On 27/05/2025 16:31, Waldek Hebisch wrote:
So, it seems that you want to represent topology via its lattice
of open sets.
Well, I'm keeping an open mind about open sets but my initial instinct
was to use:
* labeled open sets for topologies.
* set size + lattice for iso classes.
The reason for the second is that, to use labeled open sets for iso
classes, the constructor algorithm would have to:
* choose a representative for each iso class (this seems very arbitrary
and I don't know how to do it)
* choose a basis.
* make sure these choices are closed under union and intersection.
(I think I should re-read your earlier messages in case you have
explained these things already)
By the way, in you earlier email you mentioned using pre-order to code
this, this prompted me to read this page:
https://en.wikipedia.org/wiki/Specialization_(pre)order
So I am starting to understand what you meant. However it suggests that
Specialization pre order only works for T0 separation. At first I
thought this would not matter (why bother about points that can't be
distinguished by the lattice). Then I had the thought these points must
be the degenerate points. Things I've read suggest that degenerate
points are very useful (for example in calculating products of
topologies and other nice properties) so perhaps I do need T0. I am
thinking that I would need to base it on a weak pre-order? Do you know
if this speculation, on my part, is valid?
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.
OK, I will have to work through this. It just seems counter-intuitive
because there are a lot less iso classes than topologies so we should be
able to code them with less data?
Thanks,
Martin
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