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The question as written in the title is too vague to be interpreted meaningfully, so allow me to explain more precisely what I mean.

There are at least two equivalent ways of defining a topology: the standard one involves specifying a family of open sets (which satisfy certain axioms), and a dual way involves specifying a family of closed sets (which satisfy certain axioms). A family of open sets (1) is closed under arbitrary union (2) closed under finite intersections (3) the space itself and the empty set are open, while a family of closed sets is (1) closed under arbitrary intersections (2) closed under finite unions (3) the space itself and the empty set are closed.

Any family of open sets determines a family of closed sets and vice-versa via set complementation (because of De Morgan's laws).

Question: Corresponding to any topology is a family of open sets and a family of closed sets -- I say that they are "indistinguishable" if the family of open sets also satisfies the axioms for a family of closed sets (possibly for a different topology), and if the family of closed sets also satisfies the axioms for a family of open sets (possibly for a different topology).

Then for which topologies are the concepts of open sets and closed sets "indistinguishable" in the sense mentioned above?

Attempt: For any such topological space, both the family of open sets and the family of closed sets are closed under arbitrary unions and arbitrary intersections. If we can get either the family of open sets or the family of closed sets to contain all points, then we automatically have the discrete topology on the space.

A topological space is T1 if and only if all points in the space are closed. Thus, if the family of closed sets of such a space also satisfies the open set axioms, then it is actually the entire power set of the space, and by DeMorgan's Laws so is the family of open sets of such a space. Thus the topology is the discrete topology.

Thus the only T1 topologies for which the concepts of open and closed sets are indistinguishable in the sense given above are the discrete topologies.

I claim that it is also possible to have non-T1 topologies for which the concepts of open and closed sets are indistinguishable as mentioned above and yet the topology is not the discrete topology. Moreover, in some cases of these non-T1 topologies the family of open sets and closed sets are strictly different, even though both families satisfy both the open and closed set axioms.

The example in mind I have is extremely simple: consider a space $X$, and a non-empty proper subset $Y$. Then $\{ \emptyset, Y, X \}$ satisfies both the open and closed set axioms, as does $\{ \emptyset, Y^c, X \}$. Obviously these topologies are not T1, even in the case that $X$ consists of only two points.

Thus the concept of a topology for which the open and closed sets satisfy both the open and closed set axioms is only possibly interesting for non-T1 spaces, for which one is not automatically guaranteed that such topologies are the discrete topology.

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    $\begingroup$ So in short, you're looking for a topology in which all open sets are closed? $\endgroup$ – Kenny Lau Apr 27 '17 at 8:54
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    $\begingroup$ An example that slightly generalizes yours : for any subset $A$ of $X$, the collection of all sets contained in $A$ and $X$ is stable under arbitrary unions and intersections and contains the empty set and $X$, and is thus a topology with your property. These topologies have the property that $A$ is a discrete subspace. $\endgroup$ – Arnaud D. Apr 27 '17 at 9:19
  • $\begingroup$ @KennyLau Not exactly but almost, just closed in some topology, not necessarily the same topology. $\endgroup$ – Chill2Macht Apr 27 '17 at 10:18
  • $\begingroup$ All sets are closed in the discrete topology. $\endgroup$ – Asaf Karagila Apr 27 '17 at 12:15
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An Aleksandrov space is a space for which any intersection of open sets is open.

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  • $\begingroup$ Yes this is exactly the concept, thank you. Here topospaces.subwiki.org/wiki/Alexandrov_space it also mentions that T1 + Alexandrov implies discrete. And the Wikipedia article mentions a duality between partial orders and Alexandrov topologies which is much more general than the two examples I mention in my question (poset and linear partial orders with three elements). It is along the lines of the generalization given by @Arnaud D. above $\endgroup$ – Chill2Macht Apr 27 '17 at 10:23

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