I’ve been getting complaints that my lemmas have not been so lovely (or little) lately, so let’s do something a bit more down to earth. This is a story I learned from the book Counterexamples in topology by Steen and Seebach [SS].
A topological space comes equipped with various operations on its power set . For instance, there are the maps (interior), (closure), and (complement). These interact with each other in nontrivial ways; for instance .
Consider the monoid generated by the symbols (interior), (closure), and (complement), where two words in , , and are identified if they induce the same action on subsets of an arbitrary topological space .
Lemma. The monoid has 14 elements, and is the monoid given by generators and relations
Proof. The relations , , and are clear, and conjugating the first by shows that is already implied by these. Note also that and are monotone, and for all . A straightforward induction shows that if and are words in and , then . We conclude that
so (this is the well-known fact that is a regular open set). Conjugating by also gives the relation (saying that is a regular closed set).
Thus, in any reduced word in , no two consecutive letters agree because of the relations , , and . Moreover, we may assume all occurrences of are at the start of the word, using and . In particular, there is at most one in the word, and removing that if necessary gives a word containing only and . But reduced words in and have length at most 3, as the letters have to alternate and no string or can occur. We conclude that is covered by the 14 elements
To show that all 14 differ, one has to construct, for any in the list above, a set in some topological space such that . In fact we will construct a single set in some topological space where all 14 sets differ.
Call the sets for the noncomplementary sets obtained from , and their complements the complementary sets. By the arguments above, the noncomplementary sets always satisfy the following inclusions:
It suffices to show that the noncomplementary sets are pairwise distinct: this forces (otherwise ), so each noncomplementary set contains and therefore cannot agree with a complementary set.
For our counterexample, consider the 5 element poset given by
and let be the disjoint union of (the Alexandroff topology on , see the previous post) with a two-point indiscrete space . Recall that the open sets in are the upwards closed ones and the closed sets the downward closed ones. Let . Then the diagram of inclusions becomes We see that all 7 noncomplementary sets defined by are pairwise distinct.
Remark. Steen and Seebach [SS, Example 32(9)] give a different example where all 14 differ, namely a suitable subset . That is probably a more familiar type of example than the one I gave above.
On the other hand, my example is minimal: the diagram of inclusions above shows that for all inclusions to be strict, the space needs at least 6 points. We claim that 6 is not possible either:
Lemma. Let be a finite topological space, and any subset. Then and .
But in a 6-element counterexample , the diagram of inclusions shows that any point occurs as the difference for some . Since each of and is either open or closed, we see that the naive constructible topology on is discrete, so is by the remark from the previous post. So our lemma shows that cannot be a counterexample.
Proof of Lemma. We will show that ; the reverse implication was already shown, and the result for follows by replacing with its complement.
In the previous post, we saw that is the Alexandroff topology on some finite poset . If is any nonempty subset, it contains a maximal element , and since is a poset this means for all . (In a preorder, you would only get .)
The closure of a subset is the lower set , and the interior is the upper set . So we need to show that if and , then .
By definition of , we get , i.e. there exists with . Choose a maximal with this property; then we claim that is a maximal element in . Indeed, if , then so , meaning that there exists with . Then and , which by definition of means . Thus is maximal in , hence since and . From we conclude that , finishing the proof.
The lemma fails for non- spaces, as we saw in the example above. More succinctly, if is indiscrete and , then and . The problem is that is maximal, but and .
[SS] L.A. Steen, J.A. Seebach, Counterexamples in Topology. Reprint of the second (1978) edition. Dover Publications, Inc., Mineola, NY, 1995.