(1) Is there a name for such a topology in general topology literature? (2) Is there any research paper studying such type of compact, non-Hausdorff and connected chain topologies?

ra2lokBQ

ra2lokBQ

Answered question

2022-11-22

On the generalized Sierpinski space
In Sierpiński topology the open sets are linearly ordered by set inclusion, i.e. If S = { 0 , 1 }, then the Sierpiński topology on S is the collection { ϕ , { 1 } , { 0 , 1 } } such that ϕ { 0 } { 0 , 1 } we can generalize it by defining a topology analogous to Sierpiński topology with nested open sets on any arbitrary non-empty set as follows: Let X be a non-empty set and I a collection of some nested subsets of X indexed by a linearly ordered set ( Λ , ) such that I always contains the void set ϕ and the whole set X, i.e.
I = { , A λ , X : A λ X , λ Λ }
such that A μ A ν whenever μ ν.
Then it is easy to show that I qualifies as a topology on X.
My questions are:
(1) Is there a name for such a topology in general topology literature?
(2) Is there any research paper studying such type of compact, non-Hausdorff and connected chain topologies?

Answer & Explanation

Lokasckf

Lokasckf

Beginner2022-11-23Added 7 answers

Step 1
This was already pointed out in comments, but I'm collecting this into an answer. (In fact, this is closer to a longer comment than to an answer, but the details can be elaborated better here than in a short comment.)
Both examples here deal with your construction in the special case that X = Λ. And in both cases we work with the sets of the form ( , a ) = { x Λ ; x < a }, which is similar to lower topology.
Step 2
Suppose that we have X = Λ and A λ = ( , λ ) = { x Λ ; x < λ }.
This means that for each bounded subset D Λ there exists a λ such that μ D A μ = A λ . It is relatively easy to show that the last condition is equivalent to λ = sup μ D μ. I.e. the linear order would have to be complete.
A counterexample suggested by Niels is Λ = X = Q . For example if ( q n ) is an increasing sequence of rational numbers such that q n 2 , then ( , q n ) = ( , 2 ), which is not of the form A λ .
There are several possibilities how to circumvent this. E.g. you could take all downward closed sets as open. Or you could take { , X , A λ ; λ Λ } as a base for the topology you want to generate.
Step 3
Even if this is a topology, it need not necessarily be compact.
Let us take X = Λ = Z .
Again A λ = { x Z ; x < λ } = ( , λ ) Z I.e. A λ 's are down-sets of the linearly ordered set Z .
Now if C is open cover of X then, for each n Z , the cover C must contain some A λ with λ > n. his shows that every open cover is infinite.
And if we take C = { A n ; n Z } = { ( , n ) Z ; n Z }, we get an open cover which does not have a finite subcover.

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