ISSN 0001-4346, Mathematical Notes, 2018, Vol. 103, No. 2, pp. 259–270. © Pleiades Publishing, Ltd., 2018. Original Russian Text © E. S. Sukhacheva, T. E. Khmyleva, 2018, published in Matematicheskie Zametki, 2018, Vol. 103, No. 2, pp. 258–272.

On a Homeomorphism between the Sorgenfrey Line S and Its Modification SP E. S. Sukhacheva1, 2* and T. E. Khmyleva1** 1

National Research Tomsk State University, Tomsk, Russia 2

Universite´ de Rouen, France

Received February 11, 2017; in final form, April 20, 2017

Abstract—A topological space SP , which is a modification of the Sorgenfrey line S, is considered. It is defined as follows: if x ∈ P ⊂ S, then a base of neighborhoods of x is the family {[x, x + ε), ε > 0} of half-open intervals, and if x ∈ S \ P , then a base of neighborhoods of x is the family {(x − ε, x], ε > 0}. A necessary and sufficient condition under which the space SP is homeomorphic to S is obtained. Similar questions were considered by V. A. Chatyrko and I. Hattori, who defined the neighborhoods of x ∈ P to be the same as in the natural topology of the real line. DOI: 10.1134/S0001434618010273 Keywords: Sorgenfrey line, point of condensation, Baire space, nowhere dense set, homeomorphism, ordinal, spaces of the first and second category, Fσ -set, Gδ -set.

1. INTRODUCTION We use the following notation: N is the set of positive integers; Z is the set of integers; R is the space of real numbers endowed with the standard Euclidean topology τE ; and S is the Sorgenfrey line (also known as the “arrow” space), that is, the set of real numbers with the topology generated by the base {(a, b] : a, b ∈ R, a < b} and denoted by τ0 . Given sets X, Y ⊂ R, Y ⊂ X, we use XY to denote the topological space in which a base of neighborhoods of a point x is Bx = {(x − ε, x] ∩ X : ε > 0} Bx = {[x, x + ε) ∩ X : ε > 0}

if if

x∈X \Y; x ∈ Y.

For X = S and Y = P , we obtain the space SP , in which the induced topology on P is that of the “right arrow.” We denote the topology on this space by τP . If Y = ∅, then SY = S. Given any set X ⊂ R, by X we denote its closure in the space (R, τE ). Similarly, Int X denotes the interior of X in the }i∈J is a family of pairwise disjoint sets, then we denote the union of these sets by space (R, τE ). If {Ai  A rather than i i∈J i∈J Ai . All definitions and terms used in this paper can be found in [1]. The main result of this paper is the following theorem. Theorem 1. For any subset P of the real line, the following conditions are equivalent: (1) the spaces SP and S are homeomorphic; (2) there exists no set ∅ = V ⊂ P which is closed in P and satisfies the condition V = V \ V ; (3) the set P is both Fσ and Gδ in R. * **

E-mail: [email protected] E-mail: [email protected]

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The existence of a set V ⊂ P mentioned in the theorem depends on the behavior of the residues of A = P \ Int P . The definition of residues of order α can be found in [2, p. 107 (Russian transl.)]. Note that the sets P ⊂ R satisfying condition (2) are precisely the sets which Kuratowski called reducible [2, p. 102 (Russian transl.)]. A proof of the equivalence of conditions (2) and (3) modulo this fact can be found in [2, p. 428 (Russian transl.)]. In this paper, we prove the equivalence of these conditions relying on a proof of the equivalence of conditions (1) and (2). The homeomorphism of the Sorgenfrey line and its modification was considered by Chatyrko and Hattori in [3], where the base of neighborhoods of a point x ∈ P ⊂ S was replaced by a base of neighborhoods of x ∈ (R, τE ). Special cases of Theorem 1 were given in [4] and [5]. The proof of Theorem 1 is presented as follows. In Sec. 2, we prove the implication (2) ⇒ (1) (Theorem 8); in Sec. 3, we prove the implication (1) ⇒ (2) (Theorem 17); and in Sec. 4, we prove the equivalence (2) ⇔ (3). 2. PROOF OF THEOREM 1: (2) ⇒ (1) Let A be a subset of the real line R. For any ordinal α, we define sets Aα , Bα , and Cα as A0 = A,

A1 = A \ A ∩ A,

Aα = Aα−1 \ Aα−1 ∩ Aα−1 , if α is not a limit ordinal and as Aα =



Aβ ,

B1 = A \ A1 ,

Bα = Aα−1 \ Aα , Bα =

β<α



Bβ ,

C1 = A \ A, Cα = Aα−1 \ Aα−1

Cα =

β<α



Cβ

β<α

for any limit ordinal α. The sets Aα are called residues of order α of the set A. Obviously, each residue Aα is closed in Aα−1 , and Aα ⊂ Cα for any α. In the sequel, we use the following properties of these sets. Property 1. For any ordinal α, (Aα )1 = Aα+1 . Property 2. For any ordinal α, Bα ∩ Cα = ∅. / Aα = Cα ∩ Aα−1 . Therefore, x ∈ / Cα . Indeed, if α is not limit and x ∈ Bα = Aα−1 \ Aα , then x ∈  Now, suppose that α is a limit ordinal and Bβ ∩ Cβ = ∅ for all β < α. If x ∈ Bα = β<α Bβ , then  / Cβ , and hence x ∈ / Cα = β<α Cβ . x ∈ Bβ for some β < α. Therefore, by assumption, we have x ∈ Property 3. If β < α, then Aα ⊂ Aβ and Cα ⊂ Cβ . The first inclusion follows directly from the definition of the sets Aα . To prove the second inclusion, it suffices to note that Cα+1 = Aα \ Aα ⊂ Aα ⊂ Cα . Property 4. If an ordinal α is not limit and β < α, then Bα ∩ Bβ = ∅; if α is limit and β < α, then Bα ⊃ Bβ . The former assertion follows from the relations Aβ ∩ Bβ = ∅ and Bα ⊂ Aα−1 ⊂ Aβ , and the latter is obvious. Property 5. If α is a limit ordinal, then   Aβ+1 , Cα = Cβ+1 , Aα = β<α

β<α

Bα =



Bβ+1 = A \ Aα .

