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?


2 Answers 2


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 $n_1<n_2<..$ 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 $i$.


Yes, the topology you defined (the co-countable topology) has the property that all convergent sequences are eventually constant:

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.


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