# A characterization of FAC spaces

In the open problem section, I defined a FAC space as a topological space in which every closed subspace is a finite union of irreducible closed subspaces.  FAC is for “finite antichain property”, since it generalizes the following theorem: a poset has the finite antichain property (namely, all its antichains are finite) if and only if its downwards-closed subsets are finite unions of ideals.

## A bit of history.

That theorem (about posets) has a curious history.  It has been discovered over and over.  I have learned about it by reading M. Kabil and Maurice Pouzet , but it had been proved earlier by Jimmie Lawson, Mike Mislove and Hilary Priestley , by R. Bonnet , and by others.  In some of those papers, you will see that the result is credited to Erdős and Tarski in 1943 .  The first time I heard about that, I looked at the Erdős-Tarski paper, and I could not find the result or anything vaguely looking like it.  My colleague Alain Finkel once asked Maurice Pouzet about that state of affairs, and apparently the result is implicit in the Erdős-Tarski paper.

## Characterizing FAC spaces.

Anyway, I think I have finally nailed the correct topological generalization of that theorem by Erdős and Tarski.  Here it is.  I do not have much merit in proving the characterization theorem below.  This is essentially the Lawson-Mislove-Priestley proof , suitably modified.

Definition. A subset A of a topological space X is a topological antichain if and only if the subspace topology on A is discrete (all its subsets are open).

Theorem. Given a topological space X, the following are equivalent.

1. Every closed subspace of X is a finite union of irreducible closed subsets;
2. Every topological antichain of X is finite.

It therefore really makes sense to call those spaces FAC.  The purpose of this post is to give a proof of that theorem.

## An alternative view on topological antichains.

First, we give an alternative characterization of topological antichains.

Lemma 1. Let A be a subset of a topological space XA is a topological antichain if and only if, for every xA, x is not in the closure cl (A — {x}).  (We take closures in X, here.)

Proof.  If A is a topological antichain, then for every xA, {x} is open in A, so A — {x} is closed in A.  That means that A — {x} occurs as the intersection of some closed subset of X with A, and the smallest one is cl (A — {x}).  That cannot contain x since its intersection with A is A — {x}.

Conversely, for every xA, if x is not in the closure cl (A — {x}), then it lies in some open subset U of X that does not intersect A — {x}.  Then UA is open in A, but only contains the point x from A.  It follows that every one-element subset of A is open in A, whence A is discrete.  ☐

## The key lemma.

Second, we prove a useful lemma, generalizing Lemma 3 of  to the topological case.

Lemma 2. Let X be a topological space, and M be an infinite family of irreducible closed subsets of X, with the property that for every infinite subfamily N of M, cl (∪N)=cl (∪M).  (∪M denotes the union of all the elements of M.)  Then cl (∪M) is irreducible closed.

Proof.  We shall repeatedly use the fact that if an open set intersects the closure of a set E, then it intersects E.  Let U and V be two open subsets of X that intersect cl (∪M).  It suffices to show that cl (∪M) intersects UV.  (Trick 8.2.3 in the book.)  Let M’ be the subfamily of those I M such that U intersects IM’ is non-empty: since U intersects cl (∪M), it intersects ∪M, hence some I M.

If M’ were finite, then MM’ would be infinite, so by assumption cl (∪(MM’)) would be equal to cl (∪M).  However, U intersects the latter, so it would intersect the former, hence also ∪(MM’), hence also some IMM’.  That would contradict the definition of M’.  Therefore M’ is infinite.

Since M’ is infinite, we use the assumption again and we obtain cl (∪M’)=cl (∪M).  We now use the fact that V, not just U, intersects cl (∪M).  Then V must also intersect cl (∪M’), hence ∪M’, and therefore it must intersect some IM’.  By definition of M’, U also intersects I.  Since I is irreducible closed, it must therefore intersect UV.  Since I is included in cl (∪M), the latter must also intersect UV.  ☐

## Proving the hard implication 2 ⇒ 1.

We now prove the difficult implication 2 ⇒ 1 of the theorem.  More precisely, we prove its contrapositive, namely: we assume that there is a closed subset C of X that cannot be written as a finite union of irreducible closed subsets, and we shall build an infinite topological antichain in X.

Recall that the sobrification S(X) of X is sober (Corollary 8.2.23 in the book), and that every sober space is a dcpo in its specialization ordering (Proposition 8.2.34).  The supremum of a directed family of irreducible closed subsets is the closure of their union (it suffices to show that the latter is irreducible: show that if it intersects two opens U and V, then it intersects UV!).  In particular, the supremum of a directed family of irreducible closed subsets included in C is again included in C.  It follows that the poset of all irreducible closed subsets of X included in C is inductive.  Hence we can apply Zorn’s Lemma (Theorem 2.4.2): every irreducible closed subset of X included in C is contained in some maximal irreducible closed subset of X included in C.  (Exercise: an irreducible closed subset of X included in C is the same thing as an irreducible closed subset of the subspace C.  That simplifies the latter statement, and also allows for a different proof of the latter fact, working in S(C) instead of S(X).)

In particular, for every x C, ↓x is such an irreducible closed subset of X included in C.  Let M0 be the collection of all maximal irreducible closed subsets of X included in C.  We have just shown that C=∪M0.  In particular, C=cl (∪M0).

Since C is not a finite union of irreducible closed subsets, it is in particular not irreducible.  The contrapositive of Lemma 2 then implies the existence of an infinite subset M1 of M0 such that cl (∪M1) is strictly included in cl (∪M0).

