aa r X i v : . [ m a t h . GN ] A ug (Non-)contractible road spaces of trees Mathieu BaillifAugust 28, 2019
Abstract
We prove that the road space of an R -special tree is contractible and that a locally metrizable spacecontaining a copy of an uncountable ω -compact subspace of a tree is not. We also raise some questionsabout possible generalizations. Despite the word ‘contractible’ in our title, our methods and results belong to set theoretictopology and have very little in common with homotopy theory. By ‘space’ is meant a topo-logical Hausdorff space, in particular ‘regular’ and ‘normal’ imply Hausdorff. We assume somefamiliarity with set theoretical trees and countable ordinals. Our terminology is standard, def-initions are given at the end of this introduction if needed. Any tree is endowed with the ordertopology (also called the interval topology).Asking questions about the contractibility of set theoretic objects is not very common, sobefore stating our results let us explain why we think it is an interesting subject, particularly formanifolds and similar spaces. (The main reason is that we like to think about these problemsand find it oddly satisfying to try to mentally squish things, but we fear it will not be seen asvaluable.)The objects we study in this note are set theoretic trees and their road spaces, which areobtained by joining consecutive points by a line segment with a topology that makes it locallyembeddable in R (the details are given below). These spaces are a good toy model for so-called Type I non-metrizable manifolds (also defined below). Metrizable manifolds that arecontractible happen to coincide with those having vanishing homotopy groups: Milnor showedin [13] that a metrizable manifold has the homotopy type of a CW-complex and WhiteheadTheorem applies (see, e.g., [11, p. 346]). If one releases the metric assumption, this is nottrue anymore as there are simple non-metrizable non-contractible manifolds with vanishinghomotopy groups (e.g. the longray described below). Of course, any space with non-trivialhomotopy groups is non-contractible. It was at first not clear to us whether contractible non-metrizable manifolds do exist at all. We then noticed that examples were found a long time agoby Calabi and Rosenlicht; the Pr¨ufer surface and some of its variants are contractible. (Theoriginal source is [5] and a more recent account is given in [2], see also [9] for more on Pr¨ufersurface and homotopy.) This particular manifold has the property that some open Euclideanset has a non-metrizable closure, in a terminology dating to P. Nyikos [14] it is in the classof Type II manifolds. Hence, in a loose sense, what makes it non-metrizable lies just at theboundary of some perfectly nice Euclidean open contractible set. We happen to be able to‘push’ this non-metrizable stuff inside this Euclidean open set, everything at once, yieldingthe contractibility. The other class of non-metrizable manifolds are the aforementioned
Type I manifolds [14, Def. 2.10]. A space X is of Type I if and only if X = ∪ α ∈ ω X α , where X α is openand X α Lindel¨of for each α , and X α ⊂ X β whenever α < β < ω (and of Type II otherwise). In ase of manifolds, each X α is an open metrizable submanifold since Lindel¨ofness of X α impliesits metrizability. Loosely speaking again, these manifolds ‘grow slowly’ instead of jumping atonce into non-metrizability. Since the Pr¨ufer manifold is of Type II, we are interested in findingwhether there are contractible non-metrizable Type I (slowly growing) manifolds. As writtenabove, road spaces of trees are good toy models. Our first result gives a contractible examplein this class of spaces: Theorem 1.1.
The road space R T of a rooted R -special tree is contractible. We shall show in a subsequent paper that given an ω -tree T , it is possible to define a surfacewhich is is homotopy equivalent to R T (and contains it), providing the example of a contractibleType I manifold we are seeking.Looking at the problem from the other side, it seemed interesting as well to see whether thereare properties purely from general topology (which have nothing to do with homotopy/homologygroups or such) which prevent a manifold (or a ‘locally nice’ space) to be contractible. Someresults are already available, for instance the following theorem was proved by S. Deo and D.Gauld (using ideas developed by the author in [1]): Theorem 1.2 ([6, Thm 3.4]) . Any locally metrizable space X containing a copy of ω is non-contractible. This implies for instance that any tree with an uncountable branch has a non-contractibleroad space. An older and simpler result (see for instance [10, Prop. 1.22]) is that the longraydefined as inserting a line segment between consecutive ordinals in ω (more formally, ω × [0 , ω . Theorem 1.3.
