{"id":4926,"date":"2022-05-20T11:22:28","date_gmt":"2022-05-20T09:22:28","guid":{"rendered":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/?page_id=4926"},"modified":"2022-11-19T14:54:22","modified_gmt":"2022-11-19T13:54:22","slug":"compact-scattered-subsets-and-a-topological-game","status":"publish","type":"page","link":"https:\/\/projects.lsv.ens-paris-saclay.fr\/topology\/?page_id=4926","title":{"rendered":"Compact scattered subsets and a topological game"},"content":{"rendered":"\n

I have already given an argument for the non-consonance of the Sorgenfrey line\u00a0R<\/strong>\u2113<\/sub>\u00a0here<\/a>. I would like to deal with the case of the space\u00a0Q<\/strong>\u00a0of rational numbers, after Costantini and Watson [1]. This would be a bit too long to cover entirely, so the bulk of the explanation will be for another time. I will explain why the compact subsets of\u00a0Q<\/strong> are all\u00a0scattered<\/em>, a notion that I will explain below. Reading between the lines, the Costantini-Watson argument relies on a property that I will characterize through the use of a topological game\u00a0G<\/em>(K<\/em>), resembling the strong Choquet game (see Section 7.6 in the book<\/a>), in which we will see that player I has a winning strategy if\u00a0K<\/em>\u00a0is compact and scattered\u2014and that is an if and only if<\/em> in any T2<\/sub> space. I have not seen that published anywhere, but maybe I just don’t know enough. In any case, I feel that characterization of compact scattered sets through a topological game is worthy of some attention.<\/p>\n\n\n\n

Scattered compact subsets<\/h2>\n\n\n\n

I said that we would see that all the compact subsets of Q<\/strong> are scattered<\/em>. And hence, we need to understand that notion first.<\/p>\n\n\n\n

Let X<\/em> be a topological space. An isolated point<\/em> of a subset D<\/em> of X<\/em> is any point x<\/em> of D<\/em> that has an open neighborhood U<\/em> in X<\/em> such that U<\/em> \u2229 D<\/em> = {x<\/em>}; if you prefer, such that {x<\/em>} is open in D<\/em> seen as a subspace of X<\/em>. The points of D<\/em> that are not isolated are called the limit points<\/em> of D<\/em>, and those are exactly the points of D<\/em> that are limits of nets of points of D<\/em>\u2013{x<\/em>}.<\/p>\n\n\n\n

A subset D<\/em> of X<\/em> is dense-in-itself<\/em> if and only if every point of D<\/em> is a limit point of D<\/em>, namely if and only if it has no isolated point. A scattered space<\/em> is a space whose only dense-in-itself subset is the empty set.<\/p>\n\n\n\n

Q<\/strong> itself is dense-in-itself, as a subset of itself; in other words, Q<\/strong> has no isolated point. Hence Q<\/strong> is not scattered. However, we have the following, which, rather amazingly, follows from the Baire property.<\/p>\n\n\n\n

Lemma A.<\/strong> Every non-empty compact subset of Q<\/strong> has at least one isolated point. Hence every compact subset of Q<\/strong> is scattered (as a topological subspace of Q<\/strong>).<\/p>\n\n\n\n

Proof.<\/em>\u00a0Let\u00a0K<\/em>\u00a0be a compact subset of\u00a0Q<\/strong>. \u00a0We consider K<\/em> as a subspace<\/em> of Q<\/strong>. As such, it is T2<\/sub>, because Q<\/strong> is T2<\/sub>. \u00a0Being compact and T2<\/sub>, K<\/em> is locally compact (Proposition 4.8.8 in the book<\/a>). \u00a0Being T2<\/sub>, K<\/em> is sober (Proposition 8.2.12), and every locally compact sober space is Choquet-complete, hence Baire (Proposition 8.3.24). We can arrive at the same conclusion by the following, alternate route. Since K<\/em> is compact in Q<\/strong>, it is also compact in R<\/strong> (in particular, sequentially compact). The natural metric d<\/em> on R<\/strong> turns it into a complete metric space, and we know that every (sequentially) compact subspace of a complete metric space is itself complete (Proposition 3.4.6). Therefore K<\/em> is completely metrizable, hence Baire. You may use Theorem 7.6.15 plus Proposition 8.3.24 for that, but you may also have learned that every complete metric space is Baire independently.<\/p>\n\n\n\n

