I'm looking for topologies T on an infinite space X which divide the subsets of X into 2 non-empty collections: (1) sets which are both open and closed (clopen); (2) sets which are neither open nor closed.

The trivial topology is one example, with X and the null set being clopen and all other subsets neither open nor closed. But I haven't been able to find other examples. Nor have I been able to find an argument to show there can't be any others besides the trivial topology.

Picture me as an old guy wandering around in Counterexamples in Topology and scratching his head, with his hair starting to fall out. Thanks for any help.

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    $\begingroup$ The trivial topology identifies all points. You could have one that identifies only some points, such as the topology on $\{a,b,c\}$ generated by $\{a\}, \{b,c\}$. Every nontrivial equivalence relation induces such a topology. $\endgroup$ – hmakholm left over Monica Oct 6 '11 at 13:36
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    $\begingroup$ If I understand correctly you want a space with a topology consisting of clopen sets? This means that the topology is stable under arbitrary unions and intersections. Note that as soon as points are closed you necessarily have the discrete topology, so one (slightly) non-trivial example would be a disjoint union of set with the discrete topology and a set with the trivial topology. $\endgroup$ – t.b. Oct 6 '11 at 13:37
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    $\begingroup$ Following up on @t.b.'s argument, whenever you have such a topology, the collection of "intersections of all open sets that contain $x$" for all $x$ partition $X$ and generate the topology. So every such topology is induced by an equivalence relation in the sense of by first comment. $\endgroup$ – hmakholm left over Monica Oct 6 '11 at 13:45
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    $\begingroup$ @Henning - "Every nontrivial equivalence relation induces such a topology." That's the clue I needed. "Counterexamples in Topology" lists the "Partition Topology" as its fifth example. It says "each partition P of any set X into disjoint subsets, together with the null set, is a basis for a topology on X, known as a partition topology." It goes on to say "every open set is also closed" in a partition topology. So if we omit the discrete and the trivial topologies, all the remaining partition topologies on infinite sets satisfy my two conditions. (Is that true?) Thanks! $\endgroup$ – MikeC Oct 6 '11 at 16:42

Combining Henning's and my comments and expanding them slightly:

The spaces whose topology consists of clopen sets only are precisely the disjoint unions of spaces with the trivial topology.

Let $X$ be a space whose topology $\tau$ consists of clopen sets only. Note that $\tau$ is stable under arbitrary unions and complements, hence it is also stable under arbitrary intersections.

Define a relation $\sim$ on $X$ as follows: $x \sim y$ if and only if for all clopen sets $U$ we have either $\{x,y\} \subset U$ or $\{x,y\} \cap U = \emptyset$.

This is obviously a reflexive and symmetric relation. It is also transitive: Suppose $x \sim y$ and $y \sim z$. Consider an arbitrary clopen set $U$. By definition of $x \sim y$ there are only two possibilities:

  1. If $\{x,y\} \cap U = \emptyset$ then we must have $\{y,z\} \cap U = \emptyset$ since $y \sim z$, hence $\{x,z\} \cap U = \emptyset$.
  2. If $\{x,y\} \subset U$ then we must also have $\{y,z\} \subset U$ since $y \sim z$ and hence $\{x,z\} \subset U$.

Thus $x \sim z$ and the relation is transitive.

Write $[x]$ for the $\sim$-equivalence class of $x$. We have just seen that for each clopen $U$ we either have $[x] \subset U$ or $[x] \cap U = \emptyset$. Thus each clopen set $U$ is a union of equivalence classes. If $x \not \sim y$ then there exists a clopen set $V$ such that $[x] \subset V$ and $[y] \cap V = \emptyset$. Thus $[x]$ is the intersection of the clopen sets containing it and hence it is clopen as well. Since each clopen set either intersects $[x]$ trivially or must contain it, the relative topology on $[x]$ is the trivial topology. Since $X$ is partitioned into equivalence classes, it follows that $X$ is a disjoint union of spaces with the trivial topology.

Conversely, if $X$ is a disjoint union of spaces with the trivial topology then it has the desired property.

Finally, as Henning observed, every equivalence relation on $X$ endows $X$ with a topology consisting of clopen sets only: the topology generated by the equivalence classes. The identity relation gives the discrete topology and the trivial relation $X \times X$ gives the trivial topology. Every other equivalence relation gives a topology as you're asking for.

Added: From your comment I learned that this is called the partition topology by Steen and Seebach in Counterexamples in Topology and appears as Example 5 on page 43.

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  • $\begingroup$ I started my earlier comment before seeing your answer. Thanks for your reply. $\endgroup$ – MikeC Oct 6 '11 at 16:56

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