aa r X i v : . [ m a t h . GN ] J u l OPEN IMAGES OF THE SORGENFREY LINE
VLAD SMOLIN
Abstract.
We give a description of Hausdorff continuous open imagesof the Sorgenfrey line: these are precisely those spaces that have a Sor-genfrey base. Using this description we prove that no Hausdorff compactspace that contains a copy of the Sorgenfrey line is a continuous openimage of it; in particular the double-arrow space is not a continuous openimage of the Sorgenfrey line.
Keywords:
Sorgenfrey line, Souslin scheme, open map, double-arrowspace, Hausdorff compact space Introduction
A continuous map is called open if the image of an open set under this map isopen.Results in the paper arose from the questions that were posed to the author byE. G. Pytkeev and M. A. Patrakeev:
Question 1.
Is the double-arrow (two-arrow) space a continuous open image ofthe Sorgenfrey line?
Question 2.
Suppose that a compact space is a continuous open image of theSorgenfrey line. What can we say about this space?
Previously, continuous open images of the Sorgenfrey line were studied in theclass of metrizable spaces. In [7] S. A. Svetlichnyi proved that if a metrizable spaceis a continuous open image of the Sorgenfrey line, then it is a Polish space, i.e.separable completely metrizable space. In [5] and [6] N. V. Velichko and M. A.Patrakeev independently constructed a continuous open map from the Sorgenfreyline onto the real line. Velichko also proved that for each such map there existsa point with the preimage of cardinality continuum. In [3] Patrakeev proved thatcontinuous open metrizable images of the Sorgenfrey line are exactly Polish spaces.He also strengthened the result of Velichko by showing that for each continuousopen map from the Sorgenfrey line onto the metrizable space there exists a pointwith the preimage of cardinality continuum.Continuous open images of submetrizable spaces were studied by Svetlichnyi. In[10] he proved that if a paracompact space is an image of a submetrizable spaceunder continuous open compact map, then it is a submetrizable space. He alsoproved that there exists a nonmetrizable (hence not submetrizable) compact spacethat is a continuous open s-image of a submetrizable space. Since the Sorgenfreyline is a hereditarily separable submetrizable space it is natural to ask the followingquestion.
Question 3.
Is there exists a Hausdorff nonmetrizable compact space that is acontinuous open image of the Sorgenfrey line?
We give a description of Hausdorff open images of the Sorgenfrey line: theseare precisely those spaces that have a Sorgenfrey base (see Theorem 1). Using thisdescription we prove that no compact space that contains a homeomorphic copy ofthe Sorgenfrey line is an open image of it (see Corollary 4). This result answersQuestion 1 negatively. But the following question remains open.
Question 4.
Is there exists a Hausdorff compact space that contains an uncountablesubspace of the Sorgenfrey line and that is a continuous open image of the Sorgenfreyline? Notation and terminology
We use terminology from [1], [2], and [4]. Also we use the following notations:
Notation 1.
The symbol := means “equals by definition”; the symbol : ←→ is usedto show that the expression on the left side is an abbreviation for expression on theright side; • ω := the set of finite ordinals = the set of natural numbers; • ∅ ∈ ω ; • n = { , . . . , n − } for all n ∈ ω ; • s is a sequence : ←→ s is a function such that dom ( s ) ∈ ω or dom ( s ) = ω ; • if s is a sequence, then lh ( s ) := dom ( s ); • h s , . . . , s n − i := the sequence s such that lh ( s ) = n ∈ ω and s ( i ) = s i forall i ∈ n ; • hi := the sequence q such that lh ( q ) = 0; • if s = h s , . . . , s n − i , then s ˆ x := h s , . . . , s n − , x i ; • f ↾ A := the restriction of function f to A ; • if s and t are sequences, then s ⊑ t : ←→ s = t ↾ lh ( s ); • s ⊏ t : ←→ s ⊑ t and s = t ; • B A := the set of functions from B to A ; • <ω A := S n ∈ ω n A = the set of finite sequences in A . Notation 2.
Let R be a binary relation on X and x, y ∈ X . Then • x ↑ R := { z ∈ X : xRz } ; • x ↓ R := { z ∈ X : zRx } ; • ( x, y ) R := x ↑ R ∩ y ↓ R .Now we introduce several relations on <ω ω and ω ω . Notation 3.
Let a, b ∈ <ω ω ∪ ω ω . Then • a ⊳ b : ←→ ∃ n ∈ ω such that – a ↾ n = b ↾ n and – a ( n ) < b ( n ). • a E b : ←→ a ⊳ b or a = b . Notation 4.