Anyway, K<\/em> is a Baire subspace of Q<\/strong>. This means that the intersection of countably many dense open subsets of K<\/em> is dense. Taking complements, the union of countably many closed sets with empty interiors (in K<\/em>) has empty interior (in K<\/em>). We consider the sets {x<\/em>}, where x<\/em> ranges over the countably many elements of K<\/em>. Those sets are closed. For each x<\/em> \u2208 K<\/em>, {x<\/em>} has non-empty interior in K<\/em> if and only if {x<\/em>} is open in K<\/em>, if and only if {x<\/em>} = U<\/em> \u2229 K<\/em> for some open subset U<\/em> of Q<\/strong>, if and only if x<\/em> is isolated in K<\/em>. Hence, if K<\/em> has no isolated point, then all the sets {x<\/em>} with x<\/em> \u2208 K<\/em> are closed sets with empty interior, and their union is K<\/em>, whose interior (in K<\/em>) is K<\/em> itself; hence, if K<\/em> has no isolated point, then it is empty. By contraposition, if K<\/em> is non-empty, then it has an isolated point.<\/p>\n\n\n\n

Let now D<\/em> be any non-empty dense-in-itself subset of K<\/em>. The closure clK<\/sub><\/em>(D<\/em>) of D<\/em> in K<\/em> is compact, since it is a closed subset of a compact set. It is also non-empty, so it has an isolated point x<\/em>, as we have proved in the first part of the lemma. By definition of ‘isolated’, U<\/em> \u2229 clK<\/sub><\/em>(D<\/em>) = {x<\/em>} for some open neighborhood U<\/em> (in K<\/em>) of x<\/em>. In particular, U<\/em> intersects clK<\/sub><\/em>(D<\/em>); then, it must also intersect D<\/em>, and the only point at which U<\/em> and D<\/em> can intersect must be x<\/em>. It follows that x<\/em> \u2208 U<\/em> \u2229 D<\/em> \u2286 U<\/em> \u2229 clK<\/sub><\/em>(D<\/em>) = {x<\/em>}, so U<\/em> \u2229 D<\/em> = {x<\/em>}. This means that x<\/em> is isolated in D<\/em>, which is impossible since D<\/em> is dense-in-itself. \u2610<\/p>\n\n\n\n

We will also need the following pretty easy observation.<\/p>\n\n\n\n

Lemma B.<\/strong> Every subset of a scattered space is scattered.<\/p>\n\n\n\n

Proof.<\/em> Let K<\/em> be a scattered space, and A<\/em> be a subset of K<\/em>, seen as a subspace. We consider any dense-in-itself subset D<\/em> of A<\/em>. If D<\/em> had an isolated point x<\/em> in K<\/em>, then by definition there would be an open neighborhood U<\/em> of x<\/em> in K<\/em> such that U<\/em> \u2229 D<\/em> = {x<\/em>}. Then U<\/em> \u2229 A<\/em> would be an open neighborhood of x<\/em> in A<\/em>, with the property that x<\/em> \u2208 (U<\/em> \u2229 A<\/em>) \u2229 D<\/em> \u2286 U<\/em> \u2229 D<\/em> = {x<\/em>}, so (U<\/em> \u2229 A<\/em>) \u2229 D<\/em> = {x<\/em>}; hence x<\/em> would be an isolated point of D<\/em> in A<\/em>, contradicting the fact that D<\/em> is dense-in-itself in A<\/em>. Therefore D<\/em> has no isolated point in K<\/em>, meaning that D<\/em> is dense-in-itself in K<\/em>. Since K<\/em> is scattered, D<\/em> must be empty. We have shown that every dense-in-itself subset D<\/em> of A<\/em> is empty. Therefore, A<\/em> is scattered. \u2610<\/p>\n\n\n\n

