A universal coregular countable second-countable space
aa r X i v : . [ m a t h . GN ] M a r A UNIVERSAL COREGULAR COUNTABLE SECOND-COUNTABLESPACE
TARAS BANAKH AND YARYNA STELMAKH
Abstract.
A Hausdorff topological space X is called superconnected (resp. coregular ) if forany nonempty open sets U , . . . U n ⊆ X , the intersection of their closures U ∩ · · · ∩ U n isnot empty (resp. the complement X \ ( U ∩ · · · ∩ U n ) is a regular topological space). Acanonical example of a coregular superconnected space is the projective space Q P ∞ of thetopological vector space Q <ω = { ( x n ) n ∈ ω ∈ Q ω : |{ n ∈ ω : x n = 0 }| < ω } over the field ofrationals Q . The space Q P ∞ is the quotient space of Q <ω \ { } ω by the equivalence relation x ∼ y iff Q · x = Q · y .We prove that every countable second-countable coregular space is homeomorphic to asubspace of Q P ∞ , and a topological space X is homeomorphic to Q P ∞ if and only if X iscountable, second-countable, and admits a decreasing sequence of closed sets ( X n ) n ∈ ω suchthat (i) X = X , T n ∈ ω X n = ∅ , (ii) for every n ∈ ω and a nonempty open set U ⊆ X n the closure U contains some set X m , and (iii) for every n ∈ ω the complement X \ X n is aregular topological space. Using this topological characterization of Q P ∞ we find topologicalcopies of the space Q P ∞ among quotient spaces, orbit spaces of group actions, and projectivespaces of topological vector spaces over countable topological fields. Introduction
A topological space X is called functionally Hausdorff if for any distinct points x, y ∈ X there exists a continuous function f : X → R such that f ( x ) = f ( y ). It is easy to seethat each countable functionally Hausdorff space is (totally) disconnected. On the otherhand, there are many examples of countable Hausdorff spaces which are connected and evensuperconnected. A topological space X is defined to be superconnected if for any nonemptyopen sets U , . . . , U n in X the intersection of their closures U ∩ · · · ∩ U n is not empty. It iseasy to see that continuous images of superconnected spaces remain superconnected.One of standard examples of a superconnected Hausdorff space is the famous Golombspace, introduced by Brown [6] and popularized by Golomb [11], [12]. The Golomb space isthe space N of positive integer numbers, endowed with the topology generated by the baseconsisting of the arithmetic progressions N ∩ ( a + b Z ) with relatively prime numbers a, b ∈ N .Using the Chinese remainder theorem, it can be shown (see [1]) that the closure of thearithmetic progression N ∩ ( a + b Z ) in the Golomb space contains the arithmetic progression p · · · p n N , where p , . . . , p n are prime divisors of b . This implies that the Golomb space issuperconnected. Then any continuous image of the Golomb space is superconnected as well.One of such images is the Kirch space [13], which is the space N endowed with the topologygenerated by the base consisting of the arithmetic progressions N ∩ ( a + b Z ) where the numbers Mathematics Subject Classification.
Key words and phrases.
Hausdorff countable connected space, the infinite rational projective space Q P ∞ ,superconnected space, coregular space, second-countable space, topologically homogeneous space, quotientspace, orbit space, projective space. a, b are relatively prime and b is not divided by a square of a prime number. The Kirch spaceis known to be superconnected and locally connected. In [2] and [3] it was shown that theGolomb and Kirch spaces are topologically rigid, i.e, have trivial homeomorphism group.Another natural example of a superconnected countable Hausdorff space is the infiniterational projective space Q P ∞ , which is the projective space of the topological vector space Q <ω = { ( x n ) n ∈ ω ∈ Q ω : |{ n ∈ ω : x n = 0 }| < ω } over the field Q of rational numbers. Here the countable power Q ω is endowed with theTychonoff product topology. The space Q P ∞ is defined as the quotient space of the space Q <ω ◦ = Q <ω \ { } ω by the equivalence relation x ∼ y iff x = λy for some nonzero rationalnumber λ . The superconnectedness of the rational projective space Q P ∞ was first noticed byGelfand and Fuks in their paper [10]. It is easy to show (see also Theorem 6) that the infiniterational projective space Q P ∞ is topologically homogeneous (i.e., for any points x, y ∈ Q P ∞ there exists a homeomorphism h of Q P ∞ such that h ( x ) = y ). So, Q P ∞ is not homeomorphicto the Golomb or Kirch space. Another important property that distinghished the space Q P ∞ from the Golomb and Kirch spaces is the coregularity of Q P ∞ .A topological space X is called coregular if X is Hausdorff and for any nonempty open sets U , . . . , U n ⊂ X the complement X \ ( U ∩ · · · ∩ U n ) is a regular topological space. We recallthat a topological space X is regular if it is Hausdorff and for every closed set F ⊂ X andpoint x ∈ X \ F there are disjoint open sets U, V in X such that x ∈ U and F ⊆ V . It iseasy to see that a topological space containing more than one point is regular if and only ifit is coregular and not superconnected. In Theorems 2 and 5 we shall prove that the space Q P ∞ is coregular and moreover, it is contains a topological copy of each coregular countablesecond-countable space. Let us recall that a topological space X is second-countable if it hasa countable base of the topology.Observe that for a coregular topological space X with a countable base of the topology { U n } n ∈ ω , the sequence ( X n ) n ∈ ω of the sets X n = U ∩· · ·∩ U n has the following two properties:(i) for any nonempty open set U ⊆ X the closure U contains some set X n , and (ii) for every n ∈ ω the complement X \ X n is a regular topological space. This property of the sequence( X n ) n ∈ ω motivates the following definition. Definition 1.
A sequence ( X n ) n ∈ ω of closed subsets of a topological space X is called • vanishing if X = X , T n ∈ ω X n = ∅ and X n +1 ⊆ X n for every n ∈ ω ; • a coregular skeleton for X if ( X n ) n ∈ ω is vanishing and has two properties: (i) for everynonempty open set U ⊆ X the closure U contains some set X n and (ii) for every n ∈ ω the complement X \ X n is a regular topological space; • a superconnecting skeleton for X if ( X n ) n ∈ ω is vanishing and for every nonempty openset U ⊆ X there exists n ∈ ω such that ∅ 6 = X n ⊆ U ; • an inductively superconnecting skeleton for X if ( X n ) n ∈ ω is vanishing and for every n ∈ ω and nonempty open set U ⊆ X n there exists m ∈ ω such that ∅ 6 = X m ⊆ U ; • superskeleton if it is both a coregular skeleton and an inductively superconnectingskeleton; • canonical superskeleton if ( X n ) n ∈ ω is a superskeleton and for every n ∈ ω the set X n +1 is nowhere dense in X n .It is clear that a topological space X is coregular if it has a coregular skeleton, and X issuperconnected of it has a superconnecting skeleton. UNIVERSAL COREGULAR COUNTABLE SECOND-COUNTABLE SPACE 3
The principal result of this paper is the following topological characterization of the infiniterational projective space Q P ∞ . Theorem 1.
A topological space is homeomorphic to the space Q P ∞ if and only if it iscountable, second-countable and possesses a superskeleton. The proof of Theorem 1 will be presented in Section 4. Since the proof is long and technical,we postpone it till the end of the paper, and first we apply Theorem 1 to finding topologicalcopies of the space Q P ∞ among quotient spaces of topological spaces by equivalence relationsand orbit spaces of group actions.2. Topological copies of the space Q P ∞ in “nature” Let E be an equivalence relation on a topological space X . A subset A ⊆ X is called E -saturated if A coincides with its E -saturation EA = S x ∈ A Ex where Ex = { y ∈ X : ( x, y ) ∈ E } is the equivalence class of a point x ∈ X . Let X/E be the space of E -equivalence classes { Ex : x ∈ X } and q : X → X/E be the map assigning to each point x ∈ X its equivalenceclass Ex ∈ X/E . The space
X/E carries the quotient topology, consisting of all subsets U ⊆ X/E whose preimage q − ( U ) is open in X . Proposition 1.
Let E ⊆ X × X and equivalence relation on a topological space X satisfyingthe following conditions: (1) the set E is closed in X × X ; (2) for any open set U ⊆ X its E -saturation EU is open in X ; (3) X admits a vanishing sequence ( X n ) n ∈ ω of non-empty E -saturated closed sets suchthat (a) for any n ∈ ω and nonempty E -saturated open set U ⊆ X n , the closure U containssome set X m ; (b) for any n ∈ ω , E -saturated open set U ⊆ X and x ∈ U , there exists an open E -saturated neighborhood V ⊆ X of x such that V ⊆ U ∪ X n .Then the quotient space X/E possesses a superskeleton. If the space X is first-countable and X/E is countable, then
X/E is homeomorphic to Q P ∞ .Proof. Let Y be the quotient space X/E . The condition (2) implies that the quotient map q : X → Y is open. Then for any E -saturated closed set A ⊆ X its image q ( A ) = Y \ q ( X \ A )is closed in Y . In particular, for every n ∈ ω the image Y n = q ( X n ) is a closed subset of Y and hence ( Y n ) n ∈ ω is a vanishing sequence of nonempty closed sets in Y .For any n ∈ ω and nonempty open set U ⊆ Y n the preimage q − ( U ) is an open E -saturated set in X n and by condition (3a), the closure q − ( U ) contains some set X m . Then Y m = q ( X m ) ⊆ q ( q − ( U )) ⊆ q ( q − ( U )) = U . This means that ( Y n ) n ∈ ω is an inductivelysuperconnecting skeleton for the space Y . By [8, 2.4.C], the condition (1,2) imply that thethe quotient space Y = X/E is Hausdorff. Now the condition (3a),(3b) ensure that theskeleton ( Y n ) n ∈ ω is coregular and hence ( Y n ) n ∈ ω is a superskeleton for X/E .If the space X is first-countable and X/E is countable, then the openness of the quotientmap q : X → X/E implies the first-countability of the the quotient space
X/E . Beingcountable and first-countable, the space
X/E is second-countable. By Theorem 1, the space
X/E is homeomorphic to Q P ∞ . (cid:3) TARAS BANAKH AND YARYNA STELMAKH
Next, we consider orbit spaces of group actions, which are special examples of quotientspaces. By a group act we understand a topological space X endowed with an action α : G × X → X a group G . The action α satisfies the following axioms: • for every g ∈ G the map α ( g, · ) : X → X , α ( g, · ) : x gx := α ( g, x ), is a homeomor-phism of X ; • for the identity 1 G of the group G and every x ∈ X we have 1 G x = x ; • ( gh ) x = g ( hx ) for all g, h ∈ G and x ∈ X .In this case we also say that X is a G -space . We say that a G -space X has closed orbits iffor any point x ∈ X its orbit Gx = { gx : g ∈ G } is a closed subset of X . A subset A ⊆ X iscalled G -invariant if it coincides with its G -saturation GA = S x ∈ A Gx . The action of G on X induces the equivalence relation E = { ( x, gx ) : x ∈ X, g ∈ G } . The quotient space X/E by this equivalence relation is called the orbit space of the G -space and is denoted by X/G . Proposition 2.