β<α

Proof. This follows from the definition of the sets Aα , Bα , and Cα . MATHEMATICAL NOTES

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Property 6. If (a, b) ⊂ R and the endpoints of this interval do not belong to A, then (A ∩ (a, b))α = Aα ∩ (a, b) for any ordinal α. Indeed, note that M ∩ (a, b) = M ∩ (a, b) if M ⊂ R and M does not contain the endpoints of the interval (a, b). Therefore, for α = 1, we have (A ∩ (a, b))1 = A ∩ (a, b) \ (A ∩ (a, b)) ∩ (A ∩ (a, b)) = (A ∩ (a, b)) \ (A ∩ (a, b)) ∩ (A ∩ (a, b)) = (A \ A) ∩ (a, b) ∩ (A ∩ (a, b)) = A \ A ∩ (a, b) ∩ A = A1 ∩ (a, b). Now, suppose that the required assertion holds for all β < α and α is not limit. Then (A ∩ (a, b))α = ((A ∩ (a, b))α−1 )1 = (Aα−1 ∩ (a, b))1 = Aα ∩ (a, b). The last equality holds because the endpoints of (a, b) do not belong to Aα−1 . If α is limit, then   (A ∩ (a, b))β = (Aβ ∩ (a, b)) = Aα ∩ (a, b). (A ∩ (a, b))α = β<α

β<α

Property 7. For any ordinal α, A \ Aα = A \ Cα ; therefore, for any ordinal β < α, we will have Aβ \ Aα = Aβ \ Cα . / Aβ0 . The definition of the set Aβ0 implies Let x ∈ A \ Aα . Take the least ordinal β0 ≤ α for which x ∈ that β0 is not limit and x ∈ Aβ0 −1 . Taking into account the relation Aβ0 = Cβ0 ∩ Aβ0 −1 , we see that x∈ / Cβ0 , i.e., A \ Aα ⊂ A \ Cα . The reverse inclusion is obvious. Property 8. For any ordinal α, the set Bα+1 is closed in R \ Cα+1 . / Aα+1 and x ∈ Bα+1 ⊂ Aα . Assuming that x ∈ / Bα+1 , we Let x ∈ Bα+1 ∩ (R \ Cα+1 ). Then x ∈ obtain x ∈ Aα \ (Aα+1 Bα+1 ) = Aα \ Aα ⊂ Cα+1 , which is impossible, because x ∈ R \ Cα+1 . The following proposition is used to prove the continuity of homeomorphisms constructed in the sequel. ∞ Proposition 2. Suppose that {Gn }∞ n=1 = {(an , bn )}n=1 is a sequence of disjoint intervals, ∞ G = n=1 Gn , M ⊂ R, G ∩ M = ∅, F ⊂ G, and ϕ : SM ∪F → SM is a map satisfying the following conditions:

(a) ϕ(Gn ) = Gn for all n ∈ N; (b) ϕ|Gn is a homeomorphism for all n ∈ N; (c) ϕ(t) = t for t ∈ SM ∪F \ G; / G, n ∈ N, and a sequence {xnk }∞ (d) if x0 ∈ k=1 ⊂ Gn is such that lim xnk = x0

k→∞

in the space SM ∪F ,

then lim ϕ(xnk ) = x0

k→∞

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(e) if y0 ∈ / G and a sequence {ykn }∞ k=1 ⊂ Gn is such that lim (ykn ) = y0

in the space SM ,

k→∞

then lim ϕ−1 (ykn ) = y0

in the space

k→∞

SM ∪F .

Then the map ϕ : SM ∪F → SM is a homeomorphism. Proof. Note that all intervals (an , bn ) are open in (R, τE ) and, hence, in SM ∪F ; therefore,  by virtue of /G= ∞ condition (b), it suffices to prove the continuity of ϕ : SM ∪F → SM at the points x ∈ n=1 (an , bn ). / G. The following cases are possible. Let x0 ∈ Case 1: x0 ∈ M and x0 = an for some n ∈ N. If the sequence of xk converges to x0 in the space SM ∪F as k → ∞, then we have xk ∈ (an , bn ) = Gn for sufficiently large k, because the base neighborhoods of x0 have the form [x0 , x0 + ε). By conditions (c) and (d), ϕ(xk ) → x0 = ϕ(x0 ) in SM ; thus, the map ϕ is continuous at x0 .

Case 2: x0 ∈ M and x0 ∈ / {an : n ∈ N}. Consider the sequence {xk }∞ k=1 , for which we have limk→∞ xk = x0 in the space SM ∪F . Without loss of generality, we can assume that x0 < xk and either / G for all k ∈ N or xk ∈ G for all k ∈ N. In the former case, property (c) implies xk ∈ lim ϕ(xk ) = lim xk = x0

k→∞

k→∞

in SM .