Let h X, τ i be a topological space, x ∈ X , B ⊆ X , and A ⊆ R . Then • ω ω := the Baire space of weight ℵ := the countable power of the discretespace of cardinality ℵ ; • the double-arrow (two-arrow) space := h M, τ A i , where M = {h x, i : 0 Let V = h V a i a ∈ <ω ω be a Souslin scheme on a set X and p ∈ ω ω .Then • fruit ( V , p ) := T n ∈ ω V p ↾ n ; • V is covering : ←→ V hi = X and V a = S n ∈ ω V a ˆ n for all a ∈ <ω ω ; • V is complete : ←→ fruit ( V , q ) = ∅ for all q ∈ ω ω ; • V has strict branches : ←→ | fruit ( V , q ) | = 1 for all q ∈ ω ω ; • V is locally strict : ←→ V a = S n ∈ ω V a ˆ n and V a ˆ m ∩ V a ˆ k = ∅ for all a ∈ <ω ω and m = k ∈ ω . Notation 5. Let V = h V a i a ∈ <ω ω be a Souslin scheme on a set X , x ∈ X , n ∈ ω ,and q ∈ ω ω . Then • q is a branch of x in V : ←→ x ∈ fruit ( V , q ); • branches ( V , x ) := the set of branches of x in V ; • rsubtree ( q, n ) := { s ∈ ( q ↾ n ) ↑ ⊏ ∩ <ω ω : q ⊳ s } ; • rsequences ( q, n ) := { p ∈ ω ω : q ⊳ p ∧ q ↾ n = p ↾ n } ; • cut ( V , q, n ) := S { fruit ( V , p ) : p ∈ rsequences ( q, n ) } . Remark 1. Let V = h V a i a ∈ <ω ω be a covering Souslin scheme on a set X , let n, m ∈ ω , n ≤ m , and q, p ∈ ω ω . Then (i) rsequences ( q, m ) ⊆ rsequences ( q, n ) ; (ii) if p E q and p ↾ n = q ↾ n , then rsequences ( q, n ) ⊆ rsequences ( p, n ) ; (iii) p ∈ rsequences ( q, n ) iff ∃ k > n such that p ↾ k ∈ rsubtree ( q, n ) ; (iv) cut ( V , q, n ) = S { V a : a ∈ rsubtree ( q, n ) } == { y ∈ V q ↾ n : branches ( V , y ) ∩ rsequences ( q, n ) = ∅} . (cid:3) Definition 2. Let V = h V a i a ∈ <ω ω be a Souslin scheme on a set X , τ a topologyon X , x ∈ X , and q is a branch of x in V . Then • q is a τ -base branch of x in V : ←→ { cut ( V , q, m ) ∪ { x } : m ∈ ω } is an openneighborhood base at the point x in the space h X, τ i .Because of this definition it is natural to introduce the following notation. Notation 6. Let V = h V a i a ∈ <ω ω be a Souslin scheme on a set X , x ∈ X , and q ∈ ω ω . Then cutBase ( V , q, x ) := { cut ( V , q, m ) ∪ { x } : m ∈ ω } . VLAD SMOLIN Remark 2. Let V = h V a i a ∈ <ω ω be a Souslin scheme on X and τ a Hausdorfftopology on X . Then for any sequence q ∈ ω ω there is at most one point x ∈ X such that q is a τ -base branch of x in V . (cid:3) Notation 7. Let V = h V a i a ∈ <ω ω be a Souslin scheme on a set X , τ a topology on X , x ∈ X . Then • BB ( V , x, τ ) := the family of τ -base branches of x in V ; • Let q ∈ ω ω is a τ -base branch of some point y in V . Then using Remark2 we define pnt ( V , q, h X, τ i ) := the point y ∈ X such that q is a τ -basebranch of y in V . Notation 8. • S := the Souslin scheme h S a i a ∈ <ω ω on ω ω such that S a := { p ∈ ω ω : a ⊑ p } for all a ∈ <ω ω . Remark 3. branches ( S , p ) = { p } for all p ∈ ω ω . (cid:3) Let h X, τ i be a topological space. Then a Souslin scheme V = h V a i a ∈ <ω ω on theset X is called an open Souslin scheme on the space h X, τ i if V a ∈ τ for all a ∈ <ω ω . Definition 3. Let h X, τ i be a Hausdorff topological space. Then an open completecovering Souslin scheme V = h V a i a ∈ <ω ω on the space h X, τ i is called a Sorgenfreybase for h X, τ i if the following conditions hold:(S1) ∀ x ∈ X ∀ q ∈ branches ( V , x ) ∀ n ∈ ω ∃ t ∈ BB ( V , x, τ ) ( t ↾ n = q ↾ n );(S2) ∀ q ∈ ω ω ∃ z ∈ X ( q ∈ BB ( V , z, τ )). Remark 4. Let V = h V a i a ∈ <ω ω be a Sorgenfrey base for a Hausdorff topologicalspace h X, τ i . Then branches ( V , x ) ⊆ Cl ω ω ( BB ( V , x, τ )) for all x ∈ X . (cid:3) Description of open images of the Sorgenfrey line The main result of this section is the following theorem. Theorem 1. A Hausdorff space is a continuous open image of the Sorgenfrey lineiff there exists a Sorgenfrey base for this space.Proof. The theorem follows from Corollary 1, Lemma 2, and Lemma 3. (cid:3) Lemma 1. The family S { cutBase ( S , p, p ) : p ∈ ω ω } is a base for a topology on ω ω and for any point x ∈ ω ω the family cutBase ( S , x, x ) is a neighborhood base at thepoint x in this topology.Proof. Let p, q ∈ ω ω , let n, m ∈ ω , and let x ∈ ( cut ( S , p, n ) ∪ { p } ) ∩ ( cut ( S , q, m ) ∪{ q } ). We now prove that there exists a set U ∈ cutBase ( S , x, x ) such that(1) x ∈ U ⊆ ( cut ( S , p, n ) ∪ { p } ) ∩ ( cut ( S , q, m ) ∪ { q } ) . Without loss of generality, we can assume that n ≤ m . Consider the set U := cut ( S , x, m ) ∪ { x } . Let z ∈ U \ { x } . Then using (i), (ii) of Remark 1 and Remark 3,we get z ∈ rsequences ( x, m ) ⊆ rsequences ( p, n ) ∩ rsequences ( q, m ). Finally, from (i),(ii) of Remark 1 and Remark 3 it follows that z ∈ ( cut ( S , p, n ) ∪{ p } ) ∩ ( cut ( S , q, m ) ∪{ q } ). (cid:3) Notation 9. σ S := the topology on ω ω that is constructed in Lemma 1. PEN IMAGES OF THE SORGENFREY LINE 5 Remark 5. S is an open complete covering Souslin scheme on the space h ω ω, σ S i . (cid:3) Lemma 2. Let h Y, τ i be a Hausdorff space and let f : h ω ω, σ S i → h Y, τ i be an opencontinuous surjection. Denote f [ S a ] by V a for all a ∈ <ω ω . Then V := h V a i a ∈ <ω ω is a Sorgenfrey base for h Y, τ i .Proof. Since f is an open surjection and S is an open complete covering Souslinscheme, we see that V is an open complete covering Souslin scheme on the space h Y, τ i .Let us prove that(2) x ∈ branches ( V , y ) for all y ∈ Y and x ∈ f − ( y ) . Let y ∈ Y and x ∈ f − ( y ). Then for any n ∈ ω it follows that y = f ( x ) ∈ f [ S x ↾ n ] = V x ↾ n , and hence x ∈ branches ( V , y ).We now prove that(3) x ∈ BB ( V , y, τ ) for all y ∈ Y and x ∈ f − ( y ) . Let y ∈ Y and x ∈ f − ( y ). Since cutBase ( S , x, x ) is an open neighborhood baseat the poin x and f is an open map, it is enough to prove that cut ( V , x, n ) ∪ { y } = f [ cut ( S , x, n ) ∪ { x } ] for all n ∈ ω .Fix n ∈ ω . Using (iv) of Remark 1, we get f [ cut ( S , x, n ) ∪ { x } ] = f [ S { S a : a ∈ rsubtree ( x, n ) } ] ∪ { y } = S { f [ S a ] : a ∈ rsubtree ( x, n ) } ∪ { y } = S { V a : a ∈ rsubtree ( x, n ) } ∪ { y } = cut ( V , x, n ) ∪ { y } . Since f is a surjection, it follows from (3) that V satisfies (S2).We now prove that V satisfies (S1). Let y ∈ Y , q ∈ branches ( V , y ), and n ∈ ω .Then y ∈ V q ↾ n = f [ S q ↾ n ]. This means that there exists x ∈ S q ↾ n such that f ( x ) = y ,so it follows from (3) that x ∈ BB ( V , y, τ ). Also since x ∈ S q ↾ n , we see that x ↾ n = q ↾ n . (cid:3) Lemma 3. Let h Y, τ i be a Hausdorff space with a Sorgenfrey base. Then thereexists a continuous open surjection f : h ω ω, σ S i → h Y, τ i . Moreover, if h Y, τ i has alocally strict Sorgenfrey base with strict branches, then h Y, τ i is homeomorphic to h ω ω, σ S i .Proof. Let V = h V a i a ∈ <ω ω