $F$ under this property, that is the intersection of all filters on $L$ that contain $F$.

Unlike filters and filter bases, the concept of filter subbase does not seem to make sense on an arbitrary poset, but only on a semilattice.

Recall that a topology on a set $X$ is a collection $\mathcal{O}$ of subsets of $X$ such that:

- Any union of elements of $\mathcal{O}$ belongs to $\mathcal{O}$.
- $X$ itself belongs to $\mathcal{O}$.
- If $U,V \in \mathcal{O}$, then $U \cap V \in \mathcal{O}$.

More generally, any collection $\mathcal{O}$ satisfying (2,3) is a **saturated topological base**, and any collection whatsoever is a **topological subbase**.

A slightly more complicated but equivalent definition of topology is this:

- Again, any union of elements of $\mathcal{O}$ belongs to $\mathcal{O}$.
- $X$ itself is a union of elements of $\mathcal{O}$.
- If $U,V \in \mathcal{O}$, then $U \cap V$ is contained in a union of elements of $\mathcal{O}$.

Now any collection satisfying (2,3) is a **topological base** (not necessarily saturated).

Given a topological subbase $\mathcal{O}$, we generate a base $\vec{\mathcal{O}}$ by closing under (2,3) in the first list; that is, if $\mathcal{O}$ is a topological subbase on a set $X$, then

$\vec{\mathcal{O}} \coloneqq \{ \bigcap_{i=1}^n U_i \;|\; U_i \in \mathcal{O} \}$

is a topological base on $X$, which in fact is saturated. (Note that $X \in \vec{\mathcal{O}}$ follows when $n = 0$.)

Given a topological base $\mathcal{O}$, we generate a topology $\overline{\mathcal{O}}$ by closing under (1); that is, if $\mathcal{O}$ is a topological base on a set $X$, then

$\overline{\mathcal{O}} \coloneqq \{ V \;|\; \forall p \in V,\; \exists U \in \mathcal{O},\; p \in U \subseteq V \}$

is a topology on $X$.

Given a topological subbase $\mathcal{O}$, we can generate a topology by first generating a base $\vec{\mathcal{O}}$ and then generating a topology $\overline{\vec{\mathcal{O}}}$. Alternatively, we can generate the same topology by closing under (1,2,3) in the second list all at once. That is, if $\mathcal{O}$ is a topological subbase on a set $X$, then

$\overline{\mathcal{O}} \coloneqq \{ V \;|\; \forall p \in V,\; \exists U_1, \ldots, U_n \in \mathcal{O},\; p \in \bigcap_{i=1}^n U_i \subseteq V \}$

is a topology on $\mathcal{O}$. Furthermore, this is the same topology as $\overline{\vec{\mathcal{O}}}$.

As with filters, being a topology is a Moore closure property, this time on subsets of the power set $\mathcal{P}X$, and the topology generated by a topological subbase $\mathcal{O}$ is the intersection of all topologies on $X$ that contain $\mathcal{O}$.

Very analogous considerations apply to local bases for a topology and bases for pretopologies, convergence structures, gauge structures, Cauchy structures, etc.

Uniformities are a little trickier than topologies, at least in the case of subbases. For now, see the page uniform space for definitions of base and subbase for a uniformity.

Recall that a $\sigma$-algebra on a set $X$ is …

…

See basis for a Grothendieck topology.

Is there a general theory of bases? That's a good question. I don't know!

Obviously this has something to do with Moore closures (and hence monads); generating a structure from a subbase is (often) taking a Moore closure. But there's some particular property of some closure operators that makes the intermediate concept of base work out.

This paper discusses bases in the generality of algebras over a monad:

- Stefan Zetzsche?, Alexandra Silva?, Matteo Sammartino?:
*Generators and bases for algebras over a monad*(2020), (arXiv:010.10223)

Last revised on February 19, 2023 at 06:56:14. See the history of this page for a list of all contributions to it.