In the latter case, for each k ∈ N, we have xk ∈ (ank , bnk )

and

x0 < ank < xk ;

therefore, limk→∞ ank = x0 . Since different intervals (ank , bnk ) are disjoint, we have limk→∞ bnk = x0 , too. By property (a), we have ϕ(xk ) ∈ (ank , bnk ), whence lim ϕ(xk ) = lim ank = lim bnk = x0 = ϕ(x0 ).

k→∞

k→∞

k→∞

This proves the continuity of the map ϕ at the point x0 .

Case 3: x0 ∈ / M and x0 = bn for some n ∈ N. As in Case 1, the continuity of ϕ at x0 follows from conditions (c) and (d). Case 4: x0 ∈ / M and x0 ∈ / {bn : v ∈ / N}. The continuity of the map ϕ at x0 is proved in the same way as in Case 2. Replacing ϕ by ϕ−1 and using condition (e) instead of (d), we see that the map ϕ−1 is continuous as well. In Propositions 3–5, we construct homeomorphisms in particular cases. Proposition 3. Let M ⊂ R, and let (a, b) ⊂ R be an interval for which M ∩ (a, b) = ∅. Then there exists a homeomorphism ϕ : SM ∪(a,b) → SM satisfying the condition ϕ(x) = x

for

x∈ / (a, b).

∞ Proof. Consider a sequence {dn } n=−∞ ⊂ (a, b) with the properties inf n dn = a, supn dn = b, and dn < dn+1 for all n ∈ Z. Note that ∞ n=−∞ [dn , dn+1 ) = (a, b). Let us define a map ϕ : SM ∪(a,b) → SM by  −x + dn + dn+1 if x ∈ [dn , dn+1 ); ϕ([dn , dn+1 )) = x if x ∈ / (a, b).

/ (a, b), then ϕ(x) = x. Let us prove the Obviously, we have ϕ([dn , dn+1 )) = (dn , dn+1 ], and if x ∈ continuity of ϕ. Since the half-open intervals [dn , dn+1 ) and (dn , dn+1 ] are clopen in SM ∩(a,b) and SM , respectively, and the map ϕ is a homeomorphism on each of the half-open intervals [dn , dn+1 ), it suffices / (a, b). Let x0 ∈ / (a, b), and let {xk }∞ to verify the continuity of ϕ and ϕ−1 at the points x ∈ k=1 be a MATHEMATICAL NOTES

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sequence such that limk→∞ xk = x0 in the space SM ∩(a,b) . Without loss of generality, we can assume that either xk ∈ / (a, b) for all k ∈ N or xk ∈ (a, b) for all k ∈ N. In the former case, we have lim ϕ(xk ) = lim xk = x0 = ϕ(x0 ),

k→∞

k→∞

i.e., the map ϕ is continuous at x0 . Suppose that xk ∈ (a, b) for all k ∈ N. In this case, either x0 = a / M . Since xk ∈ [dnk , dnk+1 ) for each k ∈ N and limk→∞(dn+1 − dnk ) = 0, and a ∈ M or x0 = b and b ∈ it follows that lim xk = lim dnk = lim dnk+1 = x0 ,

k→∞

k→∞

k→∞

and since ϕ(xk ) ∈ (dnk , dnk+1 ], it follows that lim ϕ(xk ) = x0 = ϕ(x0 ),

k→∞

i.e., the map ϕ is continuous at x0 . The proof of the continuity of ϕ−1 is similar. Proposition 4. For any set P ⊂ R, there exists a homeomorphism ϕ : SP → SP \Int P such that ϕ(x) = x for all x ∈ / Int P .  Proof. Let us represent the set Int P , which is open in R, as a union ∞ n=1 (an , bn ). According to Proposition 3, for each n ∈ N, there exists a homeomorphism ϕn : S(an ,bn ) → S such that ϕn (x) = x for x∈ / (an , bn ). Consider the map ϕ : SP → SP \ Int P defined by  ϕn (t) if t ∈ (an , bn ), ϕ(t) = t if t ∈ / Int P. It is easy to see that ϕ satisfies all assumption of Proposition 2 and, hence, is a homeomorphism. Proposition 5. Let F be a closed subset of an interval (a, b) ⊂ R. Then there exists a homeomor/ (a, b). phism ϕ : SF → S such that ϕ(x) = x for all x ∈ Proof. Since the set F is closed in (a, b), it follows that (a, b) \ F =

∞ 

(an , bn ).

n=1

According to Proposition 3, for each n ∈ N, there exists a homeomorphism ϕn : SF → SF ∪(an ,bn ) such / (an , bn ). The map ϕ0 : SF → S(a,b) defined by that ϕn (x) = x for x ∈  ϕn (t) if t ∈ (an , bn ), ϕ0 (t) =  t if t ∈ / ∞ n=1 (an , bn ) satisfies all assumptions of Proposition 2 and, hence, is a homeomorphism. By Proposition 3, there ex/ (a, b). The composition ϕ = ϕ1 ◦ ϕ0 ists a homeomorphism ϕ1 : S(a,b) → S such that ϕ(x) = x for x ∈ is the required homeomorphism of SF onto S. Proposition 6. Suppose that a set A ⊂ R is such that Int A = ∅ and there exists no set ∅ = V ⊂ A closed in A and satisfying the condition V = V \ V . Then Int A = ∅, i.e., the set A is nowhere dense in R. Proof. Suppose that, on the contrary, there exists an interval [a, b] ⊂ A. We set V = [a, b] ∩ A and claim that V = [a, b]. Indeed, if x ∈ (a, b) ⊂ A, then there exists a sequence {xn }∞ n=1 ⊂ (a, b) ∩ A converging to x. In this case, we have x ∈ (a, b) ∩ A ⊂ [a, b] ∩ A = V and, therefore, [a, b] ⊂ V . The reverse inclusion is obvious; thus, V = [a, b]. MATHEMATICAL NOTES

