# General Topology/Miscellaneous spaces

## Noetherian spaces

[edit | edit source]**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.

**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.

## Irreducible spaces

[edit | edit source]**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 .

**Definition (sober)**:

A topological space is called **sober** iff every closed and irreducible subset of admits a unique generic point.

## Local spaces

[edit | edit source]## Exercises

[edit | edit source]- 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.