{"id":1224,"date":"2017-07-05T11:25:29","date_gmt":"2017-07-05T09:25:29","guid":{"rendered":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/?page_id=1224"},"modified":"2023-03-19T16:30:28","modified_gmt":"2023-03-19T15:30:28","slug":"the-o-functor-does-not-preserve-binary-products","status":"publish","type":"page","link":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/?page_id=1224","title":{"rendered":"The O functor does not preserve binary products"},"content":{"rendered":"

In Exercise 8.4.23 of the book<\/a>, I said:<\/p>\n

Exercise 8.4.21 may give you the false impression that the O<\/strong> functor preserves binary products. This is wrong, although an explicit counterexample seems too complicated to study here: see Johnstone (1982, 2.14).<\/p><\/blockquote>\n

My purpose here is to show that that is not that complicated after all.<\/p>\n

My initial plan was to follow Isbell [1], Theorem 2.\u00a0 The proof is only 5 lines, so that should be doable… or so I thought.\u00a0 Isbell wrote in such a terse style that an incredible amount of mathematics would have to be called in just to explain those 5 lines.\u00a0 That seems to have been part of his personality<\/a>.\u00a0 I’ll let you judge.\u00a0 The following is what Isbell writes:<\/p>\n

Theorem 2.<\/strong> If two subspaces of a Hausdorff space have no common point but a nonempty locale intersection, then their product locale is not a space.<\/p>\n

Proof. Let A<\/em> and B<\/em> be such subspaces of X<\/em>. The traces on A \u00d7 B<\/em> of the open rectangles U\u03b1<\/sub><\/em> \u00d7 V\u03b1<\/sub><\/em> of X \u00d7 X<\/em> disjoint from the diagonal D<\/em> cover the points of A \u00d7 B<\/em>. But if I<\/em> is the intersection locale A <\/em>\u2229 B<\/em>, the diagonal of I \u00d7 I<\/em> is contained in A \u00d7 B<\/em> and (in D<\/em>–<\/sup>) disjoint from \u22c1U\u03b1<\/sub><\/em> \u00d7 V\u03b1<\/sub><\/em>=W<\/em>.\u00a0 Thus W \u2229 (A \u00d7 B)<\/em> is an open proper part of\u00a0A \u00d7 B<\/em> containing all the points.<\/p><\/blockquote>\n

If you understand it, you certainly do not need to read the following.<\/p>\n

What Isbell calls a ‘part’ is a sublocale<\/a>.\u00a0 He silently equates any topological space with the corresponding locale and every subspace with corresponding sublocales.\u00a0 The so-called intersection locale is quite probably defined as a suitable pullback.\u00a0 I do not know what\u00a0D<\/em>–<\/sup> stands for.\u00a0 (I think I have seen it defined in one of this other papers, but cannot find it back.)<\/p>\n

Instead, I will give you an elementary proof.\u00a0 This is quite probably the same as Isbell’s proof, and I was certainly guided by it.\u00a0 I would be unable to say whether they are exactly the same at the moment.\u00a0 Notably, I need the open subsets U\u03b1<\/sub><\/em> and V\u03b1 <\/sub><\/em>above (which I’ll rename Ui<\/sub><\/em> and Vi<\/sub><\/em>, because I would like to reserve \u03b1 for the first component of Galois connections) to be regular<\/em> opens, although Isbell does not seem to make such an assumption.\u00a0 That won’t cost me anything, as I will claim in the ‘separation by regular open subsets’ section below.<\/p>\n

Accordingly, we shall spend some time studying the regular open subsets of a topological space, in relation with its dense subsets.\u00a0 That will be needed in building certain Galois connections later\u2014recall that the frame of Galois connections between two frames is their coproduct in Frm<\/strong>, hence their product in the category of locales.<\/p>\n

Regular open subsets<\/h2>\n

A regular open<\/em> subset of a topological space X<\/em> is a subset U<\/em> such that U<\/em>=int(cl(U<\/em>)) (see Exercise 8.1.8 in the book<\/a>).\u00a0 Of course every regular open is open: the interior operation int always gives you an open set as result.\u00a0 But not all open subsets are regular.\u00a0 For example, the union of open intervals (0, 1) \u222a (1, 2) in R<\/strong> is not regular: its closure is [0, 2], and the interior of the latter is (0, 2).<\/p>\n

An alternative description of regular opens is as the regular elements of the frame O<\/strong>(X<\/em>).\u00a0 Those were introduced in Exercise 8.1.7.\u00a0 A frame \u03a9 is a complete Heyting algebra, meaning that there is an implication (or residuation) operation \u21d2, with the defining property that u<\/em> \u21d2 v<\/em> is the largest element w<\/em> such that u<\/em> \u22c0 w<\/em> \u2264 v<\/em>.\u00a0 Equivalently, for all u<\/em>, v<\/em>, w<\/em> in \u03a9, u<\/em> \u22c0 w<\/em> \u2264 v<\/em> if and only\u00a0 if w<\/em> \u2264 u<\/em> \u21d2 v<\/em>.<\/p>\n