Decomposing compact scattered subsets, and scattering heights<\/h2>\n\n\n\n

In the following, let X<\/em> be any topological space.<\/p>\n\n\n\n

The compact scattered subsets K<\/em> of a X<\/em> can be stratified, as follows. If K<\/em> is non-empty, then we form the subset K<\/em>0<\/sub> of all isolated points of K<\/em>. For each x<\/em> \u2208 K<\/em>0<\/sub>, there is an open neighborhood Ux<\/sub><\/em> of x<\/em> in X<\/em> that intersects K<\/em> at x<\/em> only. Hence U<\/em>0<\/sub>(K<\/em>) \u225d \u222ax<\/em> \u2208 K<\/em><\/em>0<\/sub><\/sub> Ux<\/sub><\/em> is an open subset of X<\/em> whose intersection with K<\/em> is exactly K<\/em>0<\/sub>. It follows that K<\/em>\u2013K<\/em>0<\/sub> = K<\/em>\u2013U<\/em>0<\/sub>(K<\/em>) is the intersection of a compact set with a closed set, hence is itself compact. We will actually need to remember that it is obtained as K<\/em> minus some open set. (This will enable us to not make any unwarranted demands on X<\/em>, such as Hausdorfness or well-filteredness.)<\/p>\n\n\n\n

For an example, consider the compact subset {0} \u222a {1\/2n<\/sup><\/em> | n<\/em> \u2208 N<\/strong>} of Q<\/strong>. In that case, K<\/em>0<\/sub> is the set of all points of the form 1\/2n<\/sup><\/em>, and this is the simplest example of a compact (necessarily scattered) subset of Q<\/strong> whose points are not all isolated. The simplest shape that U<\/em>0<\/sub>(K<\/em>) may take here is ]0, \u221e[ \u2229 Q<\/strong>; we may also take the union of the sets ](1\u2013\u03b5)\/2n<\/sup><\/em>, (1+\u03b5)\/2n<\/sup><\/em>[ \u2229 Q<\/strong>, n<\/em> \u2208 N<\/strong>, for any fixed \u03b5>0 such that \u03b5<1.<\/p>\n\n\n\n

If K<\/em>\u2013K<\/em>0<\/sub> is empty, we stop here. (That is not the case in the example above.) Otherwise, K’<\/em>1<\/sub> \u225d K<\/em>\u2013K<\/em>0<\/sub> is another non-empty compact subset of X<\/em> (remember that is is compact because it can also be written as K<\/em>\u2013U<\/em>0<\/sub>(K<\/em>)), and therefore K<\/em>\u2013K<\/em>0<\/sub> must be scattered by Lemma B. We can now form its set K<\/em>1<\/sub> of isolated points, as above. (In the example, K<\/em>\u2013K<\/em>0<\/sub> = {0}, and now 0 is the only isolated point, so K<\/em>1<\/sub>={0}.) Just as with K<\/em>0<\/sub>, there is an open subset U<\/em>1<\/sub>(K<\/em>) \u225d \u222ax<\/em> \u2208 K<\/em><\/em>1<\/sub><\/sub> Ux<\/sub><\/em> of X<\/em> such that K<\/em>1<\/sub> = K’<\/em>1<\/sub> \u2229 U<\/em>1<\/sub>(K<\/em>), and then K’<\/em>1<\/sub>\u2013K<\/em>1<\/sub> = K’<\/em>1<\/sub>\u2013U<\/em>1<\/sub>(K<\/em>) = K<\/em>\u2013U<\/em>0<\/sub>(K<\/em>)\u2013U<\/em>1<\/sub>(K<\/em>), so K’<\/em>1<\/sub>\u2013K<\/em>1<\/sub> is compact, and in fact equal to K<\/em> minus some open set.<\/p>\n\n\n\n