be a Sorgenfrey base for h Y, τ i .Let f be the map from ω ω to Y such that f ( p ) := pnt ( V , p, h Y, τ i ) for all p ∈ ω ω . From property (S2) of V it follows that f is a surjection. Let us prove that f : h ω ω, σ S i → h Y, τ i is continuous and open.Let p ∈ ω ω . Since p ∈ BB ( V , f ( p ) , τ ), we see that cutBase ( V , p, f ( p )) is an openneighborhood base at the point f ( p ) in h Y, τ i . Hence we must only prove that(4) f [ cut ( S , p, n ) ∪ { p } ] = cut ( V , p, n ) ∪ { f ( p ) } for all n ∈ ω. Let n ∈ ω . We will prove two inclusions. VLAD SMOLIN ” ⊆ ” Let q ∈ cut ( S , p, n ). From the definition of f it follows that q ∈ branches ( V , f ( q )).From Remark 3 and (iv) of Remark 1 it follows that q ↾ n = p ↾ n and p ⊳ q .Finally, f ( q ) ∈ cut ( V , p, n ).” ⊇ ” Let x ∈ cut ( V , p, n ). From (iv) of Remark 1 it follows that there exists q ∈ branches ( V , x ) such that q ↾ n = p ↾ n and p ⊳ q . Choose m ∈ ω such that q ↾ m = p ↾ m and p ↾ ( m + 1) ⊳ q ↾ ( m + 1). From property (S1) of V it followsthat there exists t ∈ BB ( V , x, τ ) such that t ↾ ( m + 1) = q ↾ ( m + 1). Hence t ↾ n = q ↾ n = p ↾ n and p ⊳ t , so t ∈ cut ( S , p, n ) by (iv) of Remark 1. Since t ∈ BB ( V , x, τ ), we see that f ( t ) = x .Now suppose that V is locally strict and has strict branches. Then for any x ∈ Y the set branches ( V , x ) is a singleton, therefore f is a bijection because f − ( x ) ⊆ BB ( V , x, τ ) ⊆ branches ( V , x ). (cid:3) Now we show that the Sorgenfrey line has a locally strict Sorgenfrey base withstrict branches. To be precise, we show that a Lusin π -base for the Sorgenfrey linethat was constructed in [3] has all this properties. Example 1. We build a Souslin scheme V S = h V S a i a ∈ <ω ω by recursion on lh ( a ).Let V S hi := R , and let the set { V S a : lh ( a ) = 1 } equals { [ i, i + 1) : i ∈ Z } .When lh ( a ) ≥ 1, consider an interval V S a = [ i, j ). Let h x n i n ∈ ω be a sequence in R such that h x n i n ∈ ω h R ,τ R i −−−−→ j , x = i , x n +1 > x n , and x n +1 − x n ≤ lh ( a )+1 . Define V S a ˆ n := [ x n , x n +1 ).The next Corollary was first observed by Mikhail Patrakeev in private corre-spondence. Corollary 1. S ∼ = h ω ω, σ S i .Proof. The reader will easily prove that V S is a locally strict Sorgenfrey base for S that has strict branches. Then from Lemma 3 it follows that S ∼ = h ω ω, σ S i . (cid:3) Corollary 2. The Sorgenfrey line is, up to homeomorphism, the unique Hausdorfftopological space that has a locally strict Sorgenfrey base with strict branches. (cid:3) The spaces that are R-bidirected along Q Definition 4. Let h X, τ i be a topological space, Q a dense subset of h X, τ i , R abinary relation on X , and x ∈ X . Then: • An open neighborhood U of x is R -right along Q if xRy for all y ∈ ( U \{ x } ) ∩ Q ; • we say that x looks to the R -right along Q if the following conditions hold: – there exists an R -right along Q open neighborhood of x ; – for any open neighborhood U of x there exists y ∈ ( U \ { x } ) ∩ Q suchthat y ↓ R is a neighborhood of x . • An open neighborhood U of x is R -left along Q if yRx for all y ∈ ( U \{ x } ) ∩ Q ; • we say that x looks to the R -left along Q if the following conditions hold: – there exists an R -left along Q open neighborhood of x ; – for any open neighborhood U of x there exists