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Let us show that V \ V = [a, b]. Since Int A = ∅, it follows that, for each point x ∈ (a, b) ⊂ A, there is a sequence {xn } ⊂ (a, b) \ A that converges to it. Taking into account the relations [a, b] \ A = [a, b] \ V and V = [a, b], we obtain x ∈ (a, b) \ A ⊂ [a, b] \ A = [a, b] \ V = V \ V . This implies [a, b] ⊂ V \ V . On the other hand, V \ V ⊂ V = [a, b], whence V \ V = [a, b]. Thus, we have found a set V closed in A and satisfying the condition V = V \ V , which contradicts the assumption. Theorem 7. Let A be a nowhere dense subset of the line R contained in an interval I = (a, b). If Aα = ∅ for a countable ordinal α, then there exists a homeomorphism ϕA : SA → S such that / I. ϕA (t) = t for all t ∈ Proof. We prove this theorem by transfinite induction on α. Let α = 1, i.e., A1 = ∅. Then A = A1 B1 = B1 . We represent the open set I \ C1 as the union of disjoint intervals In = (an , bn ). By virtue of Property 8, the set B1 is closed in R \ C1 ; therefore, B1 ∩ In is closed in In for any n ∈ N. Proposition 5 implies the existence of a homeomorphism ϕn : SA∩In → S / In . According to Proposition 2, the map ϕA,I defined by such that ϕn (t) = t for t ∈  ϕn (t) if t ∈ In , ϕA,I (t) =  t if t ∈ / n∈N In is a homeomorphism between SA and S. Suppose that the assertion of the theorem holds for all β < α, i.e., for any nowhere dense A ⊂ I with / I. Aβ = ∅, there exists a homeomorphism ϕA,I : SA → S such that ϕA,I (t) = t for t ∈ Let us prove this assertion for α. There are two possible cases.

Case 1: α = β + 1, i.e., the ordinal α is not limit. Then we have Aβ = Aα Bα = Bα . Let us represent the open set I \ Cβ as the union of disjoint intervals In = (an , bn ). Since A is nowhere dense, it follows that there exists a sequence of disjoint intervals Ink = (akn , bkn ) ⊂ In satisfying the conditions ∞  A ∩ In ⊂ (akn , bkn ), inf akn = an , sup bkn = bn , k

k=−∞

< bk+1 akn < bkn < ak+1 n n

k

for any k ∈ Z.

For each interval Ink , the points akn and bkn do not belong to A; therefore, Property 8 implies (A ∩ Ink )β = Aβ ∩ Ink ⊂ Aβ ∩ In ⊂ Cβ ∩ In = ∅. By the induction hypothesis, there exists a homeomorphism ϕA,Ink : SA∩Ink → S such that ϕA,Ink (t) = t for t ∈ / Ink . Consider the map ϕA\Aβ : SA → SAβ defined by  ϕA,Ink (t) if t ∈ Ink , ϕA\Aβ (t) =  t if t ∈ / n∈N, k∈Z Ink . According to Proposition 2, this map is a homeomorphism. We have (Aβ )1 = Aβ+1 = ∅; applying the theorem to the set Aβ and α = 1, we see that there exists a homeomorphism ϕAβ : SAβ → S such / I. Thus, the map ϕA = ϕAβ ◦ ϕA\Aβ : SA → S is the required homeomorphism. that ϕAβ (t) = t for t ∈   Case 2: α is limit. In this case, Aα = ∅ and Aα = β<α Aβ . Let I \ Cα = ∞ n=1 (an , bn ). As in k k k Case 1, for each n ∈ N, we define a sequence of intervals In = (an , bn ) ⊂ (an , bn ) = In so that A ∩ In ⊂

∞ 

(akn , bkn ) ⊂ (an , bn ) ⊂ I \ Cα .

k=−∞

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For any n ∈ N and k ∈ Z, consider the system {[akn , bkn ] ∩ Cβ }β<α of closed subsets. If [akn , bkn ] ∩ Cβ = ∅ for all β < α, then, by virtue of Property 3, this system has the finite intersection property in the compact set [akn , bkn ]. Therefore, 

