This is the continuation of this previous post.
Although we keep bragging about not using points, we do need to connect point-free topology and point set topology. To get rid of the points of a topological space, what we do is to forget everything except the complete lattice structure we talked so much about on the previous post.
Forgetting is easy! There is nothing to brag about… 😲
Now, suppose you want to prove a certain theorem. You used the whole point-free thingy and concluded that the “point-free version” of the topological space you are looking for does actually exist. That is, all you have is a complete lattice … and as long as you can manage to find out (construct, maybe) a topological space whose family of open sets is isomorphic to as a complete lattice, you are done!
But, how do you do that? 😅
Back to the Points 🌩️⛪️⚡️🚗
Take for example, any nonempty set with the chaotic topology . There is no hope we can recover the original set solely by looking at the lattice .
It is very obvious, for example,
that dealing with
non spaces
is kind of…
… pointless! 😁
We shall investigate what properties a topological space needs to have so we can recover a point set topology from the complete lattice of its open sets.
So, we are looking at the open sets of a topology. But we are using those blinders… 🙈 we do not see sets! And we do not see points…
Given a point , in a topological space, there is the filter of neighbourhoods of : . Let’s denote by the open neighbourhoods of . In family is not a filter in the lattice of subsets of . When we consider as a lattice, however, is a filter in . Let’s call it an open filter, to remind us that we are treating it as a filter in , and not as filter in — the family of subsets of .
A topology in a set can be defined in terms of the neighbourhoods for each . In this case, it is an axiom, that is a neighbourhood basis for . In other words, the open neighbourhoods can be used to reconstruct . And that would be easy… if we only knew what point we are talking about! 😱
The challenge 💭
Pick up a filter in the lattice . Is it possible that this filter corresponds to the family of open neighbourhoods of a point ? What criteria could we use to distinguish filters that might or not correspond to the neighbourhood basis for some ?
The catch 💣
Our canonical example will be the open filters given by two families:
Although, it is true that
since we do not “see” sets… from the point of view of the lattice ,
We cannot distinguish between and just by taking meets. 😰
The trick 😎
In one hand, in , the infimum does not coincide, in general, with the intersection. Therefore, although,
it might happen that
On the other hand, the supremum is exactly the same thing as the union!!! 🎉 🎊 🎉 🎊
And one thing we know about the union is that if belongs to the union, it has to belong to one of the sets in the union:
In terms of , this means that if we are only dealing with open sets (i.e. ),
So, if is a lattice isomorphic to , in order for a filter to correspond to the open neighbourhoods of a point, it needs to satisfy
When has the above property, we say it is a completely prime filter. If the implication is known to be valid only for finitary filters, we say that is a prime filter.
The completely prime filters are the candidates for being “neighbourhood of some point”. When a topological space is such that every completely prime filter in is of the form for some , we call it sober.
This is not the definition of sober in
📚 Jorge Picado & Aleš Pultr Frames and Locales: topology without points. Springer Basel, Series Frontiers in Mathematics, Vol. 28, xx + 400 pp., 2012 (ISBN: 978-3-0348-0153-9).
There, sober is defined some other way, and in proposition 1.3.1., the equivalence of both definitions is demonstrated. But I found this way of defining it much more natural. And later on the book, in notes 1.5 the authors state:
(1) The property from 1.3.1 is often more handy than the original definition of sobriety. We will often use it instead.
I saw no good reason to define it differently. ✅
Continues… next post.