We repeat the process. If K’<\/em>2<\/sub> \u225d K<\/em>\u2013(K<\/em>0<\/sub>\u222aK<\/em>1<\/sub>) is empty, then we stop here. (That is indeed the case in our example.) Otherwise, K’<\/em>2<\/sub> is non-empty, compact, and scattered, and we form its set K<\/em>2<\/sub> of isolated points, and so on.<\/p>\n\n\n\n

For a more complicated example, take the previous example {0} \u222a {1\/2n<\/sup><\/em> | n<\/em> \u2208 N<\/strong>}, and add to it all the points in the various sequences {1\/2n<\/sup><\/em>+1\/2m<\/sup><\/em>+<\/sup>n<\/sup><\/em>+1<\/sup> | m<\/em> \u2208 N<\/strong>}, converging to 1\/2n<\/sup><\/em>, for each n<\/em> \u2208 N<\/strong>. Hence our newer example consists of 0, the points 0, 1\/2n<\/sup><\/em>, and 1\/2n<\/sup><\/em>+1\/2m<\/sup><\/em>+<\/sup>n<\/sup><\/em>+1<\/sup>, for all m<\/em> and n<\/em>. In that case, K<\/em>0<\/sub> is the collection of points of the form 1\/2n<\/sup><\/em>+1\/2m<\/sup><\/em>+<\/sup>n<\/sup><\/em>+1<\/sup>, K<\/em>1<\/sub> is the collection of points of the form 1\/2n<\/sup><\/em>, and K<\/em>2<\/sub>={0}. You probably start to see the pattern.<\/p>\n\n\n\n

We build K<\/em>0<\/sub>, K<\/em>1<\/sub>, K<\/em>2<\/sub>, …, as above. This defines a process that builds pairwise disjoint sets K<\/em>i<\/em><\/sub>, and a decreasing sequence of non-empty compact scattered subsets K’<\/em>n<\/em><\/sub> \u225d K<\/em>\u2013\u222ai<\/em>=0<\/sub>n<\/em>\u20131<\/sup> K<\/em>i<\/em><\/sub> of X<\/em>, which may stop at some stage n<\/em> (meaning that K’<\/em>n<\/em><\/sub> is empty) or not. Also, each set K’<\/em>n<\/em><\/sub> is not just compact, but can be written as K<\/em> minus some union of open sets \u222ai<\/em>=0<\/sub>n<\/em>\u20131<\/sup>U<\/em>i<\/em><\/sub>(K<\/em>).<\/p>\n\n\n\n

If the process does not stop at any finite stage, then we obtain a decreasing sequence of non-empty compact scattered subsets K’<\/em>n<\/em><\/sub> of X<\/em>, n<\/em> \u2208 N<\/strong>. If X<\/em> were well-filtered and T1<\/sub>, we could argue that their intersection K’<\/em>\u03c9<\/sub> \u225d K<\/em>\u2013\u222ai<\/em> \u2208 N<\/strong><\/sub> K<\/em>i<\/em><\/sub> is also non-empty and compact. But we can show this even without assuming anything on X<\/em>. Indeed, K’<\/em>\u03c9<\/sub> is the intersection of the sets K’<\/em>n<\/em><\/sub> = K<\/em>\u2013\u222ai<\/em>=0<\/sub>n<\/em>\u20131<\/sup> K<\/em>i<\/em><\/sub>, which are also equal to K<\/em>\u2013\u222ai<\/em>=0<\/sub>n<\/em>\u20131<\/sup> U<\/em>i<\/em><\/sub>(K<\/em>), so K’<\/em>\u03c9<\/sub> is also equal to K<\/em> minus the union of the chain of open sets \u222ai<\/em>=0<\/sub>n<\/em>\u20131<\/sup> U<\/em>i<\/em><\/sub>(K<\/em>), n<\/em> \u2208 N<\/strong>. If K’<\/em>\u03c9<\/sub> were empty, then K<\/em> would be included in that directed union of open sets, hence it would be included in one of them, since K<\/em> is compact; in other words, some K’<\/em>n<\/em><\/sub> would be empty. As we have assumed this not to be the case, K’<\/em>\u03c9<\/sub> is not empty. Also, K’<\/em>\u03c9<\/sub> is equal to K<\/em> minus the infinite union of open sets \u222ai<\/em>=0<\/sub>\u221e<\/sup> U<\/em>i<\/em><\/sub>(K<\/em>), and is therefore the intersection of K<\/em> with a closed set, hence is itself compact. By Lemma B, K’<\/em>\u03c9<\/sub> is also scattered.<\/p>\n\n\n\n

