The cocountable topology or countable complement topology on any set X consists of the empty set and all cocountable subsets of X, that is all sets whose complement in X is countable. It follows that the only closed subsets are X and the countable subsets of X.
Every set X with the cocountable topology is Lindelöf, since every nonempty open set omits only countably many points of X. It is also T1, as all singletons are closed. The only compact subsets of X are the finite subsets, so X has the property that all compact subsets are closed, even though it is not Hausdorff if uncountable.
The cocountable topology on a countable set is the discrete topology. The cocountable topology on an uncountable set is hyperconnected, thus connected, locally connected and pseudocompact, but neither weakly countably compact nor countably metacompact.
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References
The cocountable topology or countable complement topology on any set X consists of the empty set and all cocountable subsets of X, that is all sets whose complement in X is countable. It follows that the only closed subsets are X and the countable subsets of X.
Every set X with the cocountable topology is Lindelöf, since every nonempty open set omits only countably many points of X. It is also T1, as all singletons are closed.
If X is an uncountable set then any two nonempty open sets intersect, hence the space is not Hausdorff. However, in the cocountable topology all convergent sequences are eventually constant, so limits are unique. Since compact sets in X are finite subsets, all compact subsets are closed, another condition usually related to Hausdorff separation axiom.
The cocountable topology on a countable set is the discrete topology. The cocountable topology on an uncountable set is hyperconnected, thus connected, locally connected and pseudocompact, but neither weakly countably compact nor countably metacompact, hence not compact.
- Cofinite topology
- List of topologies
- Steen, Lynn Arthur; Seebach, J. Arthur Jr. [1995] [1978], Counterexamples in Topology [Dover reprint of 1978 ed.], Berlin, New York: Springer-Verlag, ISBN 978-0-486-68735-3, MR 0507446 [See example 20].
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Because $\Bbb Q$ is countable, the set $U = \Bbb R \setminus \Bbb Q$ is an open set in co-countable topology [its complement is $\Bbb Q$] and $\sqrt{3} \in U$ because it's irrational.
So $\sqrt{3}$ has an open neighbourhood that contains no points from the rationals so no sequence from the rationals can converge to $\sqrt{3}$.
In fact, something much stronger is true: suppose $[x_n]$ is any sequence in $\Bbb R$ in the co-countable topology and suppose the sequence converges to $x$.
Define $C=\{x_n: x_n \neq x\}$, which is a countable subset of $\Bbb R$ so $O:=\Bbb R \setminus C$ is open by definition.
It's clear that $x \notin C$ so $x \in O$ and so by convergence:
$$\exists N \in \Bbb N: \forall n \ge N: x_n \in O$$
But $x_n \in O$ is only possible when $x_n=x$ [otherwise $x_n \neq x$ and $x_n \in C$], so
$$\exists N \in \Bbb N: \forall n \ge N: x_n = x$$
so $x_n$ is an eventually constantly-equal-to-$x$ sequence. This proves that if $x_n \in A$ for all $n$ and $x_n \to x$ then $x \in A$, for any $A \subseteq \Bbb R$, so all subsets of $\Bbb R$ in the co-countable topology are sequentially closed. In particular no irrational sequence can converge to a rational, and vice versa.