# A sequence $(x_{n})$ is convergent in $X$ $\iff$ $(x_{n})$ is a stationary sequence. Does $X$ have to be discrete?

Let $$X$$ be a topological space so that a sequence $$(x_{n})$$ is convergent in $$X$$ $$\iff$$ $$(x_{n})$$ is a stationary sequence. Does $$X$$ have to be discrete?

Does topology $$T = \left\{ X \subset \mathbb{R} | card (X^{c}) \leq \aleph_{0} \right\}$$ work?

• Possible duplicate of Convergent sequence in co-countable topology iff sequence is eventually constant(Your proposed topology is called the cocountable topology, or the countable complement topology; Also, a more standard term for stationary sequence is eventually constant sequence) Apr 23, 2019 at 2:10
• Nitpick: You should replace $T$ with $T\cup \{\Bbb R\}.$.... BTW there is a non-empty compact Hausdorff space with no isolated points (e.g. $\beta \Bbb N \setminus \Bbb N$) in which the only convergent sequences are "eventually constant". Apr 23, 2019 at 2:57
• Apr 23, 2019 at 3:10
• The result mentioned by Daniel Wainfleet is proved in Stone-Čech compactifications and limits of sequences Apr 23, 2019 at 3:18

Yes, this topology works. Let $$x_n \to x$$. If it is not true that $$x_n=x$$ for all $$n$$ sufficiently large then there exist integers $$n_1 such that $$x_{n_i} \neq x$$ for all $$i$$. Consider the set $$U=\mathbb R \setminus \{x_{n_i}: i\geq 1\}$$. This is an open set containing $$x$$. Since$$x_n \to x$$ it must be true that $$x_n \in U$$ for all for all $$n$$ sufficiently large. This contradicts the fact that $$x_{n_i} \notin U$$ for any $$i$$.
Suppose $$(x_n)_n$$ converges to $$p \in X$$ in the co-countable topology. Define $$C=\{x_n: x_n \neq p\}$$ which is an at most countable subset of $$X$$, so $$O:=X\setminus C$$ is open in the co-countable topology, and as $$p \notin C$$ by definition, $$p \in O$$.
By the definition of convergence, there must be some index $$N \in \mathbb{N}$$ such that $$\forall n \ge N: x_n \in O$$
But it’s clear that $$x_n \in O$$ iff $$x_n \notin C$$ iff $$x_n = p$$, so that $$\forall n \ge N: x_n =p$$ which says that $$(x_n)_n$$ is eventually constant with value $$p$$.
And if $$X$$ is an uncountable set then the co-countable topology is a non-discrete topology on $$X$$ that nevertheless obeys the property that all convergent sequences are eventually constantly their limit.