Let X be a G -space with closed G -orbits, possessing a vanishing sequence ( X n ) n ∈ ω of nonempty G -invariant closed subsets such that (1) for any n ∈ ω and nonempty open G -invariant set U ⊆ X n , the closure U containssome set X m ; (2) for any n ∈ ω , point x ∈ X \ X n , and open G -invariant neighborhood U ⊆ X of x ∈ U ,there exists an open G -invariant neighborhood V ⊆ X of x such that V ⊆ U ∪ X n .Then the orbit space X/G has a superskeleton. If X is first-countable and X/G is countable,then the space
X/G is homeomorphic to Q P ∞ .Proof. The closedness of orbits implies that the orbit space Y = X/G is a T -space. Sinceeach open set U ⊆ X has open G -saturation GU = S g ∈ G gU , the quotient map q : X → X/G is open. Then for every n ∈ ω , the image Y n = q ( X n ) of the closed G -invariant set X n isa closed subset of Y and hence ( Y n ) n ∈ ω is a vanishing sequence of nonempty closed sets in Y . The condition (2) implies that for every n ∈ ω , the open subspace Y \ Y n of the T -space Y is regular and hence Hausdorff. Since T n ∈ ω Y n = ∅ , the space Y is Hausdorff and theequivalence relation E = { ( x, gx ) : x ∈ X, g ∈ G } = { ( x, y ) ∈ X × X : q ( x ) = q ( y ) } is closed in X × X . Now we can apply Proposition 1 and conclude that ( Y n ) n ∈ ω is a superskele-ton in the space X/G (and the space
X/G is homeomorphic to Q P ∞ if X is first-countableand X/G is countable). (cid:3)
Now we find topological copies of the space Q P ∞ among infinite projective spaces of singular G -spaces. Definition 2.
A topological space X endowed with a continuous action α : G × X → X ofa Hausdorff topological group G is called singular if it has the following properties:(i) the topological space X is regular and infinite;(ii) the set Fix G ( X ) = { x ∈ X : Gx = { x }} is a singleton;(iii) for every x ∈ X \ Fix G ( X ) the map α x : G → X , α x : g gx = α ( g, x ), is injectiveand open;(iv) the orbit Gx of every point x ∈ X \ Fix G ( X ) contains the singleton Fix G ( X ) in itsclosure Gx ; UNIVERSAL COREGULAR COUNTABLE SECOND-COUNTABLE SPACE 5 (v) for any points x ∈ X \ Fix G ( X ) and y ∈ X , there exists a neighborhood U ⊆ X of y such that for any neighborhood W ⊆ X of the singleton Fix G ( X ), there exists aneighborhood V ⊆ X of Fix G ( X ) such that α u ( α − x ( V )) ⊆ W for every u ∈ U . Example 1.
There are many natural examples of singular G -spaces:(1) The complex plane C endowed with the action of the multiplicative group C ∗ of non-zero complex numbers.(2) Any subfield F ⊆ C endowed with the action of the multiplicative group F ∗ = F \ { } .(3) Ahe real line R endowed with the action of the multiplicative group R ∗ of non-zeroreal numbers.(4) The real line R endowed with the action of the multiplicative group R + of positivereal numbers.(5) The closed half-line R + = [0 , ∞ ) endowed with the action of the multiplicative group R + .(6) The one-point compactification R of the space R of real numbers endowed with thenatural action of the additive group R .(7) The space Q of rationals, endowed with the action of the multiplicative group Q ∗ ofnon-zero rational numbers.(8) The space Q of rationals, endowed with the action of the multiplicative group Q + ofpositive rational numbers.(9) The one-point compactification Z = Z ∪ { + ∞} of the discrete space Z endowed withthe natural action of the additive group Z of integer numbers.(10) The one-point compactification of any non-compact locally compact topological group G , endowed with the natural action of the topological group G .Given a singular G -space X , consider the G -space X ω endowed with the Tychonoff producttopology and the coordinatewise action of the group G . Let s be the unique point of thesingleton Fix( X ; G ). We shall be interested in two special subspaces of X ω : X <ω := { x ∈ X ω : |{ n ∈ ω : x ( n ) = s }| < ω } and X <ω ◦ := X <ω \ { s } ω . The orbit space X <ω ◦ /G is called the infinite projective space of the singular G -space X andis denoted by X P ∞ .If X = F is a non-discrete topological field endowed with the action of its multiplicativegroup F ∗ , then F <ω is a topological vector space over the field F and F P ∞ is the projectivespace of the topological vector space F <ω in the standard sense. In particular, Q P ∞ is theprojective space of the topological vector space Q <ω over the topological field Q of rationalnumbers. Theorem 2.
The infinite projective space X P ∞ of any singular G -space X possesses a canon-ical superskeleton. If the singular G space X is countable and metrizable, then its infiniteprojective space X P ∞ is homeomorphic to the space Q P ∞ .Proof. Let s be the unique point of the singleton Fix G ( X ). It will be convenient to identifyelements x ∈ X <ω with sequences ( x n ) n ∈ ω .Since the group G acts by homeomorphisms on the space X <ω ◦ , for every open set U ⊆ X <ω ◦ the set GU = S g ∈ G gU is open. This implies that the quotient map q : X <ω ◦ → X P ∞ is open.Now using Proposition 2, we shall prove that X P ∞ possesses a canonical superskeleton.For every n ∈ ω consider the G -invariant subspace X n = { x ∈ X <ω ◦ : ∀ i ∈ n x i = s } TARAS BANAKH AND YARYNA STELMAKH of X <ω ◦ . Observe that X n = pr − n ( { s } n ) where pr n : X <ω ◦ → X n , pr n : x ( x , . . . , x n − ), isthe natural projection.By Definition 2(iii), for every x ∈ X \ { s } its orbit Gx is an open set in X . Consequently,the singleton { s } = X \ S x ∈ X \{ s } Gx is closed in X and { s } n is closed in X n , which impliesthat X n is closed in X .Taking into account the openness of the quotient map q : X <ω ◦ → Y and the G -invariantnessof the closed sets X n , we conclude that for every n ∈ ω the set Y n = q ( X n ) is closed in X P ∞ .Therefire, ( Y n ) n ∈ ω is a vanishing sequence of nonempty closed sets in the space X P ∞ . Sincethe singleton { s } is nowhere dense in X (by Definition 2(iv)), for every n ∈ ω the set X n +1 is nowhere dense in X n and then the set Y n +1 is nowhere dense in Y n . Claim 1.
For any point x ∈ X <ω ◦ its orbit Gx is closed in X <ω ◦ .Proof. Given any point y ∈ X <ω ◦ \ Gx , we should find an open neighborhood V of y in X <ω ◦ such that V ∩ Gx = ∅ . Since y ∈ X <ω ◦ , the set Ω = { n ∈ ω : y n = s } is finite and nonempty.If x n / ∈ Gy n for some n ∈ Ω, then Gx n ∩ Gy n = ∅ and by Definition 2(iii), Gy n is an openneighborhood of y n in X and hence V = { v ∈ X <ω ◦ : v n ∈ Gy n } is an open neighborhood of y that is disjoint with the orbit Gx of x . So, we assume that for every n ∈ Ω, the point x n belongs to the orbit Gy n and hence x n = g n y n for some g n ∈ G .If g n = g k for some numbers n, k ∈ Ω, then we can choose an open neighborhood W ⊆ G of the identity in the Hausdorff topological group G such that W g − n ∩ W g − k = ∅ . ByDefinition 2(iii), the sets W y n and W y k are open neighborhoods of y n and y k , respectively.We claim that the open neighborhood V = { v ∈ X <ω ◦ : v n ∈ W y n , v k ∈ W y k } of y doesnot intersect the orbit Gx of x . In the opposite case we can find an element g ∈ G suchthat gx n ∈ W y n and gx k ∈ W y k . Then gg n y n = gx n ∈ W y n and gg k y k = gx k ∈ W y k .The injectivity of the maps α y n and α y k guarantees that gg n ∈ W and gg k ∈ W . Then g n ∈ g − W ⊆ g k W − W and finally, g n W − ∩ g k W − = ∅ , which contradicts the choice of theneighborhood W . This contradiction shows that V ∩ Gx = ∅ .Finally, assume that g n = g k for all n, k ∈ Ω. Fix any number n ∈ Ω. Since y / ∈ Gx , thereexists m ∈ ω \ Ω such that y m = s = g n x m . Then g − n x m = s = y m and by the Hausdorffproperty of X , we can find an open neighborhood W m ⊆ X of y m = s such that g − n x m / ∈ W m .By the continuity of the map α x m : G → X , α x m : g gx m , the set F = α − x m ( W m ) is closedin G and does not contain g − n . Find an open neighborhood U ⊆ G of the identity such that F ∩ U g − n = ∅ . We claim that the open neighborhood V = { v ∈ X <ω ◦ : v n ∈ U y n , v m ∈ W m } of y does not intersect the orbit Gx . In the opposite case we can find an element g ∈ G suchthat gx n ∈ U y n and gx m ∈ W m . Then gg n y n = g n x n ∈ U y n and the injectivity of the map α y n implies that gg n ∈ U . Also the inclusion gx m ∈ W m implies g ∈ F . Then gg n ∈ U ∩ F g n ,which contradicts the choice of the neighborhood U . (cid:3) Claim 2.
For every n ∈ ω , the closure U of any nonempty G -invariant open set U ⊆ X n contains some space X m .Proof. Fix any point x ∈ U . For every m > n , identify the ordinal m with the set { , . . . , m − } and consider the projection π m : X m → X m \ n , π m : x ( x n , . . . , x m − ). By the definitionof the Tychonoff product topology on X n , there exists m ≥ n and an open neighborhood V ⊆ X m \ n of π m ( x ) such that π − m ( V ) ⊆ U . Then U = GU ⊇ G · π − m ( V ) = π − m ( GV ).We claim that { s } m \ n ⊂ GV . Given any open set W ⊆ X m \ n that contains the singleton { s } m \ n , find an open neighborhood W s ⊆ X of s such that W m \ ns ⊆ W . Take any point UNIVERSAL COREGULAR COUNTABLE SECOND-COUNTABLE SPACE 7 z ∈ X \ { s } . By Definition 2(v), for every i ∈ m \ n , there exists a neighborhood W i ⊆ X of s such that α x i ( α − z ( W i )) ⊆ W s . Definition 2(iv) ensures that the intersection Gz ∩ T i ∈ m \ n W i contains some point w = s . Then the element g = α − z ( w ) is well-defined and gx i = α x i ( g ) ∈ α x i ( α − z ( W i )) ⊆ W s . The point ( gx i ) i ∈ m \ n belongs to the intersection W m \ ns ∩ Q i ∈ m \ n gV i ⊆ W ∩ gV ⊆ W ∩ GV , witnessing that { s } m \ n ⊂ GV .Since the projection π m : X n → X m \ n is an open G -equivariant map, X m = π − m ( { s } m \ n ) ⊆ π − m ( GV ) ⊆ π − m ( GV ) ⊆ U . (cid:3)
Claim 3.