Since K’<\/em>\u03c9<\/sub> is non-empty (in case we have arrived up to this point), the process definitely does not end here. We form the subset K<\/em>\u03c9<\/sub> of isolated points of K’<\/em>\u03c9<\/sub>. Then K’<\/em>\u03c9+1<\/sub> \u225d K’<\/em>\u03c9<\/sub>\u2013K<\/em>\u03c9<\/sub> is compact and scattered, and equal to K<\/em> minus some open set. If K’<\/em>\u03c9+1<\/sub> is empty, then we stop here, otherwise we form the subset K<\/em>\u03c9+1<\/sub> of isolated points of K’<\/em>\u03c9+1<\/sub>, and so on.<\/p>\n\n\n\n

Formally, and recalling that X<\/em> is well-filtered and T1<\/sub>, we define an ordinal-indexed, antitonic sequence of compact scattered sets K’<\/em>\u03b1<\/sub> as follows: K’<\/em>0<\/sub> \u225d K<\/em>, K’<\/em>\u03b1+1<\/sub> \u225d K’<\/em>\u03b1<\/sub>\u2013K<\/em>\u03b1<\/sub> where K<\/em>\u03b1<\/sub> is the set of isolated points in K’<\/em>\u03b1<\/sub> (for any ordinal \u03b1), and K’<\/em>\u03b2<\/sub> is the filtered intersection of the sets K’<\/em>\u03b1<\/sub>, \u03b1<\u03b2, for every limit ordinal \u03b2. Also, for every ordinal \u03b1, K’<\/em>\u03b1+1<\/sub> is also equal to K’<\/em>\u03b1<\/sub>\u2013U<\/em>\u03b1<\/sub>(K<\/em>) for some open subset U<\/em>\u03b1<\/sub>(K<\/em>) of X<\/em>, from which we deduce that for every ordinal \u03b1, K’<\/em>\u03b1<\/sub> is equal to K<\/em> minus the open set \u222a\u03b3<\u03b1<\/sub> U<\/em>\u03b3<\/sub>(K<\/em>), by induction on \u03b3. We say that the sequence stops<\/em> (giving meaning to the informal expression “we stop here” or “the process ends here” that we have used informally until now) at the first ordinal \u03b1 such that K’<\/em>\u03b1<\/sub> is empty.<\/p>\n\n\n\n

That ordinal is called the scattering height<\/em> of K<\/em>.  It must exist, and here is one possible argument.  As long as the sequence has not stopped at any ordinal <\u03b1, all the sets U<\/em>\u03b3<\/sub>(K<\/em>) with \u03b3<\u03b1 are disjoint and non-empty, so the cardinality of \u222a\u03b3<\u03b1<\/sub> U<\/em>\u03b3<\/sub>(K<\/em>) is at least that of \u03b1.  Since that cardinality cannot exceed the cardinality of K<\/em>, the sequence must stop at or before any given cardinal number strictly larger than c<\/em>, for example the cardinality of the powerset of K<\/em>.<\/p>\n\n\n\n

Since the scattering height of K<\/em> exists, every point x<\/em> of K<\/em> lies in some set K<\/em>\u03b1(x<\/em>)<\/sub>. Moreover, those sets are pairwise disjoint. Hence every point x<\/em> of K<\/em> lies in some unique<\/em> set K<\/em>\u03b1(x<\/em>)<\/sub>; \u03b1(x<\/em>) is the scattering height<\/em> of x<\/em> in K<\/em>.<\/p>\n\n\n\n