 k k k k Cβ = ([akn , bkn ] ∩ Cβ ) = ∅. [an , bn ] ∩ Cα = [an , bn ] ∩ β<α

β<α

But this is impossible, because [akn , bkn ] ∩ Cα = ∅. Thus, for each interval [akn , bkn ], there exists an ordinal βnk < α for which [akn , bkn ] ∩ Cβnk = ∅. Since the endpoints of the interval [akn , bkn ] do not belong to A, it follows from Property 6 that ((akn , bkn ) ∩ A)βnk = (akn , bkn ) ∩ Aβnk ⊂ [akn , bkn ] ∩ Cβnk = ∅. By the induction hypothesis, there exists a homeomorphism ϕA,Ink : SA∩Ink → S such that ϕA,Ink (t) = t for t ∈ / Ink . By Proposition 2, the map  ϕA,Ink if t ∈ Ink , ϕA (t) =  t if t ∈ / n∈N, k∈Z Ink is the required homeomorphism. Theorem 8. Let P ⊂ R. If there exists no set ∅ = V ⊂ P closed in P and satisfying the condition V = V \ V , then the spaces SP and S are homeomorphic. Proof. By Proposition 4, the space SP is homeomorphic to the space SP \Int P . We set A = P \ Int P . Since A is closed in P , it follows that there exists no set V ⊂ A which is closed in A and satisfies the condition V = V \ V . Proposition 6 implies Int A = ∅. Consider the system {Cα ; α < ω1 } of closed sets defined for A. By Property 3, this system is decreasing, and the Baire–Hausdorff theorem [6, p. 162] implies the existence of an ordinal α0 < ω1 for which Cα0 = Cα0 +1 = · · · . Assuming that Cα0 = ∅ and applying the relation Cα0 = Cα0 +1 = Aα0 \ Aα0 ⊂ Aα0 ⊂ Cα0 , we see that Aα0 \ Aα0 = Aα0 = ∅. Setting V = Aα0 , we obtain a contradiction to the assumption of the theorem, because Aα0 is closed in A and, hence, in P . Thus, Cα0 = ∅, which implies Aα0 = ∅. Since the set A is nowhere dense in R, it follows that there exists a sequence {xn }+∞ n=−∞ satisfying the conditions inf xn = −∞, n

sup xn = +∞, n

xn < xn+1 ,

xn = A,

n ∈ Z.

Property 6 implies (A ∩ (xn , xn+1 ))α = Aα ∩ (xn , xn+1 ) = ∅. According to Theorem 7, for each n ∈ Z, there exists a homeomorphism ϕn : SA∩(xn ,xn+1 ) → S such  / In . Since SA = ∞ that ϕn (t) = t for all t ∈ n=−∞ (xn , xn+1 ] and the half-open intervals (xn , xn+1 ] are clopen subsets of SA for all n ∈ Z, it follows that the map ϕ : SA → S defined by ϕ(t) = ϕn (t) for t ∈ (xn , xn+1 ] is a homeomorphism. MATHEMATICAL NOTES

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3. PROOF OF THEOREM 1: (1) ⇒ (2) Recall that a topological  space X is said to be Baire if, given any sequence of open dense sets Gn ⊂ X, the intersection ∞ n=1 Gn is dense in X. Proposition 9. The space SP is Baire for any P ⊂ R. Proof. Let {Gn }∞ n=1 be a sequence of open dense subsets of SP . Each set Gn is a union of disjoint intervals of the forms (x1 , x2 ], [x3 , x4 ), (x5 , x6 ), and [x7 , x8 ]. Replacing the intervals of the forms (x1 , x2 ], n , which are open [x3 , x4 ), and [x7 , x8 ] by (x1 , x2 ), (x3 , x4 ), and (x7 , x8 ), respectively, we obtain sets G ∞  ∞ and dense in R. Since the space R is Baire, it follows that n=1 Gn ⊂ n=1 Gn is dense in R and, hence, in SP . All open sets and dense Gδ -sets in a Baire space are Baire [7, p. 228]. This implies the following corollary. Corollary 10. The following assertions hold: (1) the space (T, τP ) is Baire for any set P ⊂ R and any dense Gδ -set T ⊂ SP ; in particular, the space (J, τP ) of irrational points is Baire; 2) if G is an open subset of SP , then (G, τP ) is Baire. Proposition 11. Let X1 and X2 be subsets of a topological space X. If X is Baire and X = X1 X2 , then at least one of the spaces X1 and X2 is of the second category. Proof. Suppose that both spaces X1 and X2 are of the first category, i.e., X1 =

∞ 

An ,

n=1

∞ 

X2 =

Bn ,

n=1

where An and Bn are nowhere dense in X1 and X2 , respectively, for each n ∈ N. Then, obviously, the sets An and Bn are also nowhere dense in



 ∞ ∞ An

Bn . X= n=1

n=1

This means that X is a set of the first category, which is impossible, because X is Baire. The idea of the proof of the following lemma is borrowed from [8]. Lemma 12. Let X be a subset of SP contained in P or in SP \ P , and let ϕ : X → S be a homeomorphism. Then X can be represented as a countable union of closed subsets of X on each of which the map ϕ is decreasing or increasing, respectively. Proof. Let X ⊂ P . For each n ∈ N, we set

1 =⇒ ϕ(y) < ϕ(x) . Fn = x ∈ X : x < y < x + n Let us prove that the set Fn is closed in X. Indeed, if x0 ∈ X is a limit point of Fn , then there exists a sequence {xi }∞ i=1 ⊂ Fn such that xi+1 < xi for every i ∈ N and limi→∞ xi = x0 . For each point y ∈ (x0 , x0 + 1/n), there exists a number i0 ∈ N such that xi < y for all i > i0 . Since xi ∈ Fn and xi < y < x0 + 1/n < xi + 1/n, it follows that ϕ(y) < ϕ(xi ) for each i ∈ N and, therefore, ϕ(y) ≤ ϕ(x0 ). Since ϕ is a homeomorphism, wehave ϕ(y) < ϕ(x0 ), whence x0 ∈ Fn . This proves the closedness of Fn in X. It is easy to see that X = ∞ n=1 Fn . k ∞ Now, let {In }k=1 be a countable family of closed intervals shorter than 1/n which forms a cover of  the set Fn . Then the Fnk = Fn ∩ Ink are closed subsets of X, and X = n,k∈N Fnk . Let us show that MATHEMATICAL NOTES