y ∈ ( U \ { x } ) ∩ Q suchthat y ↑ R is a neighborhood of x . PEN IMAGES OF THE SORGENFREY LINE 7 Recall that a binary relation R on a set X is asymmetric if xRy ⇒ ¬ yRx for all x, y ∈ X . Definition 5. Let h X, τ i be a topological space, Q a dense subset of h X, τ i , and R an asymmetric binary relation on X . Then we say that h X, τ i is R-bidirected alongQ if there are dense subsets A l , A r of h X, τ i such that • X = A l ∪ A r and A l ∩ A r = ∅ ; • x looks to the R -right along Q for all x ∈ A r ; • x looks to the R -left along Q for all x ∈ A l . Example 2. It is easy to check that the double-arrow space is R -bidirected alongitself, where R is the strict lexicographic order on it. Lemma 4. Let h X, τ i be a Hausdorff space. Suppose that there are a dense subset Q of h X, τ i and an asymmetric binary relation R on X such that h X, τ i is R-bidirectedalong Q. Then h X, τ i is not a continuous open image of S .Proof. Assume the converse. From Theorem 1 it follows that there exists V = h V a i a ∈ <ω ω is a Sorgenfrey base for h X, τ i . Let A l and A r be the sets from Definition5. We build, by recursion on n , a sequence h x n i n ∈ ω in Q , a sequence h t n i n ∈ ω in ω ω , and an ⊏ -increasing sequence h p n i n ∈ ω in <ω ω such that(5) ∀ k ∈ ω ∃ x , x ∈ cut ( V , p, k ) ∩ Q such that x RzRx for all z ∈ fruit ( V , p ) , where p = S n ∈ ω p n .Take p := hi . Suppose we have constructed p , . . . , p n ; x , . . . , x n − ; and t , . . . , t n − . Now we consider two cases. Let n be even. Then take any point x ∈ V p n ∩ A r . From property (S1) of V it follows that there exists q ∈ BB ( V , x, τ )such that(6) p n ⊏ q. Take m > lh ( p n ) such that cut ( V , q, m ) ∪{ x } ⊆ V p n . Since x looks to the R -rightalong Q , we can take x n ∈ ( cut ( V , q, m ) \ { x } ) ∩ Q such that(7) x n ↓ R is a neighborhood of x. From (iv) of Remark 1 it follows that we can take t n ∈ branches ( V , x n ) suchthat(8) t n ∈ rsequences ( q, m ) . Consider k > m such that(9) q ↾ k ⊳ t n ↾ k. Since q ∈ BB ( V , x, τ ), we see that there exists a > k such that(10) cut ( V , q, a ) ∪ { x } ⊆ x n ↓ R . VLAD SMOLIN Let p n +1 be any element of rsubtree ( q, a ). From (6) and inequalities a > k >m > lh ( p n ) it follows that p n = q ↾ lh ( p n ) ⊏ q ↾ a ⊏ p n +1 . Since p n +1 ∈ rsubtree ( q, a ),from (iv) of Remark 1 and (10) it follows that(11) zRx n for all z ∈ V p n +1 . Now we show that(12) p n ⊏ t n and p n +1 ⊳ t n ↾ lh ( p n +1 ) . From inequality m > lh ( p n ), (6), and (8) it follows that p n ⊏ t n . To prove thesecond part of (12), we must observe that q ↾ k ⊏ p n +1 and use (9).Now let n be odd. If we argue as above by taking x ∈ V p n ∩ A l , x n such that x n ↑ R is a neighborhood of x , and t n ∈ branches ( V , x ( n )), then we can choose p n +1 such that(13) x n Rz fol all z ∈ V p n +1 ;and(14) p n ⊏ t n and p n +1 ⊳ t n ↾ lh ( p n +1 ) . We now prove (5). Let k ∈ ω , n > k , and n is even. Then p ↾ k ⊏ p n . So from(12) it follows that t n ∈ rsequences ( p, k ). Hence using (iv) of Remark 1, we get x n ∈ cut ( V , p, k ). Finally, from (11) it follows that zRx n for all z ∈ fruit ( V , p ) ⊆ V p n +1 . If we argue as above by taking m > k such that m is odd, then we can showthat x m ∈ cut ( V , p, k ) and x m Rz for all z ∈ fruit ( V , p ) ⊆ V p m +1 .Let us prove