Definition (noetherian topological space):
Let be a topological space. is called noetherian if and only if for all ascending chains of open subsets
there exists such that for all , we have .
In the latter case, we say that the ascending chain stabilizes.
Proposition (noetherian iff all open sets are compact):
Let be a topological space. is noetherian if and only if all of its open subsets are compact.
(On the condition of the axiom of dependent choice.)
Proof: Suppose first that is noetherian, and let be open. Let be an open cover of . By definition of the subspace topology, each is open in . An open cover of is constructed as thus: Pick arbitrary. Once are chosen, either we already have , or we may select such that . This process must terminate, or else, upon defining
we obtain an ascending chain
which does not stabilize. Suppose now that all open subsets of are compact. Let
be any ascending chain of open subsets of , and define
We immediately see that is an open cover of , so that by its compactness, we may extract a finite subcover for certain indices . Now set so that for
- , ie. ,
that is, the ascending chain stabilizes.
Definition (irreducible space):
Let be a topological space. is called irreducible or hyperconnected if whenever are open and nonempty, then .
Proposition (characterisation of irreducible spaces):
Let be a topological space. Then the following are equivalent:
- is irreducible
- Whenever are two closed subsets of that are both not all of , then
- Whenever is open and nonempty, it is dense
- Whenever is closed, it is nowhere dense
Proof: We prove 1. 2. 3. 4. 1. Let first be irreducible, and suppose that are two proper closed subsets of . Suppose that , and define and . Then . Suppose now that is open and nonempty and 2. holds. If is open, but not dense, is not all of , and further is closed and not all of ( was nonempty). Therefore, , the union of two proper closed subsets, which is impossible by 2. Suppose now 3. holds and is closed. Then is open and hence dense in . Let be an arbitrary open subset, and suppose that is dense in .
Definition (generic point):
Let be a topological space. A generic point is an element such that .
Proposition (a generic point is contained in every open subset):
Let be a topological space, and let be a generic point of . Whenever is open, .
That is, every open set contains every generic point.
Proof: Suppose . Then is a superset of the closure of , in contradiction to .
A topological space is called sober iff every closed and irreducible subset of admits a unique generic point.
- Suppose that is equipped with the cofinite topology. Prove that this topological space is irreducible, but does not admit a generic point.
- Prove that on the two-point space one may find a topology that makes into an irreducible space with two generic points. Generalize this example to any set.