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.