that(15) ∀ z ∈ fruit ( V , p ) : p BB ( V , z, τ ) . Let z ∈ fruit ( V , p ), we consider two cases.Case 1: z ∈ A r . From (5) it follows that for any k ∈ ω there exists x ∈ cut ( V , p, k ) ∩ Q such that x Rz , and so ¬ zRx and z = x . Hence for all k ∈ ω if cut ( V , p, k ) ∪ { z } is an open neighborhood of z , then it is not R -right along Q .And since z looks to the R -right along Q , we see that cutBase ( V , p, z ) is not anopen neighborhood base at the point z . So p BB ( V , z, τ ).Case 2: z ∈ A l . Arguing as above, we can take x ∈ cut ( V , p, k ) ∩ Q such that zRx , and so p BB ( V , z, τ ).Formula (15) contradicts property (S2) of V . The lemma is proved. (cid:3) The spaces that are not open images of the Sorgenfrey line A subset of a topological space is called co-dense if its complement is dense. Theorem 2. Suppose that h X, τ i is a Hausdorff topological space, S ⊆ X is a denseand co-dense subset of X . If h S, τ ↾ S i is homeomorphic to the Sorgenfrey line, then h X, τ i is not a continuous open image of the Sorgenfrey line. PEN IMAGES OF THE SORGENFREY LINE 9 Proof. Since [0 , S ∼ = S , without loss of generality, we can assume that S = [0 , , S = h S, τ ↾ S i .If h X, τ i is not first-countable, then it is not a continuous open image of theSorgenfrey line, because the first axiom of countability is preserved by continuousopen maps. So suppose that h X, τ i is first-countable.Using Lemma 4 it is enough to prove that there exists an asymmetric relation R on X such that h X, τ i is R -bidirected along [0 , A ⊆ X , by ˘ A wedenote A ∩ [0 , L the function with domain X \ [0 , 1) such that for all z ∈ X \ [0 , L ( z ) := { x ∈ [0 , 1] : ∃ p ∈ ω [0 , p h X,τ i −−−→ z ∧ p [0 , R −−−→ x ) } . Since h X, τ i is a first countable space and [0 , 1) is dense in it, we see that L ( z ) = ∅ for all z ∈ X \ [0 , < well-orders L ( z ) for all z ∈ X \ [0 , . Let z ∈ X \ [0 , q ∈ ω L ( z ) be such that q ( n + 1) 1) such that(17) q [0 , S −−−→ x. Let us prove that(18) U z ∩ U x = ∅ for all U x ∈ nbhds ( x, τ ) and U z ∈ nbhds ( z, τ ) . Let U x ∈ nbhds ( x, τ ) and U z ∈ nbhds ( z, τ ). Without loss of generality, we canassume that ˘ U x = [ x, x + ε ). From (17) it follows that there exists n ∈ ω such that q ( n ) ∈ ˘ U x ⊆ U x . Since q ( n ) ∈ L ( z ), there exists p ∈ ω [0 , 1) such that p h X,τ i −−−→ z and p [0 , R −−−→ q ( n ). Since q ( n ) = x , we obtain q ( n ) ∈ ( x, x + ε ), and so p is eventuallyin ( x, x + ε ) ⊆ U x . Also p is eventually in U z , hence U x ∩ U z = ∅ . Formula (18)contradicts the Hausdorff property of h X, τ i , so (16) is proved.Denote by m the function that takes each point z ∈ X \ [0 , 1) to the < -minimalelement of L ( z ). Also let M be the function that takes each point z ∈ X \ [0 , 1) tothe sup [0 , ( L ( z )). Now we prove a technical lemma about this functions. Lemma 5. Let z ∈ X \ [0 , and a, b ∈ R such that a < b . Then (i) ∃ U ∈ nbhds ( z, τ )( ˘ U < M ( z )) ; (ii) ∀ U ∈ nbhds ( z, τ )( ˘ U = [ a, b ) → M ( z ) ≤ b ) ; (iii) ∀ x ∈ [0 , x < m ( z ) → ∃ U ∈ nbhds ( z, τ )( x < ˘ U )) ; (iv) ∀ x ∈ [0 , ∀ U ∈ nbhds ( z, τ )( x < ˘ U → x < m ( z )) ; (v) ∀ U ∈ nbhds ( z, τ ) ∃ x ∈ ˘ U ( x < m ( z )) .Proof. (i). Assume the converse. Assume that for any U ∈ nbhds ( z, τ ) there exists x ∈ ˘ U such that M ( z ) ≤ x . Let p ∈ ω [0 , 1) be such that(19) ∀ n ∈ ω ( M ( z ) ≤ p ( n ))and(20) p h X,τ i −−−→ z. Take x ∈ [0 , 1] such that p ′ [0 , R −−−→ x , where p ′ is a subsequence of p . Then from(20) it follows that p ′ h X,τ i −−−→ z , and so x ∈ L ( z ). From (19) it follows that M ( z ) ≤ x .From equality M ( z ) = sup [0 , ( L ( z )) it follows that x = M ( z ), and so using (19), weget p ′ [0 , S −−−→ M ( z ). Hence p ′ h X,τ i −−−→ M ( z ). This contradicts the Hausdorff propertyof h X, τ i , so we prove (i).