First observation.<\/strong> The scattering height \u03b2 of the compact set K<\/em>, namely the least ordinal \u03b2 such that K’<\/em>\u03b2<\/sub> is empty, cannot be a limit ordinal.<\/p>\n\n\n\n

Indeed, this is a variant of the argument we have already used in order to show that K’<\/em>\u03c9<\/sub> is non-empty, assuming that no K’<\/em>n<\/em><\/sub>  was empty, for any n<\/em><\u03c9.<\/p>\n\n\n\n

Explicitly, if \u03b2 is a limit ordinal, and the sequence stops at \u03b2, then no K’<\/em>\u03b1<\/sub> with \u03b1<\u03b2 can be empty (because \u03b2 is least). Since K’<\/em>\u03b1<\/sub> = K<\/em> \u2013 \u222a\u03b3<\u03b1<\/sub> U<\/em>\u03b3<\/sub>(K<\/em>), this means that K<\/em> is included in none of the open sets \u222a\u03b3<\u03b1<\/sub> U<\/em>\u03b3<\/sub>(K<\/em>), for any \u03b1<\u03b2. If K’<\/em>\u03b2<\/sub> = K<\/em> \u2013 \u222a\u03b3<\u03b2<\/sub> U<\/em>\u03b3<\/sub>(K<\/em>) were empty, then K<\/em> would be included in \u222a\u03b3<\u03b2<\/sub> U<\/em>\u03b3<\/sub>(K<\/em>), which is also the union of the chain of open sets \u222a\u03b3<\u03b1<\/sub>U<\/em>\u03b3<\/sub>(K<\/em>), \u03b1<\u03b2, because \u03b2 is a limit ordinal. (More explicitly, for every ordinal \u03b3<\u03b2, there is an ordinal \u03b1 such that \u03b3<\u03b1 and \u03b1<\u03b2, for example \u03b3+1.) Since K<\/em> is compact, K<\/em> must then be included in \u222a\u03b3<\u03b1<\/sub> U<\/em>\u03b3<\/sub>(K<\/em>) for some \u03b1<\u03b2, and this is impossible.<\/p>\n\n\n\n

Second observation.<\/strong> The last non-empty set of isolated points we see in the sequence, namely K<\/em>\u03b1<\/sub> where \u03b1+1 is the scattering height of the compact set K,<\/em> is finite<\/em>. I.e., there are only finitely many points of maximal scattering height in K<\/em>.<\/p>\n\n\n\n

Indeed, since K’<\/em>\u03b1+1<\/sub> is empty, K’<\/em>\u03b1<\/sub> is equal to K<\/em>\u03b1<\/sub>, namely all the points of K’<\/em>\u03b1<\/sub> are isolated. Therefore, for every point x<\/em> in K’<\/em>\u03b1<\/sub>, there is an open neighborhood Ux<\/sub><\/em> of x<\/em> in X<\/em> such that Ux<\/sub><\/em> \u2229 K<\/em>\u03b1<\/sub> = {x<\/em>}; the family of those sets Ux<\/sub><\/em> is an open cover of K<\/em>\u03b1<\/sub>, and since K<\/em>\u03b1<\/sub> is compact, it must contain a finite subcover… which, in turn, contains only finitely many points from K<\/em>\u03b1<\/sub>.<\/p>\n\n\n\n

All of that is really an essential part of Costantini and Watson’s proof. This is somehow hidden, but those two observations do occur near the top of page 266 in [1], in the middle of an otherwise technical proof.<\/p>\n\n\n\n

A topological game<\/h2>\n\n\n\n

We arrive at the main topic of this post.<\/p>\n\n\n\n

What Costantini and Watson really do with those observations is something that is best described using a certain topological game, which I will define below. I believe it somehow clarifies what is happening, but I haven’t seen this in the literature. This rests on the following.<\/p>\n\n\n\n