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the map ϕ decreases on each of the sets Fnk . Take x, y ∈ Fnk , x < y. Since x, y ∈ Ink , it follows that x < y < x + 1/n, and since x ∈ Fn , it follows that ϕ(y) < ϕ(x). In the case X ⊂ SP \ P , we consider the subsets

1 Fn = x ∈ X : x − < y < x =⇒ ϕ(y) < ϕ(x) n of X and prove that the map ϕ increases on each Fnk in the same way as in the case considered above. Proposition 13. Let V be a subset of the real line satisfying the condition V = V \ V . Then the following assertions hold: a) the set V is perfect, i.e., it is closed and has no isolated points; b) all points of V are points of condensation, i.e., any neighborhood of any point x ∈ V contains uncountably many points of V ; c) if V ∩ (a, b) = ∅, then V ∩ (a, b) and the set V \ V ∩ (a, b) is dense in V ∩ (a, b);  d) if R \ V = ∞ n=1 (an , bn ) and D = {an , n ∈ N} ∪ {bn , n ∈ N}, then the set V \ D is dense in V , and for any interval (a, b), the set (V \ D) ∩ (a, b) is dense in V ∩ (a, b). Proof. Assertion (a) follows from the relation V = V \ V . The proof of (b) can be found in [6, p. 147]. Assertion (c) is easy to verify, and (d) follows from (b) and (c). Now, consider a perfect set F ⊂ R and let D denote ∞ the set of the endpoints of all intervals adjacent to F , i.e., D = {an , n ∈ N} ∪ {bn , n ∈ N}, where n=1 (an , bn ) = R \ F . Note that only points of D can be isolated in (F, τ0 ) or (F, τS ). Proposition 14. Let F be a perfect nowhere dense subset of R, and let D be the set of the endpoints of all intervals adjacent to F . Then there exists a homothety ϕ between F \ D and the set J ⊂ R of all irrational numbers, and the following assertions hold: (1) the map ϕ : (F \ D, τE ) → (J, τE ) is a homeomorphism; (2) the map ϕ : (F \ D, τP ) → (J, τϕ(P ∩(F \D)) ) is a homeomorphism for any set P ⊂ R. Proof. As is well known [6, p. 147], the set F \ D is similar to the set of all irrational points, i.e., there exists a map ϕ : F \ D → J such that ϕ(x) < ϕ(y) if and only if x < y. This implies both assertions (1) and (2), because the topologies τE , τ0 , and τS are defined by using the natural ordering of the real line and the set F \ D has no isolated points in each of these topologies. This proposition and Corollary 10 imply the following assertion. Corollary 15. For any set P ⊂ R, the space (F \ D, τP ) is Baire and dense in itself. In the sequel, we use the following notation: given a set W ⊂ R, by (W, τE ) we denote this set treated as a subspace of the topological space (R, τE ). By (W, τS ), (W, τ0 ), and (W, τP ) we mean that W is treated as a subspace of S, S0 , and SP , respectively. Proposition 16. Let P ⊂ R, and let V ⊂ P be a closed set in P satisfying the condition V = V \ V . Then there exists a (possibly infinite) interval (a, b) containing either a set W1 ⊂ V ∩ (a, b) dense in (V ∩ (a, b), τE ) for which (W1 , τP ) is a dense-in-itself space of the second category or a set W2 ⊂ (V \ V ) ∩ (a, b) dense in (V ∩ (a, b), τE ) for which (W2 , τP ) is a dense-in-itself space of the second category. MATHEMATICAL NOTES

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Proof. Consider the following three cases.

Case 1: the set V is not nowhere dense in R. This means that there exists an interval (a, b) ⊂ V . Since (a, b) is open in SP , it follows by Corollary 10 that the space ((a, b), τP ) is Baire, and since (a, b) ∩ V = ((a, b) ∩ V ) ((a, b) ∩ V \ V ), it follows by Proposition 11 that one of the sets W1 = V ∩ (a, b) and W2 = (V \ V ) ∩ (a, b) is of the second category in the topology τP . Obviously, W1 and W2 are dense subsets of V ∩ (a, b) in the topology τE . Let us show that these sets are dense in itself in the topology τP . Indeed, let x0 ∈ W1 , and let ε > 0 be chosen so that (x0 , x0 + ε) ⊂ (a, b). Since (a, b) ⊂ V , it follows that the interval (x0 , x0 + ε) contains points of the set V , which means that W1 is dense in itself in the topology τP . Similarly, if x0 ∈ W2 and ε > 0 is such that (x0 − ε, x0 ) ⊂ (a, b), then the interval (x0 − ε, x0 ) contains points of the set V \ V , because (a, b) ⊂ V = V \ V , and hence the set W2 is dense in itself in the topology τP .

Case 2: the set V is nowhere dense in R and both sets V \ D and (V \ V ) \ D are dense in (V \ D, τP ). Note that the space (V \ D, τP ) is dense in itself, and hence so are the subspaces V \ D and (V \ V ) \ D. The space (V \ D, τP ) is Baire by Corollary 15; therefore, by Proposition 11, either W1 = V \ D or W2 = (V \ V ) \ D is a space of the second category in the topology τP . By assumption, W1 and W2 are also dense in V \ D in the topology τE ; therefore, by Proposition 13 (d), W1 and W2 are dense subsets of V in the topology τE . Case 3: the set V is nowhere dense in R and at least one of the sets V \ D and (V \ V ) \ D is not dense in (V \ D, τP ). Suppose that the set (V \ V ) \ D is not dense in (V \ D, τP ). This means that there exists a point x0 ∈ V \ D and a number ε > 0 for which [x0 , x0 + ε) ∩ ((V \ V ) \ D) = ∅.

