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Willard, General Topology theorem 5.6 says that any family of subsets of a set $X$ is a subbase of the topology it generates. In particular, this family does not need to cover.

I don't quite how this works when the family is not a cover. Recall, a base of a topology is a family of opens $B$ such that the family of all unions of arbitrary subfamilies of $B$ is the given topology. Whilst a subbase of a topology is a family of opens $B$ such that the family of all intersections of finite subfamilies of $B$ is a base of the topology. Thus if we are given an arbitrary family $B$ of subsets of a set $X$ which does not cover $X$, I don't see how any union of finite intersections taken from $B$ can be $X$. And of course any topology on $X$ has to include $X$.

It did occur to me that a finite intersection should include the empty intersection and the empty intersection is the ambient set $X$. Is this what I am missing?

Mozibur Ullah
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  • "is a subbase of the topology it generates" does not require that it generates a topology of $X$. It generates a topology on whatever it covers. If it doesn't cover $X$, the topology it generates is not a topology of $X$, but it is a topology on the subset of $X$ that it covers. – Eric Towers May 30 '25 at 23:24
  • @Eric Towers: That sort of makes sense. But the topology generated by a family of subsets is a topology on the whole of the ambient set. Plus Willard doesn't make this distinction. – Mozibur Ullah May 30 '25 at 23:31
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    You have to know that the topology on $X$ generated by $B \subset \mathscr{P}\left(X\right)$ "is just" the smallest topology on $X$ that contains every elements of $B$ (which always exists) (by smallest, I mean inclusion-wise. Recall that the intersection of topologies is a topology) Here you don't have to put any condition on $B$ – Pielcq May 30 '25 at 23:32
  • To your last comment; yes. – altwoa May 30 '25 at 23:33

2 Answers2

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Good question!

Indeed it depends on the interpretation of "finite intersection". If we understand a finite intersection as the intersection of $n \ge 1$ sets, then you are right: Willard's definition would require that a subbase $\mathscr C$ covers $X$.

Some authors explicitly require this. For example, in Munkres's "Topology" one finds

Definition. A subbasis $\mathcal S$ for a topology on $X$ is a collection of subsets of $X$ whose union equals $X$. The topology generated by the subbasis $\mathcal S$ is defined to be the collection $\mathcal T$ of all unions of finite intersections of elements of $\mathcal S$.

However, Willard claims that any collection of subsets of a set $X$ is a subbase for some topology on $X$. This can't be true with the above interpretation of finite intersection. As you say, one must include the empty intersection to make it work.

The concept of an empty intersection is not unproblematic. It only makes sense if we consider a family $\mathscr A$ of subsets of a given set $X$. In fact, we can define $\bigcap^X \mathscr A = \{ x \in X \mid x \in A \text{ for all } A \in \mathscr A \}$. For non-empty $\mathscr A$ this agrees with the usual definition $\bigcap \mathscr A = \{ x \mid x \in A \text{ for all } A \in \mathscr A \}$. However, the definition $\bigcap \emptyset = \{ x \mid x \in A \text{ for all } A \in \emptyset \}$ does not make sense; it would produce the "set" of all objects $x$, and such a thing does not exist. In contrast, $\bigcap^X \emptyset = \{ x \in X \mid x \in A \text{ for all } A \in \emptyset \} = X$.

So we should better call $\bigcap^X \emptyset$ the empty intersection relative to $X$.

Some authors use a completely different definition: They call $\mathscr C$ a subbasis for a topology $\mathscr T$ if $\mathscr T$ is the coarsest topology containing $\mathscr C$. Clearly $\mathscr C$ must contain all finite non-empty intersections of elements of $\mathscr C$ and $X$.

This avoids the above discussion.

Remark.

Munkres discusses the concepts of "empty union" and "empty intersection"on p.12 in the section "Arbitrary Unions and Intersections". For any collection of sets $\mathscr A$ one can define $$\bigcup \mathscr A = \{ x \mid \text{There exists } A \in \mathscr A \text{ such that } x \in A \} .$$

This also makes sense for $\mathscr A = \emptyset$; we get $\bigcup \emptyset = \emptyset$. Note that this is relevant in Munkres's definition of a subbasis since the empty set belongs to $\mathcal T$.

Concerning $\bigcap \emptyset$ he gives the same arguments as we did above. However, he states

Not all mathematicians follow this convention, however. To avoid difficulty, we shall nor define the intersection when $\mathscr A$ is empty.

Paul Frost
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Things become clearer if you think of "generating a topology" in a different way. https://en.wikipedia.org/wiki/Subbase has a good explanation.

Given a set $X$, any collection $\mathcal C$ of subsets of $X$ generates a topology $\tau$, in the sense that $\tau$ will be the smallest topology on $X$ that contains $\mathcal C$.

There is always such a smallest topology, because there is at least one topology containing $\mathcal C$, namely the discrete topology on $X$, and the intersection of any family of topologies on $X$ is also a topology on $X$. So $\tau$ will be the intersection of all topologies on $X$ containing $\mathcal C$. It's a description of $\tau$ "from the outside" if you wish.

That works whether $\mathcal C$ covers $X$ or not. It's independent of that.

A description "from the inside" would be in terms of unions of finite intersections of elements of $\mathcal C$, with $X$ explicitly added or not, depending on whether you use the "empty intersection convention" or not.

PatrickR
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