Let be an algebraic (infinite) Galois extension with group .
We can define the Krull topology on the group .
We wish to show this topology is totally disconnected.
Let some subset of .
Then there exists a subgroup of the form where is a finite Galois extension with (this is just a property on how this topology is defined).
Then, .
Thus, is a union of two non-empty open sets.
My question is why are those sets open?
Let be a subgroup of the Galois group. Let . I wish to prove that (the topological closure).
It is sufficient to prove that is closed and .
The closed part is not the problem. It is the inclusion which is giving me a problem.
Thus, far I know that if is of form where is finite Galois and then .
And this is somehow supposed to prove that therefore .
That is what I do not see.
Since is the intersection of all closed sets which contain it is sufficient to prove that if is a closed set which contains and intersects for all (those "special" groups) then and therefore is in the intersection. This was the theorem I assumed did the trick - and which is why I asked my question above.
I understand what you did but I was thinking.
Is the theorem "a set is open if and only if it contains all its interior points" still true in general topology?
One direction is obvious but I do not think the other direction needs to holds.
The above theorem is of course true for the Krull Topology basically by definition.
And I know it is true for metric spaces - in fact it can be regarded as a definition of 'open'.
But I cannot imagine it working in general topology.
(I never studied topology before this is why I keep on getting these basic questions).