When v<\/em> = \u22a5, we obtain intuitionistic negation<\/em> \u00acu<\/em>. When \u03a9 is the frame of opens O<\/strong>(X<\/em>) of a topological space X<\/em>, \u00acU<\/em> is the interior of the complement of U<\/em> \u2014 not the complement of U<\/em>, which would in general fail to be open.<\/p>\n

The regular elements of a frame \u03a9 are the elements u<\/em> that are equal to their double negation \u00ac\u00acu<\/em>.\u00a0 Equivalently, those are exactly the elements that can be expressed as the negation \u00acu<\/em> of some u<\/em> \u2208 \u03a9.\u00a0 That is so, because intuitionistic negation has the property that \u00ac\u00ac\u00acu=<\/em>\u00acu<\/em> for every u<\/em> (although\u00a0\u00ac\u00acu\u2260<\/em>u<\/em> in general).<\/p>\n

It turns out that the regular elements of O<\/strong>(X<\/em>) are exactly the regular open subsets of X<\/em>, as Exercise 8.1.8 asks you to check.\u00a0 The idea is simply that \u00ac\u00acU<\/em>=int(cl(U<\/em>)).<\/p>\n

Owing to the definition of intuitionistic negation (through implication), we shall use the following equivalence: for all u<\/em>, v<\/em> in \u03a9, u<\/em> \u22c0 v<\/em> =\u00a0\u22a5 if and only\u00a0 if v<\/em> \u2264 \u00acu<\/em>, if and only if u<\/em> \u2264 \u00acv<\/em>.\u00a0 When \u03a9=O<\/strong>(X<\/em>), this translates to: for two open subsets U<\/em> and V<\/em>, U<\/em> and V<\/em> are disjoint if and only if V<\/em> \u2286 \u00acU<\/em>, if and only if U<\/em>\u00a0\u2286 \u00acV<\/em>.<\/p>\n

Separation by regular open subsets<\/h2>\n

Recall that a topological space X<\/em> is Hausdorff, or T2<\/sub>, if and only if for every pair of distinct points x<\/em>\u2260y<\/em>, there are disjoint open subsets U<\/em> and V<\/em> such that U<\/em> contains x<\/em> and V<\/em> contains y<\/em>.<\/p>\n

I claim that you can take both U<\/em> and V<\/em> regular open here.\u00a0 In other words,<\/p>\n

Lemma.<\/strong> A topological space X<\/em> is\u00a0T2<\/sub> if and only if, for every pair of distinct points x<\/em>\u2260y<\/em>, there are disjoint regular<\/em> open subsets U<\/em> and V<\/em> such that U<\/em> contains x<\/em> and V<\/em> contains y<\/em>.<\/p>\n

Proof.<\/em> The if direction is obvious.\u00a0 In the only if direction, imagine we have disjoint open subsets U<\/em> and V<\/em> such that U<\/em> contains x<\/em> and V<\/em> contains y<\/em>.\u00a0 Since U<\/em> and V<\/em> are disjoint, V<\/em> \u2286 \u00acU<\/em>, and since \u00acU<\/em>=\u00ac\u00ac\u00acU<\/em>, V<\/em> \u2286 \u00ac\u00ac\u00acU<\/em>.\u00a0 It follows that V<\/em> and\u00a0\u00ac\u00acU<\/em> are disjoint.\u00a0 Since\u00a0\u00ac\u00acU<\/em> and V<\/em> are disjoint, \u00ac\u00acU <\/em>\u2286 \u00acV = \u00ac\u00ac\u00acV<\/em>, so\u00a0\u00ac\u00acU<\/em> and\u00a0\u00ac\u00acV<\/em> are disjoint.<\/p>\n

We have U<\/em> \u2286 \u00ac\u00acU<\/em>, so\u00a0\u00ac\u00acU<\/em> is again an open neighborhood of x<\/em>.\u00a0 Similarly,\u00a0\u00ac\u00acV<\/em> is an open neighborhood of y<\/em>.\u00a0 They are disjoint, and regular open.\u00a0 \u2610<\/p>\n

Dense subspaces and regular opens<\/h2>\n

Fix a given dense subset A<\/em> of a topological space X<\/em>.<\/p>\n

For every open subset U<\/em> of X<\/em>, U<\/em> \u2229 A<\/em> is an open subset of A<\/em>, viewed as a subspace<\/em> of X<\/em>.\u00a0 Can we retrieve U<\/em> from the data of U<\/em> \u2229 A<\/em>?\u00a0 I.e., is the map U<\/em> \u27fc U<\/em> \u2229 A<\/em> injective?<\/p>\n

That would seem so, because A<\/em> is dense, but that is wrong in general.\u00a0 For example, if you take the real line\u00a0R<\/strong> for X<\/em>, with its usual metric topology, and the subset Q<\/strong> of rationals for A<\/em>, then (0, 5) and (0, \u03c0) \u222a (\u03c0, 5) have exactly the same intersection with A<\/em>.<\/p>\n

