On Tue, May 27, 2025 at 07:01:25AM +0100, Martin Baker wrote:
> On 26/05/2025 21:31, Waldek Hebisch wrote:
> > > * isomorphism classes are sets of _unlabeled_ sets
> > > (both with the usual meet,join top and bottom axioms)
> >
> > I do not think this will work well (at least not in a simple
> > way). In isomorphism class you allow arbitrary relabeling, so
> > at "absolute" level labels do not matter. But they
> > matter in relative sense. You need to know when two elements
> > are in fact one element (that is they are equal), you need
> > to know if for given p and q there is an open set containing
> > q but such that p is not in this set. Easiest way to
> > keep this information is to use labels. You could try to
> > encode elements of the set by their properties, but to
> > do this you need to know which combination of properties
> > are possible.
> >
> > What you wrote above looks more like you would like to represent
> > lattice of open sets as an abstract lattice. AFAICS you still
> > would have to decide when two abstrace lattices are isomorphic
> > and it is not clear if isomorphism class of lattices gives you
> > isomorphism class of topologial spaces. Mathematical version of
> > Murphy principle says that if there is no reason for some property
> > to be true, then it is false. And ATM I see no reason to
> > have nice correspondence there (of course I may be missing
> > something).
>
> Perhaps if I use an example:
> A topology might be represented like this:
> [{}, {3}, {1, 2}, {1, 2, 3}]
>
> But for a isomorphism class we want a representation that includes multiple
> topologies such as:
> [{}, {3}, {1, 2}, {1, 2, 3}]
> [{}, {1}, {2, 3}, {1, 2, 3}]
> and so on
>
> We could choose one to represent the whole class:
> [{}, {3}, {1, 2}, {1, 2, 3}]
> Alternatively we could represent it as a partial order:
> [ _|_<a , _|_ < b , a<T, b<T]
> Or we could represent it as a lattice:
> [ a/\b=_|_ , a\/b=T]
>
> It seems to me that all 3 of these represent the same structure (upto
> isomorphism).
> I have used labels for the partial order and lattice cases but these labels
> represent elements of the topology not elements of the open sets.
> (So, in this case _|_={} a={3} b={1, 2} T={1, 2, 3})
> So I think the partial order seems more compact and involves less arbitrary
> choices.
So, it seems that you want to represent topology via its lattice
of open sets. There is a trivial difficulty here: two point
space with discrete topology: [{}, {1}, {2}, {1, 2}] has
isomorphic lattice of open sets, so there are multiple
non-isomorphic topologies with "the same" lattice of open
sets. This is trivial difficulty and vanishes if you require
your space to be T_0. However, IMO your example is misleading
(like computing 2^0 and 2^1 and concuding that exponential is
quite small). Consider sligtly bigger topology with basis
[{1}, {1, 2}, {1, 3}, {1, 4}, {1, 5}, {1, 2, 3, 4, 5, 6}]
The topology has 19 open sets, so lattice of open sets is
significantly bigger than the basis above. Consider labeling
of part of the lattice of open sets corresponding to the
basis above:
a = {1}, b = {1, 2}, c = {1, 3}, d = {1, 4}, e = {1, 5},
f = {1, 2, 3, 4, 5, 6}
If you look at this you will notice that each set above
introduces exactly one new element (new in the sense that
it did not appear in earlier sets). So, in effect labeling
of open sets gives you labeling of elements of the space.
But since lattice of open sets is much bigger than the
basis you have a lot of irrelevant labels.
--
Waldek Hebisch
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