For any n ∈ ω , point x ∈ X \ X n , and G -invariant open neighborhood U ⊆ X <ω ◦ of x , there exists a G -invariant open neighborhood V ⊆ U of x such that V ⊆ U ∪ X n .Proof. By the definition of the space X <ω ◦ ∋ x , there exists an index i ∈ ω such that x i = s .By the definition of the Tychonoff product topology on the regular topological space X <ω ◦ ,there exist m > max { n, i } and an open set W ⊆ X m such that x ∈ pr − m ( W ) ⊆ pr − m ( W ) ⊆ U where pr m : X <ω ◦ → X m , pr m : y ( y , . . . , y m − ), is the projection onto the first m coordinates. By Definition 2(v), for every j ∈ m the point x j has a neighborhood W j ⊆ X such that for any neighborhood O ⊆ X of s there exists a neighborhood O ′ of s such that α u ( α − x i ( O ′ )) ⊆ O for every u ∈ W j . Replacing W j by smaller neighborhoods, we can assumethat Q j ∈ m W j ⊆ W .Now consider the “hyperplane” H = { y ∈ X m : y i = x i } in X m , the open set V i = { ( y j ) j ∈ m ∈ X m : y i ∈ Gx i } in X m , and the continuous map r : V i → H, r : ( y j ) j ∈ m (cid:0) ( α − x i ( y i )) − · y j (cid:1) j ∈ m . The map r assigns to each y ∈ V i the unique point of the intersection Gy ∩ H . The continuityof the map r follows from the continuity of the action α , the openness and injectivity ofthe map α x i , and the continuity of the inversion in the topological groups G . The preimage V m := r − ( Q j ∈ m W j ) is an open G -invariant set in X m \{ s } m and the preimage V := pr − m ( V m )is an open G -invariant neighborhood of x in X The skeleton ( Y n ) n ∈ ω contructed in the proof of Theorem 2 will be called thecanonical superskeleton of the space X P ∞ .Let F be a topological field. Three elements F ∗ x, F ∗ y, F ∗ z of the projective space F P ∞ are called collinear if the union F ∗ x ∪ F ∗ y ∪ F ∗ z is contained in some 2-dimensional vectorsubspace of F <ω .For two topological fileds F , F a map f : F P ∞ → F P ∞ is called affine if for anycollinear elements F ∗ x, F ∗ y, F ∗ z ∈ F P ∞ , the elements f ( F ∗ x ) , f ( F ∗ y ) , f ( F ∗ z ) are collinear inthe projective space F ∗ P ∞ . A bijective map f : F P ∞ → F P ∞ is called an affine isomorphism if both maps f and f − are affine. If an affine isomorphism f : F P ∞ → F P ∞ is also ahomeomorphism, then f is called an affine topological isomorphism . The projective spaces F P ∞ , F P ∞ are called affinely isomorphic (resp. affinely homeomorphic ) if there exists anaffine topological ismorphism f : F P ∞ → F P ∞ .In spite of the fact that for any countable subfields F , F ⊆ C , the infinite projective spaces F P ∞ and F P ∞ are homeomorphic (by Theorem 2), we have the following rigidity result foraffine isomorphisms between infinite projective spaces. Proposition 3. Two (topological) fields F , F are (topologically) isomorphic if and only iftheir infinite projective spaces F P ∞ , F P ∞ are affinely isomorphic (affinely homeomorphic).Proof. If σ : F → F is a (topological) isomorphism of the (topological) fields F , F , thenthe map f induces the affine (topological) isomorphism˜ σ : F P ∞ → F P ∞ , ˜ σ : F ∗ · ( x n ) n ∈ ω F ∗ · ( σ ( x n )) n ∈ ω , of the projective spaces F P ∞ and F P ∞ . This proves the “only if” part of the proposition.To prove the “if” part, assume that g : F P ∞ → F P ∞ is an affine (topological) isomorphismbetween the projective spaces F P ∞ and F P ∞ . Identify the space F with the 3-dimensionalvector subspace { ( x n ) n ∈ ω ∈ F <ω : ∀ n ≥ x n = 0) } of the topological vector space F <ω .Consider the vectors e = (1 , , e = (0 , , 0) and e = (0 , , 1) in F , and observe that theelements F ∗ e , F ∗ e , F ∗ e are not collinear in the projective space F P ∞ . Then their images g ( F ∗ e ) , g ( F ∗ e ) , g ( F ∗ e ) are not collinear in the projective space F P ∞ and hence the union g ( F ∗ e ) ∪ g ( F ∗ e ) ∪ g ( F ∗ e ) is contained in a unique 3-dimensional vector subspace L of thetopological vector space F <ω . We can choose a basis e ′ ∈ g ( F ∗ e ), e ′ = g ( F ∗ e ), e ′ ∈ g ( F ∗ e )for the space L such that F ∗ ( e ′ + e ′ + e ′ ) = g ( F ∗ ( e + e + e )). Then the affine (topological)isomorphism g induces an affine (topological) isomorphism h = g ↾ F P of the projective planes F P = ( F \ { } ) / F ∗ and F P = ( F \ { } ) / F ∗ such that h ( F ∗ e i ) = F ∗ e ′ i for i ∈ { , , } UNIVERSAL COREGULAR COUNTABLE SECOND-COUNTABLE SPACE 9 and h ( F ∗ ( e + e + e )) = F ∗ ( e ′ + e ′ + e ′ ). By a classical result of Hilbert (cf. Proposition3.11 in [9] or Lemma 2.8.2 in [14]), there exists an isomorphism σ : F → F of the fields F , F such that for any ( x , x , x ) ∈ F we have h ( F ∗ ( x e + x e + x e )) = F ∗ ( σ ( x ) e ′ + σ ( x ) e ′ + σ ( x ) e ′ ) . Therefore, the fields F , F are isomorphic.If the affine isomorphism g is a homeomorphism, then so is the map h . In this case weshall prove that the field isomorphism σ is a homeomorphism. For this observe that for every i ∈ { , } , the map f i : F i → F i P ∞ , f : x F ∗ i ( e + xe ) is a topological embedding. Since σ = f − ◦ h ◦ f and σ − = f − ◦ h − ◦ f , then maps σ , σ − are continuous and σ : F → F is a topological isomorphism of the topological fields F , F . (cid:3) By Example 1(2,5), the spaces C , R , ¯ R + endowed with suitable group actions are singular G -spaces. By Theorem 2, the infinite projective spaces C P ∞ , R P ∞ , R + P ∞ possess (canonical)superskeleta. It can be shown that each of these spaces has a countable base of the topologyconsisting of sets, homeomorphic to the space R <ω , so is a (non-metrizable) R <ω -manifold.It can be shown that the R <ω -manifolds C P ∞ , R P ∞ , R + P ∞ are pairwise non-homeomorphic(because of different homotopical properties of complements Y \ Y n of their canonical skeleta).The distinguishing topological property of the space R + P ∞ is possessing a superskeleton( Y n ) n ∈ ω such that for every n < m in ω the complement Y n \ Y m is contractible.This observation and the topological characterization of the space Q P ∞ suggests the fol-lowing topological characterization of the space R + P ∞ . Conjecture 1. A Hausdorff topological space X is homeomorphic to R + P ∞ if and only if X possesses a superskeleton ( X n ) n ∈ ω such that for every n < m in ω the set X m is a Z -set in X n and the space X n \ X m is homeomorphic to R <ω . A closed subset A of a topological space X is called a Z -set in X if the set C ( I ω , X \ A )is dense in the space C ( I ω , X ) of continuous functions from the Hilbert cube I ω = [0 , ω to X , endowed with the compact-open topology. For more information on Infinite-DimensionalTopology, see the monographs [4], [15], [5], [17]. For the topological characterization of thespace R <ω , see [16], [7], [5, § § R P ∞ , C P ∞ , R + P ∞ contain dense subspaces, homeomorphicto Q P ∞ . Problem 1. Does the Golomb (or Kirch) space contain a subspace homeomorphic to Q P ∞ ? Some properties of skeleta in topological spaces In this section we establish some properties of various skeleta in topological spaces. Firstwe fix some standard notations.For a subset A of a topological space X by ¯ A and ∂A we denote the closure and boundary of A in X . By ω we denote the smallest infinite ordinal, and by N the set ω \{ } of positive integernumbers. Ordinals are identified with the sets of smaller ordinals. So, n = { , . . . , n − } forany natural number n ∈ ω . Lemma 1. A (Hausdorff second-countable) topological space is coregular if (and only if ) ithas a coregular skeleton.Proof. To prove the “if” part, assume that a topological space X has a coregular skeleton( X n ) n ∈ ω . First we show that the space X is Hausdorff. Fix any distinct points x, y ∈ X . By Definition 1, ( X n ) n ∈ ω is a vanishing sequence of closed sets in X . Consequently, T n ∈ ω X n and we can find a number m ∈ ω such that x, y ∈ X \ X m . Since ( X n ) n ∈ ω is a coregularskeleton, the space X \ X m is regular and hence Hausdorff. Then the points x, y have disjointopen neighborhoods O x , O y in the space X \ X m . Since X \ X m is open in X , the sets O x , O y remain open in X , witnessing that X is Hausdorff.Now fix any non-empty open sets U , . . . , U k ⊆ X . By Definition 1, for every i ∈ { , . . . , k } there exists a number n i ∈ ω such that X n i ⊆ U i . Then for the number n = max i ≤ k n i wehave X n ⊆ n \ i =1 X n i ⊆ k \ i =1 U i . Since the skeleton ( X i ) i ∈ ω is coregular, the space X \ X n