If cl (∪M1) were an irreducible closed subset, since it is included in C, it would be included in some maximal irreducible closed subset of X included in C, namely in some member of M0.  In particular, ∪M1 would be included in some member of M0.  Since all elements of M0 are maximal, hence pairwise incomparable, that would imply that M1 contains only one element.  That is impossible, since M1 is infinite.

Hence we can reapply Lemma 2: there is an infinite subset M2 of M1 such that cl (∪M2) is strictly included in cl (∪M1).

Again cl (∪M2) cannot be irreducible, hence we can find an infinite subset M3 of M2 such that cl (∪M3) is strictly included in cl (∪M2), and proceed infinitely that way.  To make that clear, we have an infinite, antitone sequence of infinite subsets M0M1M2 ⊇ … ⊇ Mn ⊇ … such that cl (∪M0) ⊃ cl (∪M1) ⊃ cl (∪M2) ⊃ … ⊃ cl (∪Mn) ⊃ … (all inclusions here are strict; from which we can conclude that the inclusions M0M1M2 ⊇ … ⊇ Mn ⊇ … are strict as well).

We can therefore pick an irreducible closed subset In in Mn that is not contained in cl (∪Mn+1), for each natural number n.  Indeed, if every element I of Mn were in cl (∪Mn+1), then ∪Mn would be included in cl (∪Mn+1), so cl (∪Mn) would be included in cl (∪Mn+1) as well, contradicting cl (∪Mn) ⊃ cl (∪Mn+1).

We now claim that In cannot be included in cl (∪{Im | mn}).  Assume the contrary.  Since for every m>n, Im is in MmMn+1, In would be included in cl (I0I1 ∪ I2 ∪ … ∪ In-1 ∪ (∪Mn+1)).  Closure commutes with finite unions (not all unions!), so In would be included in I0I1 ∪ I2 ∪ … ∪ In-1 ∪ cl (∪Mn+1).  (Remember that each one of I0, I1, I2, …, In-1 is already closed.)  Since In is irreducible, it must be included in one term of the union.  It cannot be included in cl (∪Mn+1), by definition.  And it cannot be included in any Ik, with k<n, since all those irreducible closed subsets, being maximal, are pairwise incomparable.

Finally, since In is not included in cl (∪{Im | mn}), we can pick a point xn of In that is not in cl (∪{Im | mn}), for each natural number n.  Since xm is in Im, for every m, cl (∪{Im | mn}) contains cl {xm | mn}.  It follows that xn is not in cl {xm | mn}.  Lemma 1 states that the family {xn | nN} is a (necessarily infinite) topological antichain.

## The proof of the easier implication 1 ⇒ 2.

For the record, here is the proof of the much easier implication 1 ⇒ 2.  We assume an infinite topological antichain A.  We shall simply show that cl (A) cannot be written as a finite union of irreducible closed subsets of X.  Assume one could write cl (A) as the finite union I0I1 ∪ I2 ∪ … ∪ In-1, where each Ik is irreducible closed in X.

For each k, if Ik contains some point x of A, then, writing cl (A) as the closure of x, namely ↓x, union cl (A — {x}) (recall that closures commute with finite unions), and recalling that Ik is irreducible and included in cl (A), we obtain that Ik is included in ↓x or in cl (A — {x}).  The latter is impossible if Ik contains x, since A is a topological antichain: indeed, Lemma 1 states that x is not in cl (A — {x}).  Therefore Ik=↓x.

That implies that every Ik can contain at most one point x from A.  Explicitly, if it contained two distinct points x and y of A, then Ik would be equal to ↓x, and also to ↓y.  If we had assumed X to be T0, we could conclude x=y, but we haven’t made that assumption, and we must therefore work slightly harder.  Since A is a topological antichain, x is not in cl (A — {x}), hence neither in the smaller subset cl ({y})=↓y.  That shows that x is not less than or equal to y.  That contradicts ↓x=↓y.

We now have infinitely many points in A, hence in cl (A) = I0I1 ∪ I2 ∪ … ∪ In-1, but each Ik can contain at most one point from A.  That is impossible, by the pigeonhole principle (“to put infinitely many pigeons in some pigeonholes, in such a way that each pigeonhole contains at most one pigeon, you need infinitely many pigeonholes”).  ☐

## To conclude.

Hence we have a full characterization of those topological spaces in which closed subsets are finite unions of irreducible closed subsets, answering our (previously) open question.

If you want something to think about, remember that I said the following on that page: every topological space X such that every closed subspace C has a dense Noetherian subspace D, is FAC.  The following is still open: are the FAC spaces exactly those such that every closed subspace has a dense Noetherian subspace?

1. Jean Goubault-Larrecq. Non-Hausdorff Topology and Domain Theory — Selected Topics in Point-Set Topology. New Mathematical Monographs 22. Cambridge University Press, 2013.
2. M. Kabil and M. Pouzet. Une extension d’un théorème de P. Julien sur les âges de mots. Informatique théorique et applications, 26(5):449–482, 1992.
3. Jimmie D. Lawson, Michael Mislove, and Hilary Priestley.  Ordered sets with no infinite antichains.  Discrete Mathematics 63:225-230, 1987.
4. R. Bonnet.  On the cardinality of the set of initial intervals of a partially ordered set.  Colloquium of the János Bolyai Mathematical Society (Colloq. Math. Soc. János Bolyai) 10:189-198, 1973.
5. Pál Erdős and Alfred Tarski.  On families of mutually exclusive sets.  Annals of Mathematics 44:315-329, 1943.

Jean Goubault-Larrecq 