Lemma C.<\/strong>  Let K<\/em> be a non-empty compact scattered subset of a topological space X<\/em>.<\/p>\n\n\n\n

    \n
  1. The scattering height of K<\/em> is a successor ordinal \u03b1+1, and K<\/em>\u03b1<\/sub> is finite.<\/li>\n\n\n\n
  2. For every open subset U<\/em> of X<\/em>, for every ordinal \u03b3, (K<\/em>\u2013U<\/em>)‘<\/em>\u03b3<\/sub> is included in K’<\/em>\u03b3<\/sub>.<\/li>\n\n\n\n
  3. For every open subset U<\/em> of X<\/em>, the scattering height of K<\/em>\u2013U<\/em> is less than or equal to the scattering height, \u03b1+1, of K<\/em>.<\/li>\n\n\n\n
  4. If additionally, U<\/em> contains K<\/em>\u03b1<\/sub>, then the scattering height of K<\/em>\u2013U<\/em> is strictly less than that of K<\/em>, hence is smaller than or equal to \u03b1.<\/li>\n<\/ol>\n\n\n\n

    Proof.<\/em>  Since K<\/em> is non-empty, its scattering height is not 0. By the first observation above, it is not a limit ordinal, so it must be a successor ordinal \u03b1+1. By the second observation, K<\/em>\u03b1<\/sub> is a finite set. This proves item 1.  Also, by definition of the scattering height, K’<\/em>\u03b1<\/sub>=K<\/em>\u03b1<\/sub>.<\/p>\n\n\n\n

    Let us prove item 2.  We claim that (K<\/em>\u2013U<\/em>)‘<\/em>\u03b3<\/sub> is included in K’<\/em>\u03b3<\/sub> for every ordinal \u03b3; by definition, (K<\/em>\u2013U<\/em>)‘<\/em>0<\/sub>=K<\/em>\u2013U<\/em>, (K<\/em>\u2013U<\/em>)‘<\/em>\u03b3+1<\/sub> = (K<\/em>\u2013U<\/em>)‘<\/em>\u03b3<\/sub>\u2013(K<\/em>\u2013U<\/em>)\u03b3<\/sub> where (K<\/em>\u2013U<\/em>)\u03b3<\/sub> is the set of isolated points in (K<\/em>\u2013U<\/em>)‘<\/em>\u03b3<\/sub>, and (K<\/em>\u2013U<\/em>)‘<\/em>\u03b2<\/sub> = \u2229\u03b3<\u03b2<\/sub> (K<\/em>\u2013U<\/em>)‘<\/em>\u03b3<\/sub> for all limit ordinals \u03b2. This is by induction on \u03b3. The only interesting case occurs when showing that (K<\/em>\u2013U<\/em>)‘<\/em>\u03b3+1<\/sub> is included in K’<\/em>\u03b3+1<\/sub>=K’<\/em>\u03b3<\/sub>\u2013K<\/em>\u03b3<\/sub>, where K<\/em>\u03b3<\/sub> is the set of isolated points of K’<\/em>\u03b3<\/sub>, knowing by induction hypothesis that (K<\/em>\u2013U<\/em>)‘<\/em>\u03b3<\/sub> is included in K’<\/em>\u03b3<\/sub>. Since (K<\/em>\u2013U<\/em>)‘<\/em>\u03b3<\/sub> \u2286 K’<\/em>\u03b3<\/sub>, we have (K<\/em>\u2013U<\/em>)‘<\/em>\u03b3+1<\/sub> = (K<\/em>\u2013U<\/em>)‘<\/em>\u03b3<\/sub>\u2013(K<\/em>\u2013U<\/em>)\u03b3<\/sub> \u2286 K’<\/em>\u03b3<\/sub>, and it remains to show that (K<\/em>\u2013U<\/em>)‘<\/em>\u03b3+1<\/sub> does not intersect K<\/em>\u03b3<\/sub>. In other words, we wish to show that (K<\/em>\u2013U<\/em>)‘<\/em>\u03b3<\/sub>\u2013(K<\/em>\u2013U<\/em>)\u03b3<\/sub> does not intersect K<\/em>\u03b3<\/sub>, or equivalently, that (K<\/em>\u2013U<\/em>)‘<\/em>\u03b3<\/sub>\u2229 K<\/em>\u03b3<\/sub> is included in (K<\/em>\u2013U<\/em>)\u03b3<\/sub>.<\/p>\n\n\n\n