is regular and so is its subspace X \ ( U ∩ · · · ∩ U k ). This completes the proof of the coregularity of X .To prove the “only if” part, assume that the space X Hausdorff, second-countable, andcoregular. If X is regular, then put X = X , X n = ∅ for all n ∈ N , and observe that ( X n ) n ∈ ω is a coregular skeleton for X . So, we assume that X is not regular and hence infinite.Fix a countable base { U n } n ∈ N of the topology of X such that U n = ∅ for all n ∈ N . Let X = X and X n = U ∩ · · · ∩ U n for every n ∈ N .To show that T n ∈ ω X n = ∅ , fix any point x ∈ X . Since the space X is infinite andHausdorff, there exists n ∈ ω such that x / ∈ U n and hence x / ∈ X n . This shows that thesequence ( X n ) n ∈ ω is vanishing. By the coregularity of X , for every n ∈ ω the space X \ X n is regular. Also observe that every nonempty open set U ⊆ X contains some set U n and then X n ⊆ U n ⊆ U . Therefore, the vanishing sequence ( X n ) n ∈ ω is a coregular skeleton for thespace X . (cid:3) Lemma 2. A (Hausdorff infinite second-countable) topological space is superconnected if (andonly if ) it possesses a superconnecting skeleton.Proof. To prove the “if” part, assume that a topological space X has a superconnectingskeleton ( X n ) n ∈ ω . By Definition 1, ( X n ) n ∈ ω is a vanishing sequence in X . To see that X is superconnected, fix any non-empty open sets U , . . . , U k ⊆ X . By Definition 1, for every i ∈ { , . . . , k } there exists a number n i ∈ ω such that ∅ 6 = X n i ⊆ U i . Then for the number n = max i ≤ k n i we have ∅ 6 = X n ⊆ n \ i =1 X n i ⊆ k \ i =1 U i , witnessing that U ∩ · · · ∩ U k = ∅ and X is superconnected.To prove the “only if” part, assume that an infinite Hausdorff second-countable space X is superconnected and fix a countable base { U n } n ∈ N of the topology of X such that U n = ∅ for all n ∈ N . Let X = X . By the superconnectedness of X , for every n ∈ N the closed set X n = U ∩ · · · ∩ U n is not empty.To show that T n ∈ ω X n = ∅ , fix any point x ∈ X . Since the space X is infinite andHausdorff, there exists n ∈ ω such that x / ∈ U n and hence x / ∈ X n . This shows that thesequence ( X n ) n ∈ ω is vanishing.To see that ( X n ) n ∈ ω is a superconnecting skeleton for X , take any nonempty open set U ⊆ X and find n ∈ ω such that U n ⊆ U . Then ∅ 6 = X n = U ∩ · · · ∩ U n ⊆ U n ⊆ U . UNIVERSAL COREGULAR COUNTABLE SECOND-COUNTABLE SPACE 11 (cid:3) Lemma 3. Let X be an infinite Hausdorff topological space and ( X n ) n ∈ ω be its superconnect-ing skeleton. Then the space X is crowded and for every n ∈ ω , the space X n is infinite.Proof. First we show that for every n ∈ ω the set X n is infinite. To derive a contradiction,assume that for some n ∈ ω the space X n is finite. We can assume that n is the smallestnumber with this property. Then for every i ∈ n the space X i is infinite and hence containssome point x i . Since X is infinite and Hausdorff, there exists a nonempty set U ⊆ X whose closure U does not intersect the finite set X n ∪ { x i } i ∈ n . Since the skeleton ( X k ) k ∈ ω issuperconnecting, the closure U contains some set X i = ∅ . Assuming that i ≥ n we conclude ∅ 6 = X i = X i ∩ U ⊆ X n ∩ U = ∅ , which is a contradiction. So, i < n and then x i ∈ X i ⊆ U ,which contradicts the choice of U . This contradiction witnesses that the spaces X n are infinite.Assuming that the space X is not crowded, we can find an isolated point x in X . Then U = { x } is an open subspace of X such that U = { x } by the Hausdorff property of X . Since( X n ) n ∈ ω is a superconnecting skeleton of X , there exist n ∈ ω such that X n ⊆ U = { x } andhence X n is finite, which is a desired contradiction. (cid:3) Lemma 4. Let X be an infinite topological space and ( X n ) n ∈ ω be its superskeleton. Then forevery n ∈ ω the space X n is crowded.Proof. By Definition 1, for every n ∈ ω the sequence ( X m ) ∞ m = n is a superconnecting coregularskeleton for the space X n . Then the space X = X is coregular and hence Hausdorff. ByLemma 3, for every n ∈ ω the set X n is infinite. Applying Lemma 3 to the superconnectingskeleton ( X m ) ∞ m = n for the infinite Hausdorff space X n , we conclude that X n croweded. (cid:3) Lemma 5. Let X be an infinite Hausdorff space and ( X n ) n ∈ ω be its superconnecting orcoregular skeleton. Then for some n ∈ ω the set X n is nowhere dense in X = X .Proof. Being infinite and Hausdorff, the space X contains two disjoint nonempty open sets U, V . By Definition 1, the closures U , V contain some set X n . Then the set X n ⊆ U ∩ V ⊆ U ∩ X \ U = ∂U is nowhere dense in X . (cid:3) Lemma 6. If ( X n ) n ∈ ω is a superskeleton for an infinite topological space X , then for someincreasing number sequence ( n k ) k ∈ ω the sequence ( X n k ) k ∈ ω is a canonical superskeleton for X .Proof. Applying Lemma 5, construct inductively an increasing number sequence ( n k ) n ∈ ω suchthat n = 0 and for every k ∈ ω the set X n n +1 is nowhere dense in X n k . (cid:3) A subset of a topological space is called regular open if it coincides with the interior of itsclosure. A topological space X is called semiregular if it is Hausdorff and has a base consistingof regular open sets. Lemma 7. Each coregular topological space X is semiregular.Proof. If the space X is finite, then it is discrete (being Hausdorff) and hence regular andsemiregular. So, we assume that X is infinite.To show that X is semiregular, fix any point x ∈ X and an open neigborhood U of x in X . Taking into account that X is infinite and Hausdorff, we can replace U by a smallerneighborhood of x and assume that X contains a non-empty open set W , which is disjointwith U . Then U ∩ W = ∅ . Since X is coregular, the space X \ W is regular. Then the point x has an open neighborhood V ⊆ X \ W such that V ⊆ U and V ∩ ( X \ W ) ⊆ U . Let O be the interior of the set V in X . Observe that O ∩ W ⊆ V ∩ W ⊆ U ∩ W = ∅ and hence O ∩ W = ∅ . Then O ⊆ V \ W ⊆ U . Taking into account that the set O is regular open, weconclude that the space X is semiregular. (cid:3) Lemma 8. A topological space X is regular if and only if its square X × X is coregular.Proof. The “only if” part is trivial. To prove the “if” part, assume that the space X × X iscoregular. Then X × X is Hausdorff and so is the space X . If X is finite, then X is discreteand hence regular. So, assume that X is infinite. Then we can fix any point x ∈ X and finda non-empty open set U ⊆ X such that x / ∈ U . By the coregularity of X × X the compement X × X \ U × U is a regular space and so is its subspace { x } × X and the space X . (cid:3) Main Results In this section we prove a difficult Theorem 3 implying Theorem 1 and many other im-portant properties of the space Q P ∞ . Let us recall that a function f : X → Y betweentopological spaces X, Y is called a topological embedding if f is a homeomorphism between X and the subspace f ( X ) of Y . Theorem 3. Let X be a countable second-countable space, ( X n ) n ∈ ω be a coregular skeleton in X , and A be a nowhere dense closed set in X . Let Y be a countable second-countable space, ( Y n ) n ∈ ω be a canonical superskeleton in Y , and B be a subset of Y such that for every n ∈ ω the intersection ¯ B ∩ Y n is nowhere dense in Y n . Let f : A → B be a homeomorphism suchthat f − ( Y n ) = A ∩ X n for all n ∈ ω . Then there exist a topological embedding ¯ f : X → Y such that ¯ f ↾ A = f and ¯ f − ( Y n ) = X n for all n ∈ ω . If the sequence ( X n ) n ∈ ω is a canonicalsuperskeleton in X , the set B is closed in Y , and for every n ∈ ω the set A ∩ X n is nowheredense in X n , then ¯ f ( X ) = Y and ¯ f is a homeomorphism.Proof. For constructing the topological embedding ¯ f : X → Y we should make some prelim-inary work with the spaces X and Y .We start with the space X endowed with a coregular skeleton ( X n ) n ∈ ω . Let ℓ X : X → ω be the function assigning to each x ∈ X the largest number n such that x ∈ X n (such thenumber n exists because T i ∈ ω X i = ∅ ). The function ℓ X will be called the level map of X .Denote by τ X the topology of X . Using the first-countability of X , for each point x ∈ X ,fix a neighborhood base { O Xn ( x ) } n ∈ ω at x such that O Xn +1 ( x ) ⊆ O Xn ( x ) ⊆ X \ X ℓ X ( x ) forevery n ∈ ω . Fix a well-order (cid:22) X on the set X ′ := X \ A such that for any x ∈ X ′ the set ↓ x := { z ∈ X ′ : z (cid:22) X x } is finite. For a nonempty