Zulip Chat Archive
Stream: Is there code for X?
Topic: totally disconnected of discrete
Adam Topaz (Nov 08 2020 at 19:09):
apply_instance
fails for the following. Is this instance missing?
import topology.subset_properties
variables (α : Type*) [topological_space α] [discrete_topology α]
example : totally_disconnected_space α := sorry
Mario Carneiro (Nov 08 2020 at 19:10):
probably
Adam Topaz (Nov 08 2020 at 19:24):
Okay, I have an ugly proof. I'm sure someone who is more familiar with the discrete_topology and/or totally_disconnected_space api can come up with a 1-liner:
import topology.subset_properties
variables (α : Type*) [topological_space α] [discrete_topology α]
open_locale classical
example : totally_disconnected_space α :=
begin
constructor,
intros T hT h1,
constructor,
intros x y,
by_contradiction contra,
specialize h1 {x} (set.univ \ {x}) (is_open_discrete _) (is_open_discrete _) _ _ _,
rcases h1 with ⟨t,ht⟩,
finish,
finish,
use x,
finish,
use y,
simp,
intro c,
apply contra,
ext,
symmetry,
assumption,
end
Mario Carneiro (Nov 08 2020 at 19:27):
I see that in the same file there is also totally_separated
, which also seems like it would be true for a discrete space and implies totally_disconnected
Adam Topaz (Nov 08 2020 at 19:28):
I have no idea what totally_separated
means
Mario Carneiro (Nov 08 2020 at 19:28):
the FOL description is there in the definition
Mario Carneiro (Nov 08 2020 at 19:29):
I don't have any deep understanding of the term but those symbols look true
Mario Carneiro (Nov 08 2020 at 19:30):
and more to the point it looks a bit easier to prove those symbols than the ones in totally_disconnected
Mario Carneiro (Nov 08 2020 at 19:33):
example : totally_separated_space α :=
⟨λ a _ b _ h, ⟨{b}ᶜ, {b}, is_open_discrete _, is_open_discrete _, by simpa⟩⟩
Adam Topaz (Nov 08 2020 at 19:36):
Funny, I used {a} and {a}\^c, instead of {b}^\c and {b}, and ended up with a messy proof again!
Mario Carneiro (Nov 08 2020 at 19:37):
yep I started with that but then you have to throw in commutativity because a \in {b}^c
unfolds to b != a
Adam Topaz (Nov 08 2020 at 19:37):
Anyway, can you PR this?
Adam Topaz (Nov 08 2020 at 19:41):
Oooh, that doesn't even require classical!
Mario Carneiro (Nov 08 2020 at 19:42):
it does, you have to prove {b}^c \cup {b} = univ
Adam Topaz (Nov 08 2020 at 19:43):
Oh right
Last updated: Dec 20 2023 at 11:08 UTC