(1)

We set W1 = (x0 , x0 + ε) ∩ (V \ D) = (x0 , x0 + ε) ∩ (V \ D). By Corollary 15, the space ((V \ D), τP ) is Baire and dense in itself; therefore, its open subset W1 ⊂ ((V \ D), τP ) is Baire (by Corollary 10) and dense in itself as well. Let us show that W1 is a dense subset of V ∩ (x0 , x0 + ε) in the topology τE . Let us choose a point x0 ∈ (V ∩ (x0 , x0 + ε). By Proposition 13 (d), there exists a sequence {xn }∞ n=1 ⊂ (V \ D) ∩ (x0 , x) + ε) such that xn → x in τE . Since (V \ D) ∩ (x0 , x + ε) = ((V \ D) ∩ (x0 , x + ε)) (((V \ V ) \ D) ∩ (x0 , x + ε)), it follows by virtue of (1) that xn = (V \ D) ∩ (x0 , x + ε) = W1 , and hence the set W1 is dense in the set (V ∩ (x0 , x + ε), τE ). If the set V \ D is not dense in (V \ D, τP ), then there exists a point x0 ∈ (V \ V ) \ D such that [x0 , x0 + ε) ∩ (V \ D) = ∅ for some ε > 0. An argument similar to that used above proves that the set W2 = (x0 , x0 + ε) ∩ ((V \ V ) \ D) = (x0 , x0 + ε) ∩ (V \ D) is as required. Theorem 17. Let P ⊂ R. If there exists a set ∅ = V ⊂ P closed in P and satisfying the condition V = V \ V , then the spaces SP and S are not homeomorphic. Proof. Suppose that there exists a homeomorphism ϕ : SP → S. In view of Proposition 16, we divide the proof of the theorem into two cases.

Case 1: there exists a set W1 ⊂ V ∩ (a, b) satisfying allconditions in Proposition 16. By Proposition 12, there exists closed sets Fn ⊂ W1 such that W1 = n∈N Fn and the map ϕ|Fn is decreasing for each n ∈ N. Since the space (W1 , τP ) is of the second category, it follows that Fn is nowhere dense MATHEMATICAL NOTES

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for some n ∈ N. Hence there exists an x0 ∈ Fn for which W1 ∩ [x0 , x0 + ε) ⊂ Fn , so that the map ϕ is decreasing on W1 ∩ [x0 , x0 + ε). Since the set (W1 , τP ) is dense in itself, it follows that there exist points x1 , x2 ∈ W1 ∩ [x0 , x0 + ε) for which x0 = x1 < x2 . The equality V = V \ V implies the existence of points y1 , y1 ∈ (V \ V ) ∩ (x0 , x0 + ε),

y1 = y2 ,

y1 < y2 .

/ P , because the set V is closed in P . Since W1 is dense in V ∩ (a, b) in the Note that y1 , y2 ∈ topology τE , it follows that there exists a sequence {yni }∞ n=1 ⊂ W1 ∩ (x0 , x0 + ε) converging to yi in τE for i = 1, 2. Without loss of generality (passing to subsequences if needed), we can assume that the 2 ∞ 1 2 sequences {yn1 }∞ n=1 and {yn }n=1 are monotone and satisfy the condition yn < yn for any n, k ∈ N. 2 ∞ If one of the sequences {yn1 }∞ n=1 and {yn }n=1 is increasing, i.e., converges to the corresponding i point yi in τP , then the sequence {ϕ(yn )}∞ n=1 converges to ϕ(yi ) in the topology τ0 , because the map ϕ : SP → S is continuous. On the other hand, since ϕ decreases on the set W1 ∩ [x0 , x0 + ε), it follows that the sequence {ϕ(yni )}∞ n=1 is decreasing and, therefore, divergent in the space (S, τ0 ). This contradiction proves that the map ϕ is discontinuous. 2 ∞ Now, suppose that both sequences {yn1 }∞ n=1 and {yn }n=1 are decreasing, i.e., divergent in SP . 2 ∞ Since ϕ is a homeomorphism, it follows that the sequences {ϕ(yn1 )}∞ n=1 and {ϕ(yn )}n=1 are divergent in S. On the other hand, the map ϕ decreases on the set W1 ∩ [x0 , x0 + ε); therefore, the 2 ∞ 2 1 sequences {ϕ(yn1 )}∞ n=1 and {ϕ(yn )}n=1 are increasing, and ϕ(yn ) < ϕ(yn ) for any n ∈ N. Thus, the 2 ∞ increasing sequence {ϕ(yn )}n=1 , which is bounded above, is divergent in (S, τ0 ); we have obtained a contradiction.