subset S ⊆ X ′ by min S we denote thesmallest element of the set S with respect to the well-order (cid:22) X .Next, do the same for the space Y endowed with a canonical superskeleton ( Y n ) n ∈ ω . Denoteby τ Y the topology of Y . Let ℓ Y : Y → ω be the function assigning to each y ∈ Y the largestnumber n such that x ∈ Y n . Using the first countability of Y , for every point y ∈ Y choosea neighborhood base { O Yn ( y ) } n ∈ ω at y such that O Yn +1 ( y ) ⊆ O Yn ( y ) ⊆ Y \ Y ℓ Y ( y ) for all n ∈ ω . Fix a well-order (cid:22) Y on the set Y ′ = Y \ B such that for any y ∈ Y the set ↓ y := { z ∈ Y : z (cid:22) Y y } is finite.If the set X ′ = X \ A is finite, then let ¯ f : X → Y be any injective function such that¯ f ↾ A = f and f ( x ) ∈ ℓ − Y ( ℓ X ( x )) \ ¯ B for any x ∈ X ′ . The choice of ¯ f is possible since for every n ∈ ω the sets ¯ B ∩ Y n and Y n +1 are nowhere dense in Y n . Then ¯ f is a required extension of f . UNIVERSAL COREGULAR COUNTABLE SECOND-COUNTABLE SPACE 13 So, we assume that the open subspace X ′ = X \ A of X is infinite. In this case we shallconstruct the topological embedding ¯ f by induction over the index set Γ = ω ∪ ( ω × ω )endowed with the strict well-order ≺ uniquely defined by the following conditions:a) for numbers n, m ∈ ω we have n ≺ m iff n < m ;b) for a number n ∈ ω and a pair ( i, j ) ∈ ω × ω we have ( i, j ) ≺ n iff i + j < n ;c) for a pair ( i, j ) ∈ ω × ω and a number n ∈ ω we have n ≺ ( i, j ) iff n ≤ i + j ;d) for two pairs ( i, j ) , ( n, m ) ∈ ω × ω we have ( i, j ) ≺ ( n, m ) iff either i + j < n + m or i + j = n + m and i < n .The initial elements of the well-ordered set Γ are:0 , (0 , , , (1 , , (0 , , , (2 , , (1 , , (0 , , , (3 , , (2 , , (1 , , (0 , , , . . . For every element γ ∈ Γ let ↓ γ = { α ∈ Γ : α ≺ γ } . Writing k ∈ ↓ γ (resp. ( i, j ) ∈ ↓ γ ) we shallunderstand that k ∈ ω ∩ ↓ γ (resp. ( i, j ) ∈ ( ω × ω ) ∩ ↓ γ ).Write the set ω as the union Ω ∪−→ Ω ∪←− Ω of pairwise disjoint sets Ω , −→ Ω , ←− Ω such that | Ω | = | A | , |−→ Ω | = | X ′ | = ω , 0 ∈ −→ Ω , and |←− Ω | ∈ { , ω } . We choose the set ←− Ω to be infinite iff the skeleton( X n ) n ∈ ω is a canonical superskeleton for X , the set B is closed in Y , and for every n ∈ ω theset A ∩ X n is nowhere dense in X n . Let ξ : Ω → A be a bijective function.Now we are ready to start the inductive construction of the topological embedding ¯ f : X → Y extending the homeomorphism f .Inductively we shall construct sequences of points { x n } n ∈ ω ⊆ X , { y n } n ∈ ω ⊆ Y , a doublesequences of open sets { U n,k } n,k ∈ ω ⊆ τ X , { V n,k } k,n ∈ ω ⊆ τ Y , and a function ℓ : Γ → ω suchthat for any γ ∈ Γ the following conditions are satisfied:(1) If γ = n for some number n ∈ ω , then(1a) ℓ ( γ ) = ℓ X ( x n ) = ℓ Y ( y n );(1b) x n / ∈ { x k } k ∈↓ γ and y n / ∈ { y k } k ∈↓ γ ;(1c) { ( i, j ) ∈ ↓ γ : x n ∈ U i,j } = { ( i, j ) ∈ ↓ γ : y n ∈ V i,j } ;(1d) { ( i, j ) ∈ ↓ γ : x n ∈ U i,j } = { ( i, j ) ∈ ↓ γ : y n ∈ V i,j } ;(1e) If n ∈ Ω, then x n = ξ ( n ) and y n = f ( x n );(1f) If n ∈ −→ Ω , then x n = min( X ′ \ { x k } k ∈↓ γ ) and y n / ∈ B ;(1g) If n ∈ ←− Ω , then y n = min( Y ′ \ { y k } k ∈↓ γ ) and x n / ∈ A .(2) If γ = ( n, k ) for some n, k ∈ ω , then2a) ℓ ( γ ) ≥ { ℓ ( α ) : α ∈ ↓ γ } ;2b) for any m ∈ ω ∩ ↓ γ with m = n , we have x m / ∈ U n,k and y m / ∈ V n,k ;2c) x n ∈ U n,k ⊆ O Xk ( x n ) ⊆ X \ X ℓ ( n ) and y n ∈ V n,k ⊆ O Yk ( x n ) ⊆ Y \ Y ℓ ( n ) ;2d) { ( i, j ) ∈ ↓ γ : U n,k ⊆ U i,j } = { ( i, j ) ∈ ↓ γ : x n ∈ U i,j } and { ( i, j ) ∈ ↓ γ : V n,k ⊆ V i,j } = { ( i, j ) ∈ ↓ γ : y n ∈ V i,j } ;2e) { ( i, j ) ∈ ↓ γ : U n,k ∩ U i,j = ∅} = { ( i, j ) ∈ ↓ γ : x n / ∈ U i,j } and { ( i, j ) ∈ ↓ γ : V n,k ∩ V i,j = ∅} = { ( i, j ) ∈ ↓ γ : y n / ∈ V i,j } ;2f) X ℓ ( γ ) = ∂U n,k and Y ℓ ( γ ) = ∂V n,k ⊆ V n,k ∩ Y ℓ ( n ) ;2g) if n ∈ Ω, then f ( U n,k ∩ A ) = V n,k ∩ B ;2h) If n / ∈ Ω, then U n,k ∩ A = ∅ = V n,k ∩ ¯ B ;2i) If ←− Ω = ∅ , then X ℓ ( γ ) = ∂U n,k ⊆ U n,k ∩ X ℓ ( n ) .0. We start the inductive construction letting x be the smallest point of the well-orderedset ( X ′ , (cid:22) X ). Since the set Y ℓ X ( x ) is nowhere dense in Y ℓ X ( x ) , the set Y ℓ X ( x ) \ Y ℓ X ( x ) is not empty and hence contains some point y . Such choice of y guarantees that the condition(1) is satisfied for γ = 0 ∈ −→ Ω .Now assume that for some γ ∈ Γ, we have defined the function ℓ on the set ↓ γ andconstructed points x n , y n and open sets U i,j , V i,j for all n ∈ ω ∩ ↓ γ and ( i, j ) ∈ ( ω × ω ) ∩ ↓ γ so that the inductive conditions (1)–(2) are satisfied.To fulfill the inductive step, consider two possible cases.1. First assume that γ = n for some number n ∈ ω . This case has three subcases.1 ′ . If n ∈ Ω, then put x n = ξ ( n ) ∈ A and y n = f ( x n ) ∈ B . Our assumption on themap f ensures that ℓ X ( x n ) = ℓ Y ( y n ). So we can put ℓ ( γ ) := ℓ X ( x n ) = ℓ Y ( y n ) and see thatthe inductive conditions (1a), (1b) are satisfied. To see that (1c) is satisfied, take any pair( i, j ) ≺ γ = n .First we assume that x n ∈ U i,j . If x i / ∈ A , then i / ∈ Ω and we obtain a contradiction x n ∈ A ∩ U i,j = ∅ applying the inductive condition (2h). This contradiction shows that x i ∈ A . Then y n = f ( x n ) ∈ f ( A ∩ U i,j ) = B ∩ V i,j ⊆ V i,j by the inductive condition (2g). Byanalogy we can show that y n ∈ V i,j implies x n ∈ U i,j . This means that the condition (1c) issatisfied.Now assume that x n ∈ U i,j . If x n ∈ U i,j , then y n ∈ V i,j ⊆ V i,j by the (already proved)inductive condition (1c). So, we assume that x n ∈ U i,j \ U i,j = ∂U i,j = X ℓ ( i,j ) (for thelast equality, see the inductive condition (2f)). Then ℓ Y ( y n ) = ℓ X ( x n ) ≥ ℓ ( i, j ) and hence y n ∈ Y ℓ ( i,j ) ⊆ V i,j by the condition (2f). By analogy we can prove that y n ∈ V i,j implies x ∈ U i,j . This means that the condition (1d) is satisfied. It is clear that the conditions(1e)–(1g) are satisfied, too.1 ′′ . Next, assume that n ∈ −→ Ω . This case requires much more work. Define the point x n by the formula (1f) and put ℓ ( n ) = ℓ X ( x n ). It remains to find a point y n ∈ Y satisfying theconditions (1a)–(1d). Lemma 9. For any nonempty set J ⊆ ( ω × ω ) ∩↓ γ , integer number l < min { ℓ ( i, j ) : ( i, j ) ∈ J } ,and nonempty open set W ⊆ Y such that W ∩ T ( i,j ) ∈ J Y ℓ ( i,j ) − = ∅ , there exists a point y ∈ W ∩ Y l \ Y l +1 such that y / ∈ S ( i,j ) ∈ J V i,j .Proof. Write the set J as { ( i , j ) , . . . , ( i m , j m ) } for some pairs ( i m , j m ) ≺ · · · ≺ ( i , j ). If J is empty, then m = 0. The inductive condition (2a) ensures that ℓ ( i k , j k ) + 2 ≤ ℓ ( i k − , j k − )for any k ∈ { , . . . , m } .Let v be any point in the set W ∩ Y ℓ ( i ,j ) − \ Y ℓ ( i ,j ) . Such point exists since the intersection W ∩ Y ℓ ( i ,j ) − is nonempty and Y ℓ ( i ,j ) is nowhere dense in Y ℓ ( i ,j ) − .By the inductive conditions (2c),(2f),(2a) we have V i ,j = V i ,j ∪ ∂V i ,j ⊆ ( Y \ Y ℓ ( i ) ) ∪ Y ℓ ( i ,j ) and ℓ ( i ) + 2 ≤ ℓ ( i , j ). Hence, the set Y ℓ ( i ,j ) − \ Y ℓ ( i ,j ) is disjoint with V i ,j . Then v / ∈ V i ,j and we can choose an open neighborhood W ⊆ W of v such that W ∩ V i ,j = ∅ .Inductively we shall construct a sequence of points v , . . . , v m ∈ Y and a sequence of opensets W , . . . , W m in Y such that for every k ∈ { , . . . , m } the following conditions are satisfied:(i) v k ∈ W k − ∩ Y ℓ ( i k ,j k ) − \ Y ℓ ( i k ,j k ) ;(ii) v k ∈ W k ⊆ W k − \ V i k ,j k . UNIVERSAL COREGULAR COUNTABLE SECOND-COUNTABLE SPACE 15 Assume that for some k ∈ { , . . . , m } we have constructed points v , . . . , v k − and an openset W , . . . , W k − satisfying the conditions (i), (ii). Since v k − ∈ Y ℓ ( i k − ,j k − ) − ⊆ Y ℓ ( i k ,j k ) and the set Y ℓ ( i k ,j k ) is nowhere dense in Y ℓ ( i k ,j k ) − , we can choose a point v k ∈ W k − ∩ Y ℓ ( i k ,j k ) − \ Y ℓ ( i k ,j k ) . By the inductive conditions (2c) and (2f), V i k ,j k = V i k ,j k ∪ ∂V i k ,j k ⊆ ( Y \ Y ℓ ( i k ) ) ∪ Y ℓ ( i k ,j k ) and ℓ ( i k ) + 1 ≤ ℓ ( i k , j k ) − 1. Consequently, the set Y ℓ ( i k ,j k ) − \ Y ℓ ( i k ,j k ) is disjoint with V i k ,j k . So, v k / ∈ V i k ,j k and we can choose an open neighborhood W k ⊆ W k − of v k such that W k ∩ V i k ,j k = ∅ . This completes the inductive step.After completing the inductive construction, consider the point v m ∈ Y ℓ ( i m ,j m ) − and itsneighborhood W m ⊆ W . The inductive condition (ii) guarantees that W m ∩ m [ k =1 V i k ,j k = m [ k =1 ( W m ∩ V i k ,j k ) ⊆ m [ k =1 ( W k ∩ V i k ,j k ) = ∅ . Taking into account that l < min { ℓ ( i, j ) : ( i, j ) ∈ J } ≤ ℓ ( i m , j m ), we conclude that v m ∈ Y ℓ ( i m ,j m ) − ⊆ Y l . Since the set Y l +1 is nowhere dense in Y l , there exists a point y ∈ W m ∩ Y l \ Y l +1 . Since W m is disjoint with S ( i,j ) ∈ J V i,j , the point y n does not belong to S ( i,j ) ∈ J V i,j . (cid:3) Now we are able to find a y n satisfying the conditions (1a)–(1d).Consider the sets I ( x n ) = { ( i, j ) ∈ ↓ γ : x n ∈ U i,j } and J ( x n ) = { ( i, j ) ∈ ↓ γ : x n / ∈ U i,j } . Claim 4. (1) For any ( i, j ) ∈ I ( x n ) ∪ J ( x n ) we have ℓ X ( x n ) < ℓ ( i, j ) . (2) For any ( i, j ) ∈ I ( x n ) we have ℓ X ( x n ) ≤ ℓ ( i ) .Proof. 