# General Topology/Miscellaneous spaces

## Noetherian spaces

Definition (noetherian topological space):

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

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

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

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

Proposition (noetherian iff all open sets are compact):

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

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

we obtain an ascending chain

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

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

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

be any ascending chain of open subsets of $X$ , and define

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

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

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

that is, the ascending chain stabilizes. $\Box$

## Irreducible spaces

Definition (irreducible space):

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

Proposition (characterisation of irreducible spaces):

Let $X$  be a topological space. Then the following are equivalent:

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

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

Definition (generic point):

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

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

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

That is, every open set contains every generic point.

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

Definition (sober):

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

## Exercises

1. Suppose that $\mathbb {N}$  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 $X=\{0,1\}$  one may find a topology that makes $X$  into an irreducible space with two generic points. Generalize this example to any set.