(ii). Assume the converse. Suppose that there exists U ∈ nbhds ( z, τ ) such that˘ U = [ a, b ) and b < M ( z ). Since M ( z ) = sup [0 , ( L ( z )), we see that there exists q ∈ ω L ( z ) such that q [0 , R −−−→ M ( z )and q ( n ) ≤ M ( z ) for all n ∈ ω. Hence there exists m ∈ ω such that b < q ( m ) ≤ M ( z ). q ( m ) ∈ L ( z ), so thereexists p ∈ ω [0 , 1) such that(21) p h X,τ i −−−→ z and(22) p [0 , R −−−→ q ( m ) . Let ε > b < ( q ( m ) − ε, q ( m )+ ε ). Then ( q ( m ) − ε, q ( m )+ ε ) ∩ U = ∅ and p is eventually in ( x m − ε, x m + ε ). This contradicts formula (21), so we prove(ii).(iii). Assume the converse. Suppose that there exists x ∈ [0 , 1) such that x < m ( z ) and ∀ U ∈ nbhds ( z, τ ) there exists y ∈ U such that y ≤ x . Take p ∈ ω [0 , p h X,τ i −−−→ z and(24) p ( n ) ≤ x for all n ∈ ω. Consider y ∈ [0 , 1] such that p ′ [0 , R −−−→ y , where p ′ is a subsequence of p . Thenfrom (23) it follows that p ′ h X,τ i −−−→ z , and so y ∈ L ( z ). From (24) it follows that y ≤ x < m ( z ), this contradicts m ( z ) is the < -minimal element of L ( z ), so we prove(iii).(iv). Assume the converse. Suppose that there exists x ∈ [0 , 1) and U ∈ nbhds ( z, τ ) such that x < ˘ U and m ( z ) ≤ x . Since m ( z ) ∈ L ( z ), we see thatthere exists p ∈ ω [0 , 1) such that(25) p h X,τ i −−−→ z and(26) p [0 , R −−−→ m ( z ) . Now let us consider two cases. Case 1: x = m ( z ). Since x < ˘ U , from (25) itfollows that there exists n ∈ ω such that x < p ( k ) for all k > n . Hence using (26) PEN IMAGES OF THE SORGENFREY LINE 11 and equality x = m ( z ), we get p [0 , S −−−→ x , and so p h X,τ i −−−→ x . This contradicts theHausdorff property of h X, τ i .Case 2: x > m ( z ). Take ε > m ( z ) − ε, m ( z ) + ε ) < x . Since x < ˘ U ,we see that ( m ( z ) − ε, m ( z ) + ε ) ∩ U = ∅ . From (26) it follows that p is eventuallyin ( m ( z ) − ε, m ( z ) + ε ), but this contradicts formula (25), so we prove (iv).(v). Assume the converse. Suppose that there exists U ∈ nbhds ( z, τ ) such that(27) m ( z ) ≤ x for all x ∈ ˘ U . From m ( z ) ∈ L ( z ) it follows that there exists p ∈ ω [0 , 1) such that(28) p h X,τ i −−−→ z and(29) p [0 , R −−−→ m ( z ) . From (28) it follows that p is eventually in U , and so from (27) and (29) it followsthat p [0 , S −−−→ m ( z ), hence p h X,τ i −−−→ m ( z ). This contradicts the Hausdorff propertyof h X, τ i , so we prove (v). (cid:3) Define xRy ⇐⇒ x < y ∨ y ∈ X \ [0 , ∧ x < M ( y ) ∨ x ∈ X \ [0 , ∧ M ( x ) ≤ y. It is not hard to prove that R is asymmetric. Define A r := [0 , 1) and A l := X \ [0 , X = A r ∪ A l and A r ∩ A l = ∅ . Let us prove that(31) x looks to the R -right along [0 , 1) for all x ∈ A r . Let x ∈ A r = [0 , W ∈ nbhds ( x, τ ) such that ˘ W = [ x, W is R -right along [0 , U be an arbitrary element of nbhds ( x, τ ). Take y ∈ ˘ U such that x < y and