1. If ( i, j ) ∈ I ( x n ) ∪ J ( x n ), then x n / ∈ ∂U i,j = X ℓ ( i,j ) and hence ℓ X ( x n ) < ℓ ( i, j ).2. Now assume that ( i, j ) ∈ I ( x n ). Then x n ∈ U i,j ⊆ X \ X ℓ X ( x i ) and hence ℓ X ( x n ) < ℓ X ( x i ) = 1 + ℓ ( i ) according to (1a). (cid:3) Choose a minimal subset I ⊆ I ( x n ) such that for every ( i, j ) ∈ I ( x n ) there exists ( p, q ) ∈ I such that U p,q ⊆ U i,j . It is clear that T ( i,j ) ∈ I ( x n ) U i,j = T ( i,j ) ∈ I U i,j . Claim 5. T ( i,j ) ∈ I ( x n ) V i,j = T ( i,j ) ∈ I V i,j .Proof. It suffices to show that for any ( i, j ) ∈ I ( x n ) there exists ( p, q ) ∈ I such that V p,q ⊆ V i,j .Given any pair ( i, j ) ∈ I ( x n ), find a ( p, q ) ∈ I such that x p ∈ U p,q ⊆ U i,j (such a pair exists bythe choice of the set I ). If ( i, j ) ≺ p , then y p ∈ V i,j by the condition (1c) and then V p,q ⊆ V i,j by the condition (2d).If p ≺ ( i, j ), the the condition (2b) implies that p = i and condition (2d) ensures that V p,q ⊆ V i,j . (cid:3) Claim 6. For any pairs ( i, j ) ≺ ( p, q ) in I we have ℓ ( p ) ≥ ℓ ( i, j ) > ℓ ( i ) .Proof. First we show that x p ∈ ∂U i,j . Assuming that x p / ∈ ∂U i,j , we conclude that x p ∈ U i,j or x p / ∈ U i,j . If x p ∈ U i,j , then the inductive condition (2d) guarantees that x p ∈ U p,q ⊆ U i,j andthe minimality of I ensures that ( i, j ) / ∈ I , which contradicts our assumption. So, x p / ∈ U i,j .In this case, U p,q ∩ U i,j = ∅ by the condition (2e), but this contradicts x n ∈ U i,j ∩ U p,q .Therefore, x p ∈ ∂U i,j = X ℓ ( i,j ) and ℓ ( p ) = ℓ X ( x p ) ≥ ℓ ( i, j ) > ℓ X ( x i ) = ℓ ( i ). (cid:3) Write the set I as { ( i , j ) , . . . , ( i m , j m ) } for some pairs ( i m , j m ) ≺ · · · ≺ ( i , j ). If I isempty, then m = 0. Claims 6 and 4 imply that ℓ ( i , j ) > ℓ ( i ) ≥ ℓ ( i , j ) > ℓ ( i ) ≥ · · · ≥ ℓ ( i m , j m ) > ℓ ( i m ) ≥ ℓ X ( x n ) . This chain of inequalities allows us to write the set J ( x n ) as the union J ( x n ) = (cid:16) m [ k =1 J k (cid:17) ∪ (cid:16) m [ k =0 J ′ k (cid:17) of the sets J k = { ( i, j ) ∈ J ( x n ) : ℓ ( i k , j k ) > ℓ ( i, j ) > ℓ ( i k ) } for k ∈ { , . . . , m } , J ′ = { ( i, j ) ∈ J ( x n ) : ℓ ( i, j ) > ℓ ( i , j ) } ,J ′ k = { ( i, j ) ∈ J ( x n ) : ℓ ( i k ) > ℓ ( i, j ) > ℓ ( i k +1 , j k +1 ) } for k ∈ { , . . . , m − } , J ′ m = { ( i, j ) ∈ J ( x n ) : ℓ ( i m ) > ℓ ( i, j ) > ℓ X ( x n ) } . Since the sets { ℓ ( i, j ) : ( i, j ) ∈ J ( x n ) } and { ℓ ( i k ) , ℓ ( i k , j k ) } mk =1 are disjoint, the union S mk =1 J k ∪ S mk =0 J ′ k is indeed equal to J ( x n ).By Lemma 9, there exists a point v ′ ∈ Y ℓ ( i ,j ) such that v ′ / ∈ S ( i,j ) ∈ J ′ V i,j . Then W ′ := Y \ S ( i,j ) ∈ J ′ V i,j is an open neighborhood of v ′ .Inductively we shall construct a sequence of points v , v ′ , , . . . , v m , v ′ m and a sequence ofopen sets W ⊇ W ′ ⊇ · · · ⊇ W m ⊇ W ′ m in Y such that for every k ∈ { , . . . , m } the followingconditions are satisfied:(a) v k ∈ W ′ k − ∩ V i k ,j k ∩ Y ℓ ( i k ) ;(b) v k ∈ W k = W ′ k − ∩ V i k ,j k ;(c) W k ∩ S ( i,j ) ∈ J k V i,j = ∅ ;(d) v ′ k ∈ W ′ k ⊆ W k \ S ( i,j ) ∈ J ′ k V i,j ;(e) if k < m , then v ′ k ∈ Y ℓ ( i k +1 ,j k +1 ) ;(f) v ′ m ∈ Y ℓ X ( x n ) \ Y ℓ X ( x n ) .To make an inductive step, assume that for some k ∈ { , . . . , m } a point v ′ k − and an openset W ′ k − with v ′ k − ∈ W k − ∩ Y ℓ ( i k ,j k ) have been constructed. By the inductive condition (2f), Y ℓ ( i k ,j k ) = ∂V i k ,j k ⊆ V i k ,j k ∩ Y ℓ ( i k ) . Consequently, there exists a point v k ∈ W ′ k − ∩ V i k ,j k ∩ Y ℓ ( i k ) .Put W k := W k − ∩ V i k ,j k . It is clear that the inductive conditions (a), (b) are satisfied. Claim 7. V i k ,j k ∩ S ( i,j ) ∈ J k V i,j = ∅ .Proof. To derive a contradiction, assume that V i k ,j k ∩ V i,j = ∅ for some ( i, j ) ∈ J k . Thedefinition of the set J k ∋ ( i, j ) yields ℓ ( i k , j k ) > ℓ ( i, j ) > ℓ ( i k ) and hence ( i, j ) ≺ ( i k , j k ) bythe condition (2a). It follows from (2e) and V i k ,j k ∩ V i,j = ∅ that y i k ∈ V i,j and hence x i k ∈ U i,j according to the condition (1d). Assuming that x i k ∈ U i,j , we obtain U i k ,j k ⊆ U i,j by theinductive condition (2d). Then x n ∈ U i k ,j k ⊆ U i,j , which contradicts the inclusion ( i, j ) ∈ J k .Therefore, x i k / ∈ U i,j and hence y i k / ∈ V i,j by condition (1c). Then y i k ∈ V i,j \ V i,j = ∂V i,j andhence ℓ ( i k ) = ℓ Y ( y i k ) ≥ ℓ ( i, j ), which contradicts the inclusion ( i, j ) ∈ J k . (cid:3) Claim 7 and the condition (b) imply the condition (c).Since v k ∈ W k ∩ Y ℓ ( i k ) ⊆ W k ∩ T ( i,j ) ∈ J ′ k Y ℓ ( i,j ) − , we can apply Lemma 9 and find a point v ′ k ∈ W k and a neighborhood W ′ k of v ′ k satisfying the inductive conditions (d),(e),(f).After completing the inductive construction, we conclude that the open subset W ′ m ∩ Y ℓ ( n ) \ Y ℓ ( n ) of the crowded space Y ℓ ( n ) contains the point v ′ m and hence is not empty. Since the UNIVERSAL COREGULAR COUNTABLE SECOND-COUNTABLE SPACE 17 space Y ℓ ( n ) is crowded (see Lemma 4) and the set ¯ B ∩ Y ℓ ( n ) is nowhere dense in Y ℓ ( n ) , thereexists a point y n ∈ W ′ m ∩ ( Y ℓ ( n ) \ Y ℓ ( n ) ) \ ( ¯ B ∪ { y k } k ∈↓ γ ) . The inductive conditions (a),(c),(d) and Claim 5 imply that I ( x n ) ⊆ I ( y n ) and J ( x n ) ⊆ J ( y n ),where I ( y n ) = { ( i, j ) ∈ ↓ γ : y n ∈ V i,j } and J ( y n ) = { ( i, j ) ∈ ↓ γ : y n / ∈ V i,j } . The condition (1c) will follow as soon as we show that I ( x n ) = I ( y n ). Assuming that I ( x n ) = I ( y n ), we can find a pair ( i, j ) ∈ I ( y n ) \ I ( x n ). The inclusion J ( x n ) ⊆ J ( y n ) ⊆ ↓ γ \ I ( y n )implies that ( i, j ) / ∈ J ( x n ) and hence x n ∈ U i,j \ U i,j = X ℓ ( i,j ) . Then ℓ Y ( y n ) = ℓ X ( x n ) ≥ ℓ ( i, j )and hence y n ∈ Y ℓ ( i,j ) = ∂V i,j and hence y n / ∈ V i,j and ( i, j ) / ∈ I ( y n ), which contradicts thechoice of ( i, j ). This completes the proof of condition (1c).To prove the condition (1d), assume that J ( x n ) = J ( y n ) and find a pair ( i, j ) ∈ J ( y n ) \ J ( x n ). Then x n ∈ U i,j . Assuming that x n ∈ U i,j , we conclude that ( i, j ) ∈ I ( x n ) = I ( y n )and hence y n ∈ V i,j ⊆ V i,j , which contradicts ( I, j ) ∈ J ( y n ). Therefore, x n ∈ U i,j \ U i,j = ∂U i,j = X ℓ ( i,j ) and ℓ Y ( y n ) = ℓ X ( x n ) ≥ ℓ ( i, j ) and then the condition (2f) ensures that y n ∈ Y ℓ ( i,j ) = ∂V i,j and hence ( i, j ) / ∈ J ( y n ), which contradicts the choice of the pair ( i, j ).This contradiction completes the proof of the condition (1d). It is clear that the conditions(1e)–(1g) holds.1 ′′′ . n ∈ ←− Ω . In this case the set ←− Ω is not empty and hence ( X n ) n ∈ ω is a canonicalsuperskeleton for X , the set B is closed in Y and for every n ∈ ω the set A ∩ X n is nowheredense in X n . In this case we put y n = min( Y ′ \ { y k } k ∈↓ γ ) and repeating the argument fromthe case 1 ′′ , can find a point x n ∈ X satisfying the conditions (1a)–(1g).2. Now consider the second case: γ = ( n, k ) for some ( n, k ) ∈ ω × ω . Since n ≺ ( n, k ), thepoints x n , y n have been already defined. So, we can choose open sets U ⊆ X and V ⊆ Y suchthat • x k / ∈ U = X and y k / ∈ V = Y for every k ∈ ↓ γ \ { x n } , • x n ∈ U ⊆ O Xk ( x n ) ∩ T ( i,j ) ∈ I ( x n ) U i,j \ S ( i,j ) ∈ J ( x n ) U i,j , and • y n ∈ V ⊆ O Yk ( y n ) ∩ T ( i,j ) ∈ I ( y n ) V i,j \ S ( i,j ) ∈ J ( y n ) V i,j ,where I ( x n ) = { ( i, j ) ∈ ↓ γ : x n ∈ U i,j } , I ( y n ) = { ( i, j ) ∈ ↓ γ : y n ∈ V i,j } J ( x n ) = { ( i, j ) ∈ ↓ γ : x n / ∈ U i,j } , J ( y n ) = { ( i, j ) ∈ ↓ γ : y n / ∈ V i,j } . If n / ∈ Ω, then x n / ∈ A , y n / ∈ B (by the inductive conditions (1f), (1g)) and we can (and will)additionally assume that