    Let x<\/em> be a point of (K<\/em>\u2013U<\/em>)‘<\/em>\u03b3<\/sub> \u2229 K<\/em>\u03b3<\/sub>. By definition of K<\/em>\u03b3<\/sub>, x<\/em> is isolated in K’<\/em>\u03b3<\/sub>, namely there is an open neighborhood U<\/em> of x<\/em> in X<\/em> such that U<\/em> \u2229 K’<\/em>\u03b3<\/sub> = {x<\/em>}. Since x<\/em> \u2208 (K<\/em>\u2013U<\/em>)‘<\/em>\u03b3<\/sub> \u2286 K’<\/em>\u03b3<\/sub>, we also have U<\/em> \u2229 (K<\/em>\u2013U<\/em>)‘<\/em>\u03b3<\/sub> = {x<\/em>}. Therefore x<\/em> is isolated in (K<\/em>\u2013U<\/em>)‘<\/em>\u03b3<\/sub>, namely x<\/em> is in (K<\/em>\u2013U<\/em>)\u03b3<\/sub>.  This finishes the proof of item 2.<\/p>\n\n\n\n

    Item 3 follows from item 2 directly: we know that K’<\/em>\u03b1+1<\/sub> is empty, so by item 2, (K<\/em>\u2013U<\/em>)‘<\/em>\u03b1+1<\/sub> is empty as well.<\/p>\n\n\n\n

    As for item 4, we note that for every ordinal \u03b3, (K<\/em>\u2013U<\/em>)‘<\/em>\u03b3<\/sub> is included in (K<\/em>\u2013U<\/em>)‘<\/em>0<\/sub>=K<\/em>\u2013U<\/em>, since the sets (K<\/em>\u2013U<\/em>)‘<\/em>\u03b3<\/sub> are smaller and smaller as \u03b3 increases. In particular, no (K<\/em>\u2013U<\/em>)‘<\/em>\u03b3<\/sub> intersects U<\/em>. Taking \u03b3\u225d\u03b1, we obtain that (K<\/em>\u2013U<\/em>)‘<\/em>\u03b1<\/sub> is included in K’<\/em>\u03b1<\/sub>, namely in K<\/em>\u03b1<\/sub>, and does not intersect U<\/em>. Hence it is included in K<\/em>\u03b1<\/sub>\u2013U<\/em>. But that set is empty, so (K<\/em>\u2013U<\/em>)‘<\/em>\u03b1<\/sub> is empty.<\/p>\n\n\n\n

    It follows that the scattering height of K<\/em>\u2013U<\/em>, namely the least ordinal \u03b3 such that (K<\/em>\u2013U<\/em>)‘<\/em>\u03b3<\/sub> is empty, is less than or equal to \u03b1. It is therefore strictly smaller than the scattering height \u03b1+1 of K<\/em>. \u2610<\/p>\n\n\n\n

    Here is the game we will play. It involves two players, I and II. I will use a relatively informal, but traditional, way of describing it.<\/p>\n\n\n\n

    The game G(K).<\/strong> Given a topological space K<\/em>, we start the game G<\/em>(K<\/em>) with C<\/em>0<\/sub> \u225d K<\/em>. In round n<\/em>+1 (n<\/em> \u2208 N<\/strong>), we have a set Cn<\/sub><\/em> (we have just said what C<\/em>0<\/sub> is), and player I wins immediately if Cn<\/sub><\/em> is empty. Otherwise,<\/p>\n\n\n\n