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Let $X$ be a topological space. Is there $N\in \mathbb{N}$ under which for every $\cdots \subset B_n \subset B_{n-1}\subset \cdots \subset B_2 \subset B_1$ of closed subsets of $X$ we have $B_i=B_{i+1}$ for all $i>N$?

M.Ramana
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  • This is clearly false in $\Bbb R$, take closed balls with the same center and decreasing radius. Or a countable discrete subspace and remove one point at a time – Alessandro Codenotti Apr 25 '22 at 05:39

2 Answers2

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For a general topological space this is false as you can see for example with $\Bbb R$ and the subsets $D_n=\{x\mid \vert x \vert \leq 1/n\}$.

However, spaces which satisfy the condition every descending chain of closed subsets stabilizes are interesting. These noetherian spaces appear in algebraic geometry for example.

Jonas Linssen
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Let $(X, \tau)$ be a topological space. For $x \in X$ let $\mathfrak{U}(x)$ be the set of all open sets containing $x$.

The following are equivalent:

  1. $X$ fulfills the condition of the question.
  2. Each chain of open sets of $X$ is finite.
  3. X is Alexandroff (i.e., each $x \in X$ has a smallest neighborhood) and the set $\{$ min $\mathfrak{U}(x): x \in X\}$ is finite.
  4. $\tau$ is finite.

Proof. $1 \Rightarrow 2$ and $4 \Rightarrow 1$ are obvious.
$2 \Rightarrow 3$: By Zorn's lemma, $X$ is Alexandroff. Let $\mathfrak{B} := \{$ min $\mathfrak{U}(x): x \in X\}$. Of course, each chain in $\mathfrak{B}$ is finite. But also each antichain $\mathfrak{A} \subset \mathfrak{B}$ (i.e. elements of $\mathfrak{A}$ are pairwise incomparable by "$\subset$") is finite: Since $X$ is noetherian, $\bigcup \mathfrak{A}$ is compact. Hence there is a finite subset $\mathfrak{A}^\prime$ of $\mathfrak{A}$ such that $\bigcup \mathfrak{A}^\prime = \bigcup \mathfrak{A}$. It follows $\mathfrak{A}^\prime = \mathfrak{A}$. (Let $U=$ min $\mathfrak{U}(x) \in \mathfrak{A}$. Then there is a $V \in \mathfrak{A}^\prime$ containing $x$. Hence, by minimality of $U$, it is $U \subset V$, hence $U = V$, since $\mathfrak{A}$ is an antichain.) Hence, by a well-known theorem about partially ordered sets, $\mathfrak{B}$ is finite, see, for instance, here.
$3 \Rightarrow 4$: Since $X$ is Alexandroff, $\{$ min $\mathfrak{U}(x): x \in X\}$ is a basis for the topology. Hence $\tau$ is finite.

Ulli
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