U ∩ A = ∅ = V ∩ ¯ B. Since ( X i ) i ∈ ω is a coregular skeleton for the space X and ( Y i ) i ∈ ω is a superskeleton for thespace Y , there exists a number l ≥ { ℓ ( α ) : α ∈ ↓ γ } such that X l ⊆ U ∩ X \ U ⊆ ∂U and Y l ⊆ V ∩ Y ℓ ( n ) ∩ Y \ V ⊆ ∂V. Since the skeleton ( Y i ) i ∈ ω is coregular, the complement Y \ Y l is a regular topological space.Being second-countable, the regular space Y \ Y l is metrizable (by the Urysohn MetrizationTheorem [8, 4.2.9]). Being countable, the metrizable space Y \ Y l is zero-dimensional. Thenwe can find a closed-and-open neighborhood V ′ ⊆ Y \ Y l of the point y n such that V ′ ⊆ V .Then ∂V ′ ⊆ Y l . By analogy we prove that the space X \ X l is metrizable and zero-dimensional. By The-orem [8, 7.1.11], the countable zero-dimensional space X \ X l is strongly zero-dimensional(which means that any disjoint closed sets in X \ X l can be separated by closed-and-openneighborhoods). Observe that the sets f − ( V ′ ) and f − (( Y \ Y l ) \ V ′ ) are two closed disjointsets in A \ X l and X \ X l . By the strong zero-dimensionality of X \ X l , there exists a closed-andopen set U ′ in X \ X l such that { x n } ∪ f − ( V ′ ) ⊆ U ′ ⊆ U \ f − (( Y \ Y l ) \ V ′ ) . Then ∂U ′ ⊆ X l and f ( A ∩ U ′ ) = B ∩ V ′ .Since ( X i ) i ∈ ω is a coregular skeleton for X and ( Y i ) i ∈ ω is a superconnecting skeleton for Y , there exists a number p > l such that X p ⊆ ∂U ′ and Y p ⊆ ∂V ′ . Lemma 10. There exists closed-and open subset V n,k ⊆ Y \ Y p such that V ′ ⊆ V n,k ⊆ V , ( V n,k \ V ′ ) ∩ ¯ B = ∅ and ∂V n,k = Y l ⊆ V n,k ∩ Y ℓ ( n ) .Proof. By the Urysohn Metrization Theorem [8, 4.2.9], the second-countable regular space Y \ Y p is metrizable. So, we can find a metric d generating the topology of Y . Since the set Y l \ Y p ⊆ Y is countable and nonempty, there exists a function ~ : ω → Y l \ Y p such that forevery y ∈ Y l \ Y p the preimage ~ − ( y ) is infinite. Since Y l \ Y p ⊆ V ∩ Y ℓ ( n ) and the set ¯ B ∩ Y ℓ ( n ) is nowhere dense in Y ℓ ( n ) , for every m ∈ ω we can find a point v m ∈ V ∩ Y ℓ ( n ) \ ¯ B such that d ( v m , ~ ( m )) < − m . Since the space Y \ Y p is zero-dimensional, the point v m has a closed-and-open neighborhood W m in Y \ ( Y p ∪ ¯ B ) such that W k ⊆ V ∩ { y ∈ Y \ Y p : d ( y, v m ) < − m } .Then the boundary ∂W m of W m in Y is contained in Y p . We claim that the open neighborhood V n,k = V ′ ∪ [ m ∈ ω W m ⊆ V of y n has the required property: X l = ∂V n,k ⊆ V n,k ∩ Y ℓ ( n ) .First we show that Y l ⊆ V n,k ∩ Y ℓ ( n ) . Given any point a ∈ Y l and open neighborhood O a ⊆ Y of a , use the nowhere density of Y p in Y l and find a point b ∈ O a ∩ ( Y l \ Y p ). Sincethe metric d generates the topology of the space Y \ Y p , there exists a number q ∈ ω suchthat the ball B ( b ; 2 − q ) = { y ∈ Y \ Y l : d ( y, b ) < − q } is contained in O a . Since the set ~ − ( b )is infinite, there exists m > q such that ~ ( m ) = b . Then d ( b, v m ) = d ( ~ ( m ) , v m ) < − m < − q and hence v m ∈ O a . On the other hand, v m ∈ V n,k ∩ Y ℓ ( n ) , witnessing that O a ∩ ( V ∩ Y ℓ ( n ) ) = ∅ and hence a ∈ V n,k ∩ Y ℓ ( n ) ⊆ V n,k .On the other hand, a ∈ Y l ⊆ Y p ⊆ X \ V ⊂ Y \ V n,k and hence a ∈ V n,k ∩ Y \ V n,k = ∂V n,k .Therefore, Y l ⊆ ∂V n,k . Assuming that Y l = ∂V n,k , we can find a point z ∈ ∂V n,k \ Y l . Choose s ∈ ω such that the ball B ( z ; 2 − s ) = { y ∈ Y : d ( z, y ) < − s } does not intersect the closedsubset Y l \ Y p of Y \ Y p . We claim that for every m ≥ s + 2, the ball B ( z ; 2 − s − ) does notintersect the set W m . Assuming that B ( z ; 2 − s − ) ∩ W m contains some point w , we concludethat d ( z, ~ ( m )) ≤ d ( z, w ) + d ( w, v m ) + d ( v m , ~ ( m )) < − s − + 2 − m + 2 − m = 2 − s − + 2 − m +1 ≤ − s − + 2 − s − = 2 − s , which contradicts the choice of s . Since z ∈ ∂V n,k , z / ∈ V n,k and hence z / ∈ V ′ ∪ S m< s W m .It follows from ∂V ′ ∪ S m< s ∂W m ⊆ Y l z that z / ∈ V ′ ∪ S m< s W m . Then we can find a UNIVERSAL COREGULAR COUNTABLE SECOND-COUNTABLE SPACE 19 neighborhood O z ⊆ B ( z ; 2 − s − ) such that O z ∩ V ′ ∪ S m< s W m = ∅ and hence O z ∩ V n,k ⊆ (cid:0) O z ∩ V ′ (cid:1) ∪ (cid:16) [ m< s O z ∩ W m (cid:17) ∪ (cid:16) [ m ≥ s B ( s ; 2 − s − ) ∩ W m (cid:17) = ∅ , which contradicts z ∈ ∂V n,k . This contradiction shows that ∂V n,k = Y l ⊆ V n,k ∩ Y ℓ ( n ) . Thechoice of the sets W m ⊆ Y \ ¯ B ensures that( V n,k \ V ) ∩ ¯ B ⊆ [ m ∈ ω ( W m ∩ ¯ B ) = ∅ . (cid:3) By analogy we can prove the following lemma. Lemma 11. There exists closed-and-open subset U n,k ⊆ X \ X p such that U ′ ⊆ U n,k ⊆ U , ( U n,k \ U ′ ) ∩ A = ∅ and ∂U n,k = X l . Moreover, if ←− Ω = ∅ , then X l ⊆ U n,k ∩ X ℓ ( n ) . Finally, observe that the number ℓ ( n, k ) := l and the sets U n,k and V n,k constructed inLemmas 10 and 11 satisfy the conditions (2a)–(2i). This completes the inductive step.After completing the inductive construction, observe that the inductive condition (1f)implies that X = { x n } n ∈ ω . So, we can consider the map ¯ f : X → Y such that ¯ f ( x n ) = y n forevery n ∈ ω . The inductive condition (1e) ensures that ¯ f ↾ A = f . We claim that the map ¯ f is a topological embedding.To see that f is continuous, take any n ∈ ω and any neighborhood O ( y n ) of the point y n = f ( x n ). Find k ∈ ω such that O Yk ( y n ) ⊆ O ( y n ). The inductive condition (2b) guaranteesthat V n,k ⊆ O Yk ( y n ). We claim that ¯ f ( U n,k ) ⊆ V n,k . Indeed, for any x m ∈ U n,k \ { x n } , theinductive condition (2b) ensures that ( n, k ) ≺ m . Then y m ∈ V n,k by the condition (1c).Therefore, f ( U n,k ) ⊆ V n,k ⊆ O Yk ( y n ) ⊆ O ( y n ), witnessing that the map f is continuous. Byanalogy we can prove the continuity of the map f − : f ( X ) → X .If ( X n ) n ∈ ω is a canonical supserskeleton for X n , the set B is closed in Y , and for every n ∈ ω the set A ∩ X n is nowhere dense in X n , then the set ←− Ω is infinite and the inductivecondition (1g) implies that ¯ f ( X ) = { y n } n ∈ ω = Y and hence the topological embedding ¯ f is ahomeomorphism. (cid:3) Now we deduce some corollaries of Theorem 1. Corollary 1. Let X, Y be two countable second-countable topological spaces, ( X n ) n ∈ ω and ( Y n ) n ∈ ω be canonical superskeleta in the spaces X, Y , and A, B be closed nowhere dense setsin the spaces X, Y , respectively. Let h : A → B be a homeomorphism such that h ( A ∩ X n ) =( B ∩ Y n ) for every n ∈ ω . Then there exists a homeomorphism ¯ h : X → Y such that ¯ h ↾ A = h and h ( X n ) = Y n for all n ∈ ω . Corollary 2. Let X, Y be two countable second-countable topological spaces and ( X n ) n ∈ ω and ( Y n ) n ∈ ω be canonical superskeleta in the spaces X, Y , respectively. Then there exists ahomeomorphism h : X → Y such that h ( X n ) = Y n for all n ∈ ω . Lemma 5 and Corollary 2 imply another corollary. Corollary 3. Let X, Y be two countable second-countable topological spaces and ( X n ) n ∈ ω and ( Y n ) n ∈ ω be superskeleta in the spaces X, Y , respectively. Then there exists an increasingnumber sequence ( n k ) k ∈ ω and a homeomorphism h : X → Y such that h ( X n k ) = Y n k for all k ∈ ω . Now we are able to prove Theorem 1 reformulating it as follows. Theorem 4 (Characterization of Q P ∞ ) . A topological space X is homeomorphic to the space Q P ∞ if and only if X is countable second-countable and admits a vanishing sequence ( X n ) n ∈ ω of nonempty closed sets that has two properties: (1) for every n ∈ ω and a nonempty open set U ⊆ X n the closure U contains some set X m ; (2) for every n ∈ ω the complement X \ X n is a regular topological space.Proof. The “only if” part follows from Theorem 2. To prove the “if” part, assume that thespace X is countable, second-countable and X has a vanishing sequence of nonempty closedsets ( X n ) n ∈ ω satisfying the conditions (1),(2). By Definition 1, ( X n ) n ∈ ω is a superskeleton for X . By Theorem 2, the space Q P ∞ also is countable, second-countable and has a superskeleton.By Corollary 3, the spaces X and Q P ∞ are homeomorphic. (cid:3) Now we prove a universality property of the space Q P ∞ . Theorem 5 (Universality of Q P ∞ ) . Each countable second-countable coregular space X ishomeomorphic to a subspace of Q P ∞ .Proof. By Lemma 1, the space X admits a coregular skeleton ( X n ) n ∈ ω . By Theorem 2, thespace Q P ∞ has a canonical superskeleton ( Y n ) n ∈ ω . Applying Theorem 1 with A = B = ∅ , weobtain a