# What exactly is the topology generated by the union of a family of topologies?

I recall reading somewhere that the supremum of a family of topologies on a set is simply the topology generated by the union of all topologies in the family. My question is, what does "generated" mean in this case? For instance, I know when dealing with filter bases, the generated filter is the set of all subsets containing some set in the filter base. But for topologies, how do we know what subsets to include in the generated topology? I would like to be able to visualize what kind of subsets are contained in this supremum. Thanks for any insight.

-
Does this not just mean "all the possible open subsets you can get from applying the usual operations to the union of these topologies"? I suppose "supremum" is referring to the poset of topologies on the set. – Aaron Mazel-Gee Nov 23 '10 at 8:01
So when you take the union of a family of topologies $\mathcal{T}$, it's not necessarily true that it will be closed under finite intersections or unions of any subfamily, correct? Are you then saying that the generated topology will be the topology not only containing the union $\cup\mathcal{T}$ of some family of topologies, but also all possible subsets that we could obtain by taking unions or finite intersections of any subfamily of $\cup\mathcal{T}$? – yunone Nov 23 '10 at 8:43
Yes, that's what it sounds like to me. – Aaron Mazel-Gee Nov 23 '10 at 10:30
See also mathoverflow.net/questions/15841/… for further discussion of one broad natural context here, the lattice of topologies on a set. – JDH Jun 23 '11 at 21:36

The following short piece of lecture notes is almost entirely devoted to an answer of your question:

http://math.uga.edu/~pete/TopSection4.pdf

Unfortunately, for reasons that now escape me, I decided to break the file here, so the important remark that the statements "$\tau$ is the topology generated by the family of sets $\mathcal{F}$" and "The family of sets $\mathcal{F}$ is a subbase for the topology $\tau$" are equivalent does not appear until the beginning of the next little piece of notes:

http://math.uga.edu/~pete/TopSection5.pdf

-
Thanks for the links! – yunone Nov 30 '10 at 2:42

The topology generated by any family of subsets $F$ of a set $X$ is the intersection of all topologies on $X$ containing $F$; $F$ then forms a subbase for the resulting topology. In practice this means that you take all finite intersections of the elements of $F$ to get a base $F'$, then take arbitrary unions of the elements of $F'$.

-
I understand that this works, but why exactly does it work? – Anthony Peter Feb 18 '14 at 2:24

It might also be useful to view your object as the ("projective") limit of the set topologized variously. The indexing is by the topologies, with the finer ones mapping to the coarser ones. A minor virtue of this is that it characterizes the result without necessarily giving a construction or description of the opens.

-

You can also do this. The intersection of any family of topologies on a set is itself a topology. The discrete topology (all subsets open) is a topology for any set. So, take the set of all topologies containing the family of topologies (remember: the discrete one is here) and you will get a topology. It will be the smallest topology containing all topologies in the family.

This is integral to the notion of weak topology. The first answer has a second correct characterization of this topology. See the book {\it Topology} by Wilansky.

-
This is a fine characterization, but it doesn't give an explicit description of the actual elements of the corresponding topology. (For example, the corresponding construction for sigma-algebras is totally nonexplicit: to get a really explicit description of the actual elements of a generated sigma-algebra one has to induct up to some uncountable ordinal or something. It is a nontrivial fact that one doesn't have to take all finite intersections, then all arbitrary unions of those, then all finite intersections of those... in the case of topologies.) – Qiaochu Yuan Nov 28 '10 at 3:52
This is because the arbitrary unions of elements in the set form a sub-basis for the topology. – ncmathsadist Apr 30 '11 at 1:05