X is an infinite set. X connected means that X is not the union of two disjoint closed subsets.
François Couchot Dear Francois, I don't understand why you define X connected iff X is not the union of two disjoint closed subsets, because the well known traditional definition of connectedness is: X connected iff X is not the union of two disjoint open subsets!
Your approach seems to be somehow weird, probably you need such approach and you force the things( using closed subsets) to obtain what you wish!
If we consider your approach, using closed subsets, then in Hausdorff spaces every finite set is not connected, because every finite set is the union of two disjoint finite subsets, and in Hausdorff spaces every finite subset is closed.
Peter Breuer Yes, you are right, but however, if A,B are disjoint open, then cl(A), cl(B) are closed, but not necessarly disjoint, the border of A can contain points from the border of B!
Dinu Teodorescu
In this case cl(A) = A and cl(B) = B because X = A U B and they are both open and disjoint. SO they are also closed. So it is equivalent.
François Couchot
Do you know an example of a space in which only the singletons and the space are connected? In the right hand or left hand topology only the singletons are connected sets but not the space itself.
Gabriel T. Prajitura I think such space, if it exists, can not be Hausdorff.
Peter Breuer Yes, the cofinite topology on an infinite set X seems to satisfy a part of requirements. No compact, no Hausdorff
I propose another example satisfying a part of requirements, more restrictive as the case of cofinite topology. Let N={1,2,3,....} be the space, N, phi and Ek={k+1,k+2,...} ( k>=1) be the open sets, N\Ek={1,2,...,k} are the closed sets.
Here we have compacity of N, every subset seems to be connected, but not Hausdorff and the singletons are not closed! Am I wrong?
Sorry.....stupid my example, taking into account requirements of FC!
Finally I found an answer to this question today. See the attached file to read this answer. Best regards.
@Peter Breuer: The proof assumes that the space is connected. Yours is not.
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