# General Topology/Miscellaneous spaces

## Noetherian spaces

Definition (noetherian topological space):

Let ${\displaystyle X}$ be a topological space. ${\displaystyle X}$ is called noetherian if and only if for all ascending chains of open subsets

${\displaystyle U_{1}\subseteq U_{2}\subseteq \cdots \subseteq U_{n}\subseteq \cdots }$

there exists ${\displaystyle N\in \mathbb {N} }$ such that for all ${\displaystyle n\geq N}$, we have ${\displaystyle U_{n}=U_{N}}$.

In the latter case, we say that the ascending chain ${\displaystyle U_{1}\subseteq U_{2}\subseteq \cdots \subseteq U_{n}\subseteq \cdots }$ stabilizes.

Proposition (noetherian iff all open sets are compact):

Let ${\displaystyle X}$ be a topological space. ${\displaystyle X}$ 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 ${\displaystyle X}$ is noetherian, and let ${\displaystyle U\subseteq X}$ be open. Let ${\displaystyle (V_{\alpha })_{\alpha \in A}}$ be an open cover of ${\displaystyle U}$. By definition of the subspace topology, each ${\displaystyle V_{\alpha }}$ is open in ${\displaystyle X}$. An open cover of ${\displaystyle U}$ is constructed as thus: Pick ${\displaystyle \alpha _{1}}$ arbitrary. Once ${\displaystyle \alpha _{1},\ldots ,\alpha _{k}}$ are chosen, either we already have ${\displaystyle U=V_{\alpha _{1}}\cup \cdots \cup V_{\alpha _{k}}}$, or we may select ${\displaystyle \alpha _{k+1}\in A}$ such that ${\displaystyle V_{\alpha }\not \subseteq V_{\alpha _{1}}\cup \cdots \cup V_{\alpha _{k}}}$. This process must terminate, or else, upon defining

${\displaystyle W_{k}:=V_{\alpha _{1}}\cup \cdots \cup V_{\alpha _{k}}}$,

we obtain an ascending chain

${\displaystyle W_{1}\subsetneq W_{2}\subsetneq \cdots \subsetneq W_{k}\subsetneq \cdots }$

which does not stabilize. Suppose now that all open subsets of ${\displaystyle X}$ are compact. Let

${\displaystyle U_{1}\subseteq U_{2}\subseteq \cdots \subseteq U_{n}\subseteq \cdots }$

be any ascending chain of open subsets of ${\displaystyle X}$, and define

${\displaystyle U_{\infty }:=\bigcup _{n\in \mathbb {N} }U_{n}}$.

We immediately see that ${\displaystyle (U_{n})_{n\in \mathbb {N} }}$ is an open cover of ${\displaystyle U_{\infty }}$, so that by its compactness, we may extract a finite subcover ${\displaystyle U_{n_{1}},\ldots ,U_{n_{k}}}$ for certain indices ${\displaystyle n_{1},\ldots ,n_{k}\in \mathbb {N} }$. Now set ${\displaystyle N:=\max\{n_{1},\ldots ,n_{k}\}}$ so that for ${\displaystyle n\geq N}$

${\displaystyle U_{\infty }\subseteq U_{N}\subseteq U_{n}\subseteq U_{\infty }}$, ie. ${\displaystyle U_{n}=U_{\infty }=U_{n}}$,

that is, the ascending chain stabilizes. ${\displaystyle \Box }$

## Irreducible spaces

Definition (irreducible space):

Let ${\displaystyle X}$ be a topological space. ${\displaystyle X}$ is called irreducible or hyperconnected if whenever ${\displaystyle U,V\subseteq }$ are open and nonempty, then ${\displaystyle U\cap V\neq \emptyset }$.

Proposition (characterisation of irreducible spaces):

Let ${\displaystyle X}$ be a topological space. Then the following are equivalent:

1. ${\displaystyle X}$ is irreducible
2. Whenever ${\displaystyle A,B\subseteq X}$ are two closed subsets of ${\displaystyle X}$ that are both not all of ${\displaystyle X}$, then ${\displaystyle A\cup B\neq X}$
3. Whenever ${\displaystyle U\subseteq X}$ is open and nonempty, it is dense
4. Whenever ${\displaystyle A\subseteq X}$ is closed, it is nowhere dense

Proof: We prove 1. ${\displaystyle \Rightarrow }$ 2. ${\displaystyle \Rightarrow }$ 3. ${\displaystyle \Rightarrow }$ 4. ${\displaystyle \Rightarrow }$ 1. Let first ${\displaystyle X}$ be irreducible, and suppose that ${\displaystyle A,B}$ are two proper closed subsets of ${\displaystyle X}$. Suppose that ${\displaystyle A\cup B=X}$, and define ${\displaystyle U:=X\setminus A}$ and ${\displaystyle V:=X\setminus B}$. Then ${\displaystyle U\cap V=X\setminus (A\cup B)=\emptyset }$. Suppose now that ${\displaystyle U\subseteq X}$ is open and nonempty and 2. holds. If ${\displaystyle U\subseteq X}$ is open, but not dense, ${\displaystyle A:={\overline {U}}}$ is not all of ${\displaystyle X}$, and further ${\displaystyle B:=X\setminus U}$ is closed and not all of ${\displaystyle X}$ (${\displaystyle U}$ was nonempty). Therefore, ${\displaystyle X=A\cup B}$, the union of two proper closed subsets, which is impossible by 2. Suppose now 3. holds and ${\displaystyle A\subseteq X}$ is closed. Then ${\displaystyle U:=X\setminus A}$ is open and hence dense in ${\displaystyle X}$. Let ${\displaystyle V\subseteq X}$ be an arbitrary open subset, and suppose that ${\displaystyle A\cap V}$ is dense in ${\displaystyle V}$. ${\displaystyle \Box }$

Definition (generic point):

Let ${\displaystyle X}$ be a topological space. A generic point is an element ${\displaystyle x_{0}\in X}$ such that ${\displaystyle {\overline {\{x_{0}\}}}=X}$.

Proposition (a generic point is contained in every open subset):

Let ${\displaystyle X}$ be a topological space, and let ${\displaystyle x_{0}}$ be a generic point of ${\displaystyle X}$. Whenever ${\displaystyle U\subseteq X}$ is open, ${\displaystyle x_{0}\in U}$.

That is, every open set contains every generic point.

Proof: Suppose ${\displaystyle x_{0}\notin U}$. Then ${\displaystyle X\setminus U}$ is a superset of the closure of ${\displaystyle \{x_{0}\}}$, in contradiction to ${\displaystyle {\overline {\{x_{0}\}}}=X}$. ${\displaystyle \Box }$

Definition (sober):

A topological space ${\displaystyle X}$ is called sober iff every closed and irreducible subset of ${\displaystyle X}$ admits a unique generic point.

## Exercises

1. Suppose that ${\displaystyle \mathbb {N} }$ is equipped with the cofinite topology. Prove that this topological space is irreducible, but does not admit a generic point.
2. Prove that on the two-point space ${\displaystyle X=\{0,1\}}$ one may find a topology that makes ${\displaystyle X}$ into an irreducible space with two generic points. Generalize this example to any set.