topological embedding f : X → Y such that f − ( Y n ) = X n for all n ∈ ω . (cid:3) Now we prove a (rather strong) homogeneity property of the space Q P ∞ .A subset A of a topological space X is called • deep if for any non-empty open sets U , . . . , U n ⊆ X the set A \ ( U ∩ · · · ∩ U n ) is finite. • shallow if there exist non-empty open sets U , . . . , U n ⊆ X such that A ∩ ( U ∩ · · · ∩ U n ) = ∅ .This definition implies that for any deep (resp. shallow) set A in a topological space X andany homeomorphism h : X → X the set h ( A ) is deep (resp. shallow). Observe also thatany infinite set in a second-countable space contains an infinite subset which is either deep orshallow. The definition implies that any finite set in a Hausdorff space is shallow. Theorem 6 (Dychotomic Homogeneity of Q P ∞ ) . Let A, B be two closed discrete subsets of Q P ∞ . If the sets A, B are either both deep or both shallow, then any bijection f : A → B extends to a homeomorphism h of Q P ∞ such that h ( A ) = B .Proof. By Theorem 2, the space Q P ∞ has a canonical superskeleton ( X n ) n ∈ ω . If both sets A, B are shallow, then we can find a number m ∈ ω such that X m is disjoint with the set A ∪ B . Let Y = X and Y n = X m + n for n ∈ N . Observe that ( Y n ) n ∈ ω is a canonicalsuperskeleton for the space Q P ∞ such that A ∪ B ⊆ Y \ Y . Since the space Q P ∞ is crowded,the sets A, B are nowhere dense in X . Applying Corollary 1, we can find a homeomorphism h : X → X such that h ↾ A = f and h ( Y n ) = Y n for all n ∈ ω .The case of deep sets A, B is more tricky. Let ℓ : X → ω be the function assigning to eachpoint x ∈ X the unique number n ∈ ω such that x ∈ X n \ X n +1 . The deepness of A, B impliesthat for every n ∈ ω the set ( A ∪ B ) \ X n is finite. Choose an increasing sequence ( n k ) k ∈ ω such that n = 0 and for every k ∈ ω the following conditions hold: • n k +1 > ℓ ( f ( a )) for any a ∈ A \ X n k +1 ; • n k +1 > ℓ ( f − ( b )) for any b ∈ B \ X n k +1 . UNIVERSAL COREGULAR COUNTABLE SECOND-COUNTABLE SPACE 21 For every k ∈ ω consider the finite sets A k = A ∩ X n k \ X n k +1 , B k = B ∩ X n k \ X n k +1 ,A = k = { a ∈ A k : f ( a ) ∈ X n k \ X n k +1 } , B = k = { b ∈ B k : f − ( b ) ∈ X n k \ X n k +1 } ,A + k = { a ∈ A k : f ( a ) ∈ X n k +1 } , B + k = { b ∈ B k : f − ( b ) ∈ X n k +1 } ,A − k = { a ∈ A k : f ( a ) / ∈ X n k } , B − k = { b ∈ B k : f − ( b ) / ∈ X n k } . Claim 8. For any k ∈ ω we have f ( A = k ) = B = k , f ( A + k ) = B − k +1 , f − ( B + k ) = A − k +1 , and f ( A − k +1 ) = B + k . Proof. The equality f ( A = k ) = B = k follows from the definition of the sets A = k and B = k . To showthat f ( A + k ) = B − k +1 , take any a ∈ A + k . Then f ( a ) ∈ B ∩ X n k +1 by the definition of A + k . Thedefinition of the number n k +2 guarantees that ℓ ( f ( a )) < n k +2 and hence f ( a ) ∈ B ∩ X n k +1 \ X n k +2 = B k +1 . Since a = f − ( f ( a )) / ∈ X n k +1 , the point f ( a ) belongs to B − k +1 . Therefore, f ( A + k ) ⊆ B − k +1 .Now take any point b ∈ B − k +1 and observe that the point a = f − ( b ) dos not belong to X n k +1 bythe definition of the set B − k +1 . Assuming that a / ∈ X n k , we conclude that ℓ ( b ) = ℓ ( f ( a )) < n k +1 and hence b / ∈ X n k +1 , which contradicts the choice of b . This contradiction shows that a ∈ A ∩ X n k \ X n k +1 . Since b = f ( a ) ∈ X n k +1 , the point a belongs to the set A + k and hence b ∈ f ( A + k ). Therefore, f ( A + k ) = B − k +1 . By analogy we can prove that f − ( B + k ) = A − k +1 andhence f ( A − k +1 ) = B + k . (cid:3) Claim 9. For every k ∈ ω we have A + k ∪ B + k ⊆ X n k +1 \ X n k +1 .Proof. Assuming that A + k X n k +1 , we can find a point a ∈ A + k \ X n k +1 . The choice of n k +1 ensures that n k +1 > ℓ ( f ( a )) and hence f ( a ) / ∈ X n k +1 , which contradicts the inclusion a ∈ A + k . This contradiction shows that A + k ⊆ X n k +1 \ X n k +1 . By analogy we can prove that B + k ⊆ X n k +1 \ X n k +1 . (cid:3) The coregularity of the skeleton ( X n ) n ∈ ω guarantees that for every k ∈ ω the countablespace X n k +1 \ X n k +1 is regular and hence metrizable and zero-dimensional. Then we can finda closed-and-open sets U k and V k in X n k +1 \ X n k +1 such that A k ∩ U k = A + k and B k ∩ V k = B + k .Let Y = Z = X and for every k ∈ N let Y k := U k ∪ X n k +1 and Z k := V k ∪ X n k +1 and observe that Y k = U k ∪ X n k +1 ⊆ X n k +1 ⊂ X n k ⊆ Y k − . The nowhere density of the set X n k +1 in X n k implies the nowhere density of Y k in X n k and also in Y k − . By analogy we canshow that Z k is nowhere dense in Z k − . It is easy to check that ( Y k ) k ∈ ω and ( Z k ) k ∈ ω arecanonical superskeleta for X such that A ∩ Y k \ Y k +1 = A − k +1 ∪ A = k +1 ∪ A + k and B ∩ Z k \ Z k +1 = B + k ∪ B = k +1 ∪ B − k +1 for every k ∈ ω . This implies that f ( A ∩ Y k ) = B ∩ Z k for every k ∈ ω . Since the spaces A, B are discrete, the intersections A ∩ Y k and B ∩ Z k are nowhere dense in the crowded spaces Y k , Z k , respectively. Applying Theorem 1, we can find a homeomorphism h : X → Y suchthat h ↾ A = f and h ( Y k ) = Z k for all k ∈ ω . (cid:3) Since any finite set in a Hausdorff topological space is shallow, Theorem 6 implies that thefollowing finite homogeneity property of the space Q P ∞ . Corollary 4 (Finite homogeneity of Q P ∞ ) . Any bijective function f : A → B between twofinite subsets A, B of the space Q P ∞ can be extended to a homeomorphism of Q P ∞ . Remark 2. By Lemma 8, the product Q P ∞ × Q P ∞ is not coregular and hence cannot behomeomorphic to Q P ∞ . Nonetheless, Q P ∞ × Q P ∞ contains a dense subspace homeomorphicto Q P ∞ , see [18]. 5. Acknowledgement The first author would like to thank MathOverflow user Fedor Petrov who turned hisattention to the infinite projective space Q P ∞ and informed the authors about the paper [10]of Gelfand and Fuks. References [1] T. Banakh, J. Mioduszewski, S. Turek, On continuous self-maps and homeomorphisms of the Golomb space ,Comment. Math. Univ. Carolin. :4 (2018) 423–442.[2] T. Banakh, D. Spirito, S. Turek, The Golomb space is topologically rigid , preprint(https://arxiv.org/abs/1912.01994).[3] T. Banakh, Y. Stelmakh, S. Turek, The Kirch space is topologically rigid , in preparation.[4] C. Bessaga, A. Pe lczy´nski, Selected Topics in Infinite-Dimensional Topology , MM , Polish Sci. Publ.,Warsaw, 1975.[5] T. Banakh, T. Radul, M. Zarichnyi, Absorbing sets in infinite- dimensional manifolds , Math. Studies,Monog. Ser. , VNTL Publ., Lviv, 1996.[6] M. Brown, A countable connected Hausdorff space , Bull. Amer. Math. Soc. (1953), 367. Abstract Some applications of the topological characterizations of the sigma-compact spaces l f and Σ, Trans. Amer. Math. Soc. :2 (1984) 837–846.[8] R. Engelking, General Topology , Heldermann Verlag, Berlin, 1989.[9] R. Hartshorne, Foundations of projective geometry , Harvard University. W. A. Benjamin, Inc., New York,1967.[10] I.M. Gelfand, D.B. Fuks, The topology of noncompact Lie groups , Funct. Analysis and Its Appl. :4 (1967)285–295.[11] S. Golomb, A connected topology for the integers , Amer. Math. Monthly, (1959), 663–665.[12] S. Golomb, Arithmetica topologica , in: General Topology and its Relations to Modern Analysis and Algebra(Proc. Sympos., Prague, 1961), Academic Press, New York; Publ. House Czech. Acad. Sci., Prague (1962) 179–186; available at https://dml.cz/bitstream/handle/10338.dmlcz/700933/Toposym 01-1961-1 41.pdf ).[13] A.M. Kirch, A countable, connected, locally connected Hausdorff space , Amer. Math. Monthly (1969),169–171.[14] A. Kryftis, A constructive approach to affine and projective planes , Ph.D. Dissertation, Univ. of Cam-bridge, 2015 ( https://arxiv.org/pdf/1601.04998.pdf ).[15] J. van Mill, Infinite-Dimensional Topology, Prerequisites and Introduction , North-Holland Math. Library , Elsevier Sci. Publ. B.V., Amsterdam, 1989.[16] J. Mogilski, Characterizing the topology of infinite-dimensional σ -compact manifolds , Proc. Amer. Math.Soc. :1 (1984) 111–118.[17] K. Sakai, Topology of infinite-dimensional manifolds , Springer (to appear).[18] Ya. Stelmakh, Almost coregular topological spaces , in preparation. T.Banakh: Ivan Franko National University of Lviv (Ukraine), and Jan Kochanowski Univer-sity in Kielce (Poland) E-mail address : [email protected] Y. Stelmakh: Ivan Franko National University of Lviv (Ukraine) E-mail address : [email protected]