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The Euclidean topology on $\Bbb R$ is well-understood. It is the one generated by the open intervals (or even just the open intervals with rational end-points). To some extent, we also understand the co-countable topology, which is generated by the sets whose complement is countable.

Easily we can see that the Euclidean topology is neither a subset nor a superset of the cocountable topology:

  • $(0,1)$ is open in the Euclidean topology, but its complement is uncountable.
  • $\Bbb Q$ is countable, but it is not closed in the Euclidean topology.

So we can consider $\tau$ to be the intersection of these two topologies. Namely, $A\subseteq\Bbb R$ is a member of $\tau$ (i.e., open) if and only if it is empty or it is both co-countable and a countable union of intervals.

So, for example, neither $(0,1)$ nor the irrational numbers are open in this topology, as remarked above. On the other hand, consider $A=\left\{\frac1{2^n}\mathrel{}\middle|\mathrel{} n\in\Bbb N\right\}\cup\{0\}$, then $\Bbb R\setminus A$ is open.

Questions.

  1. Is there a nice way to describe $\tau$? Does it have a name?
  2. Are there any nice (non-trivial) properties of this topology?
Asaf Karagila
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    Perhaps a more transparent description: the nontrivial closed sets are exactly the countable sets which are closed and nowhere dense in the Euclidean topology. This suggests a generalization: given any topology $\tau$ on a set $X$ and any cardinal $\kappa$, let $\tau_\kappa^{\mathsf{NDC}}$ be the topology on $X$ generated by the complements of the $\tau$-nowhere-dense-closed sets of cardinality $<\kappa$. So this would be $\mathsf{euc_{\omega_1}^{NDC}}$. – Noah Schweber Oct 18 '21 at 22:16
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    @NoahSchweber all countable closed sets are nowhere dense – Alessandro Codenotti Oct 18 '21 at 22:21
  • @AlessandroCodenotti Yes, but that breaks down for higher cardinalities which I think is a relevant generalization to consider. So a bit of redundancy seemed appropriate. – Noah Schweber Oct 18 '21 at 22:39
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    I used this topology as an example to answer a question here (specifically, using the fact that it has the same convergent sequences as the usual topology). – Eric Wofsey Oct 18 '21 at 22:55
  • I've seen the topology generated by their union (so their join in lattice terms rather than their meet) in some papers being used as an example. AFAIK not the meet/intersection.. Maybe because I don't travel in non-Hausdorff circles... – Henno Brandsma Oct 19 '21 at 05:56

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