But that works with regular opens:<\/p>\n

Lemma.<\/strong> Let A<\/em> be a dense subset of a topological space X<\/em>.\u00a0 There is at most one regular open subset U<\/em> of X<\/em> such that U<\/em> \u2229 A<\/em> is equal to a given open subset V<\/em> of A<\/em>, and that is int(cl(V<\/em>)).\u00a0 (Interior and closure are taken in X<\/em>.)
\nEquivalently, for every regular open subset U<\/em> of X<\/em>, U<\/em>=int(cl(U<\/em> \u2229 A<\/em>)).<\/p>\n

Proof.<\/em> We first claim that if U<\/em> is any open subset of X<\/em>, then U<\/em> is included in int(cl(U<\/em> \u2229 A<\/em>)).\u00a0 To prove this, let x<\/em> be an arbitrary element of U<\/em>.\u00a0 For every open neighborhood U’<\/em> of x<\/em>,\u00a0U<\/em> \u2229 U’<\/em> is open, non-empty (it contains x<\/em>), so its intersects A<\/em>, because A<\/em> is dense: (U<\/em> \u2229 U’<\/em>) \u2229 A<\/em> is non-empty, or equivalently, U’<\/em> \u2229 (U<\/em> \u2229 A<\/em>) is non-empty.\u00a0 This shows that every open neighborhood U’<\/em> of x<\/em> intersects U<\/em> \u2229 A<\/em>, hence that x<\/em> is in cl(U<\/em> \u2229 A<\/em>).\u00a0 (Otherwise, take the complement of cl(U<\/em> \u2229 A<\/em>) for U’<\/em>.)\u00a0 Since x<\/em> is an arbitrary element of U<\/em>, U<\/em> is included in cl(U<\/em> \u2229 A<\/em>).\u00a0 Because U<\/em> is open, it follows that U<\/em> is included in the largest open set included in cl(U<\/em> \u2229 A<\/em>), namely int(cl(U<\/em> \u2229 A<\/em>)).<\/p>\n

Conversely, int(cl(U<\/em> \u2229 A<\/em>)) is always included in int(cl(U<\/em>)), and that is equal to U<\/em> if U<\/em> is regular open.<\/p>\n

This show that if U<\/em> is regular open and U<\/em> \u2229 A<\/em> = V<\/em>, then U<\/em> = int(cl(V<\/em>)), showing that such a U<\/em> is unique.\u00a0 \u2610<\/p>\n

Corollary.<\/strong> Let A<\/em> be a dense subset of a topological space X<\/em>.\u00a0 For two regular open subsets U<\/em> and V<\/em> of X<\/em>, U<\/em> \u2286 V<\/em> if and only if U<\/em> \u2229 A<\/em> \u2286 V<\/em> \u2229 A<\/em>.<\/p>\n

Proof.<\/em> If U<\/em> \u2229 A<\/em> \u2286 V<\/em> \u2229 A<\/em>, then (U<\/em>\u00a0\u222a V<\/em>) \u2229 A<\/em> = (U<\/em> \u2229 A<\/em>)\u00a0\u222a (V<\/em> \u2229 A<\/em>) = V<\/em> \u2229 A<\/em>.\u00a0 The previous Lemma implies that U<\/em>\u00a0\u222a V<\/em> = V<\/em>, that is, U<\/em> \u2286 V<\/em>. \u00a0(Note added on March 19th, 2023: this argument would hold only if\u00a0 U<\/em>\u00a0\u222a V<\/em> is regular. \u00a0Instead, we realize that\u00a0U<\/em> \u2229\u00a0V<\/em> is regular, see\u00a0Exercise 8.1.7 in the book<\/a>. If U<\/em> \u2229 A<\/em> \u2286 V<\/em> \u2229 A<\/em>, then (U<\/em> \u2229\u00a0V<\/em>) \u2229 A<\/em> =\u00a0U<\/em> \u2229 A<\/em>. \u00a0The previous Lemma implies that U<\/em> \u2229\u00a0V =\u00a0U<\/em>, that is, U \u2286 V<\/em>.) \u00a0\u2610<\/p>\n

Locale products<\/h2>\n

Let us start moving towards our goal.<\/p>\n

Given two frames \u03a91<\/sub> and \u03a92<\/sub>, the frame Gal(\u03a91<\/sub>, \u03a92<\/sub>) of Galois connections between\u00a0\u03a91<\/sub> and \u03a92<\/sub> is their coproduct in the category Frm<\/strong> of frames.\u00a0 (See Exercise 8.4.28 in the book<\/a>.)<\/p>\n

Among those Galois connections, the so-called (u<\/em>1<\/sub> \u00d7\u00a0u<\/em>2<\/sub>)-rectangles play a fundamental role.\u00a0 (Definition 8.4.20.)\u00a0 Those are the Galois connections (\u03b1, \u03b3) such that \u03b1 maps:<\/p>\n