V ∈ nbhds ( x, τ ) such that ˘ V = [ x, y ). Let us prove that V ⊆ y ↓ R . Let z ∈ V . If z ∈ ˘ V , then z < y , and so zRy . Suppose that z ∈ V \ [0 , M ( z ) ≤ y , and so by definition of R , we get zRy .Now we shall prove that(32) z looks to the R -left along [0 , 1) for all z ∈ A l . Let z ∈ A l = X \ [0 , U ∈ nbhds ( z, τ ) such that ˘ U < M ( z ), and so by definition of R , we get xRz for all x ∈ ˘ U , hence U is an R -left along [0 , Now let U be an arbitrary element of nbhds ( z, τ ). From (v) of Lemma 5 it followsthat there exists x ∈ ˘ U such that(33) x < m ( z ) . From (33) and (iii) of Lemma 5 it follows that there exists V ∈ nbhds ( z, τ ) suchthat(34) x < ˘ V . Let us prove that V ⊆ x ↑ R . Let y ∈ V . If y ∈ ˘ V , then from (34) it follows that x < y , and so xRy . Let y ∈ V \ [0 , x < m ( y ) ≤ M ( y ),and so by definition of R , we get xRy .From (30), (31), and (32) it follows that h X, τ i is R -bidirected along [0 , (cid:3) Corollary 3. If b S is a Hausdorff compactification of the Sorgenfrey line, then itis not a continuous open image of the Sorgenfrey line.Proof. By b S ∗ denote b S \ S . From [9, Ch. 4, Pr. 43] it follows that b S ∗ is a densesubset of b S . Then from Theorem 2 it follows that b S is not a continuous openimage of the Sorgenfrey line. (cid:3) Lemma 6. The property of being a continuous open image of the Sorgenfrey lineis preserved by closed subspaces without isolated points.Proof. Let h X, τ i be a topological space such that there exists a continuous opensurjection f : S → h X, τ i . Let F ⊆ X be a closed subset of h X, τ i without isolatedpoints and Z := f − [ F ]. Since f is a continuous function, we have(35) Z is a closed subset of S . From [9, Ch. 2, Pr. 337] it follows that f ↾ Z : Z S → h F, τ ↾ F i is a continuousopen surjection. Since h F, τ ↾ F i has no isolated points and f ↾ Z is an open map, wesee that(36) Z has no isolated points . From (35), (36) and [8, (iii) of Theorem 4.6] it follows that Z S ∼ = S . And so h F, τ ↾ F i is a continuous open image of the Sorgenfrey line. (cid:3) Corollary 4. Suppose that the Sorgenfrey line is embeddable in a Hausdorff com-pact space h X, τ i . Then h X, τ i is not a continuous open image of the Sorgenfreyline.Proof. Let A ⊆ X be such that h A, τ ↾ A i ∼ = S . Let S := Cl h X,τ i ( A ), then h S, τ ↾ S i isa compactification of S . So from Corollary 3 it follows that(37) h S, τ ↾ S i is not a continuous open image of the Sorgenfrey line . From Lemma 6 and (37) it follows that h X, τ i is not a continuous open image ofthe Sorgenfrey line. (cid:3) PEN IMAGES OF THE SORGENFREY LINE 13 References [1] A. S. Kechris, Classical Descriptive Set Theory , Graduate Texts in Mathematics, 156 NewYork: Springer-Verlag, 1995.[2] K. P. Hart, J. Nagata, and J. E. Vaughan, eds., Encyclopedia of general topology , Elsevier,Amsterdam, 2004.[3] M. Patrakeev, Metrizable images of the Sorgenfrey line , Topology Proc. (2015), 253–269.[4] K. Kunen, Set Theory: An Introduction to Independence Proofs , Studies in Logic and theFoundations of Mathematics, 102 Amsterdam-New York: North-Holland Publishing Co., 1980.[5] N. V. 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Vlad SmolinKrasovskii Institute of Mathematics and Mechanics,Sofia Kovalevskaya street, 16,620990, Ekaterinburg, Russia E-mail address ::