Case 2: there exists a set W2 ⊂ (V \ V ) ∩ (a, b) satisfying all conditions in Proposition 16. In this case, the proof is similar to that in Case 1. 4. PROOF OF THEOREM 1: (2) ⇔ (3) (3) ⇒ (2). Let P be an Fσ - and Gδ -set in R. Suppose that there exists a set ∅ = V ⊂ P which is closed in P and satisfies the condition V = V \ V . Then V is an Fσ - and Gδ -set in R. Since V is Fσ , it  F follows that V = ∞ n=1 n , where the Fn ⊂ R are closed in R for all n ∈ N. Since V is Gδ , it follows that it is completely metrizable, which implies the existence of a number n ∈ N for which IntV Fn = ∅, i.e., such that [x0 − ε, x0 + ε] ∩ V ⊂ Fn for some ε > 0. We see that the set [x0 − ε, x0 + ε] ∩ V = [x0 − ε, x0 + ε] ∩ V ∩ Fn = [x0 − ε, x0 + ε] ∩ Fn is compact. Let y0 ∈ (V \ V ) ∩ (x0 − ε, x0 + ε). Then there exists a sequence {yn }∞ n−1 such that yn ∈ V ∩ (x0 − ε, x0 + ε)

and

lim yn = y0 .

n→∞

/ V ,. This contradicts the compactness of [x0 − ε, x0 + ε] ∩ V , because y0 ∈ (2) ⇒ (3). Since the set A = P \ Int P is closed in P , it follows that A contains no subsets V = ∅ which are closed in A and satisfy the condition V = V \ V . According to Theorem 8, there exists an ordinal α < ω1 for which Cα = ∅. Hence   (Aβ \ Aβ+1 ) = Bβ+1 . A= 0≤β<α

0≤β<α

By virtue of Property 8, the set Bβ+1 is closed in R \ Cβ+1 for each ordinal β and, therefore, is an Fσ -set in R. Any countable union of Fσ -sets is Fσ ; thus, the sets A and, hence, P are Fσ in R. Let us show that the set A is Gδ . To this end, we prove by transfinite induction that A \ Cγ is a Gδ -set for each ordinal γ. If γ = 1, then the set A \ C1 = A \ A1 = B1 is closed in R \ C1 ; hence this is a Gδ -set. MATHEMATICAL NOTES

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Suppose that the set A \ Cβ is Gδ for all β < γ. If the ordinal γ is not limit, then, according to Property 7, the set A \ Cγ = (A \ Aγ−1 ) ∪ (Aγ−1 \ Aγ ) = (A \ Cγ−1 ) ∪ Bγ is Gδ , because A \ Cγ−1 is Gδ by the induction hypothesis and Bγ is closed in R \ Cγ and, hence, is Gδ as well. Now, suppose that the ordinal γ is limit. As is known, a set in a separable metric space is Gδ if it is Gδ at each of its points [2, p. 366 (Russian transl.)]. Take x ∈ A \ Cγ . Since  Bβ+1 , A \ Cγ = A \ Aγ = 0≤β<γ

it follows that x ∈ Bβ0 +1 for some β0 < γ, and since Bβ0 +1 ∩ Cβ0 +1 = ∅, it follows that there exists an ε > 0 for which (x − ε, x + ε) ∩ Cβ0 +1 = ∅. For β ≥ β0 + 1, we have Bβ+1 ⊂ Aβ ⊂ Cβ ⊂ Cβ0 +1 and, therefore,

(x0 − ε, x0 + ε) ∩





Bβ+1

= ∅.

β0 +1≤β<γ

This implies (x − ε, x + ε) ∩ (A \ Cγ ) = (x − ε, x + ε) ∩



Bβ+1

= (x − ε, x + ε) ∩ (A \ Cβ0 +1 ).

β0 +1≤β<γ

By the induction hypothesis, the sets A \ Cβ0 +1 , and, therefore, (x − ε, x + ε) ∩ (A \ Cβ0 +1 ) are Gδ in R. Thus, A \ Cγ is a Gδ -set at each of its points in R. It was proved in [9] that a set X ⊂ S is homeomorphic to S if and only if X is a dense-in-itself Fσ and Gδ -set in S. This implies the following assertion. Corollary 18. Let P ⊂ R be a set for which the spaces SP and S are homeomorphic. Then the following assertions hold: (1) if P is a dense-in-itself subset of (R, τS ), then P and S are homeomorphic; (2) if S \ P is a dense-in-itself subset of S, then S \ P and S are homeomorphic. ACKNOWLEDGMENTS The authors thank Professor A. Bouziad from Rouen University for useful comments. REFERENCES 1. R. Engelking, General Topology (PWN, Warsaw, 1977; Mir, Moscow, 1986). 2. K. Kuratowski, Topology (Academic, New York–London; PWN, Warsaw, 1966; Mir, Moscow, 1966), Vol. 1 [in Russian]. 3. V. A. Chatyrko and Y. Hattori, “A poset of topologies on the set of real numbers,” Comment. Math. Univ. Carolin. 54 (2), 189–196 (2013). 4. E. C. Sukhacheva and T. E. Khmyleva, “On linearly ordered topological spaces homeomorphic to the Sorgenfrey line,” Vestnik Tomsk. Gos. Univ., Mat. Mekh., No. 5 (31), 63–68 (2014). 5. T. E. Khmyleva, “On a homeomorphism between the Sorgenfrey line and it modification SQ ,” Vestnik Tomsk. Gos. Univ., Mat. Mekh., No. 1 (39), 53–56 (2016). 6. P. S. Aleksandrov, Introduction to Set Theory and General Topology (Nauka, Moscow, 1977) [in Russian]. 7. V. V. Tkachuk, A Cp -Theory Problems Book. Topological and Function Spaces (Springer, New York, 2011). 8. E. K. van Douwen, “Retracts of the Sorgenfrey line,” Compositio Math. 38 (2), 155–161 (1979). 9. D. K. Burke and J. T. Moore, “Subspaces of the Sorgenfrey line,” Topology Appl. 90 (1-3), 57–68 (1998). MATHEMATICAL NOTES

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