Is there any characterization for a topological space under which every convergent sequence is stationary? (e.g. discrete topology)
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$\begingroup$ By "stationary" do you simply mean "eventually constant"? $\endgroup$– Joseph Van NameCommented Sep 8, 2015 at 1:15
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$\begingroup$ For a Stone space it is equivalent to the condition that every closed infinite subset is non-metrizable. So at the Boolean algebra level it means that the Boolean algebra has no infinite countable quotient. There are many such examples, notably all Boolean algebras with uncountable cofinality (as studied notably by Koppelberg). $\endgroup$– YCorCommented Apr 23, 2019 at 11:21
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Let me answer this question by assuming that by "stationary" you mean "eventually constant."
One may characterize the topological spaces where every convergent sequence is eventually constant in terms of sequentially closed spaces. Suppose that $(X,\mathcal{T})$ is a topolofical space. Then we say that a set $A\subseteq X$ is sequentially closed in $\mathcal{T}$ if whenever $x_{n}\in A$ for all $n$ and $x_{n}\rightarrow x$, then $x\in A$ as well. The complement of a sequentially closed set shall be called a sequentially open set. Let $\mathcal{T}_{s}$ be the collection of all sequentially open sets in $X$. Then $(X,\mathcal{T}_{s})$ is a topological space. It turns out that $(X,\mathcal{T}_{s})$ is discrete if and only if the space $(X,\mathcal{T})$ has no non-trivial convergent sequences (see these notes for more details).
Several well-known classes of spaces consist solely of spaces with no non-trivial convergent sequences. A completely regular space $X$ is said to be an $F$-space if whenever $U,V$ are disjoint cozero sets, then there is a continuous function $f:X\rightarrow[0,1]$ where $U\subseteq f^{-1}[\{0\}],V\subseteq f^{-1}[\{1\}]$. No $F$-space has a convergent sequence consisting of distinct points. For example, every $P$-space is an $F$-space and $\beta\mathbb{R}\setminus\mathbb{R}$ is an $F$-space. Furthermore, the notion of an $F$-space can be weakened in several ways so that there is no convergent sequence consisting of non-trivial points (for example, consider the spaces $X$ so that whenever $A,B$ are disjoint countable subsets of $X$, then there are disjoint open sets $U,V$ with $A\subseteq U,B\subseteq V$).