Let T be a tree, S ⊂ T be uncountable and ω -compact in the subspace topology,and let X be a locally metrizable space. Then there is no continuous h : S × [0 , → X suchthat h ( · ,
1) : S → X has uncountable image and h ( · , is constant. In particular, if X containsa copy of S then X is not contractible. Recall that a space is ω -compact if and only if any closed discrete subset is at most count-able. Notice that an uncountable ω -compact subset of a tree cannot be metrizable, since ω -compactness is equivalent to Lindel¨ofness in metrizable spaces (see, e.g., [8, Thm 4.1.15]),and a Lindel¨of subset of a tree is countable. We note that we could add the assumption that T has height ω without loss of generality, because if S contains an uncountable branch E , then E is ω -compact in the subspace topology and thus homeomorphic to a stationary subset of ω , and we may apply Theorem 1.5 below. Theorem 1.3 gives us another motivation for thesehomotopy questions: to find where is the line among the set theoretic or topological propertiesof trees between contractibility and non-contractibility of their road spaces (see Question 1.7 be-low). A quick corollary, which is another exhibit of the similarities between ω and Suslin trees,is the following. Its proof is immediate since the road space of a tree satisfying the assumptionsis locally metrizable and a Suslin tree is ω -compact (see Lemma 2.1 below). Corollary 1.4.
Let T be a tree of height ω . If T contains an uncountable subset S which is ω -compact in the subspace topology (in particular, if T is a Suslin tree), then its road space R T is non-contractible. Our second generalization of Deo and Gauld’s theorem is to weaken the local metrizability ofthe target space X . First countability is not enough, as the cone Cω = ω × [0 , / ( x, ∼ ( y, ω itself. (First countabilityfollows from the easily proved fact that an open set in ω × [0 ,
1] containing ω ×{ } must containa small strip ω × ( a, heorem 1.5. Let X be a regular space with G δ points such that each point has a closed [ ℵ , ℵ ] -compact neighborhood. Let S be a stationary subset of ω endowed with the subspace topology.Then there is no continuous h : S × [0 , → X such that h ( · ,
1) : S → X has an uncountableimage and h ( · , is constant. In particular, if X contains a copy of S then X is not contractible. Recall that a space is [ ℵ , ℵ ] -compact if and only if any cover of X of size ≤ ℵ has acountable subcover. This is equivalent to the property that given any uncountable subspace E of X there is x ∈ X which is a condensation point of E , that is, any neighborhood of x contains uncountably many points of E . We do not know whether this theorem holds if S is aSuslin tree but our main tool (any continuous map from S to X must be constant on the pointsabove x ∈ S if x is high enough, see Lemma 3.4 below) does not, as we will show in anotherpaper [3]. Notice that Theorem 1.5 holds if X is locally metrizable. Indeed, a stationary subsetof ω is ω -compact, ω -compactness is preserved by continuous functions and is equivalent toLindel¨ofness in metrizable spaces; hence the image of S × [0 ,
1] in X is (locally) Lindel¨of.We give two proofs of Theorem 1.3. One is entirely topological and takes a few pages, ingreat part because we take the time to first show that many well known properties of Suslintrees also hold for ω -compact subsets of trees of height ω . These generalizations are probablypart of the folklore and no more than standard exercices, but we found convenient to gatherthe proofs of most of them in Section 2. The property that is central to our proofs is thatany continuous map from S to a locally metrizable space must be constant on the points above x ∈ S if x is high enough in the tree (see Lemma 2.2 (i) below). This is a slight generalizationof a result of Stepr¯ans [17]. Our other proof is by forcing, and is based on another argument ofStepr¯ans in the same paper. It is quite short but uses classical theorems on ccc forcing, some ofthe properties in Section 2 and (a weaker version of) Theorem 1.5. Since the proof of Theorem1.5 is very similar to the topological proof of Theorem 1.3, we decided to present our argumentsin the following order: first the topological proof of Theorem 1.3, then that of Theorem 1.5 andfinally the forcing argument for Theorem 1.3. Both proofs of Theorem 1.3 use heavily the treestructure. We do not know whether the result can be generalized for arbitrary non-metrizable ω -compact spaces. Question 1.6.
Is there a locally metrizable contractible space containing a non-metrizable ω -compact subspace ? Is there a manifold with these properties ? We will show in another paper that if X is countably compact, non-compact and TypeI, then X is not contractible. A consequence is that Quesion 1.6 has a negative answer for ω -compact locally compact spaces under the proper forcing axiom PFA .Looking only at road spaces of trees, we do not know whether contractibility entails R -specialness (but actually have a strange feeling of having overlooked something simple): Questions 1.7.
Let T be a tree and R T be its road space.(a) Does the contractibility of R T implies that T is R -special ?(b) Does the non-contractibility of R T imply that T contains a Suslin subtree ? Notice that it is not possible for (a) and (b) to have both a positive answer since there aremodels of set theory with a non- R -special tree T that does not contain a Suslin subtree, see[16]. We do not know whether these trees have contractible road spaces.We end this introduction with definitions and notations. Any function in this note is assumedto be continuous otherwise stated. A contraction is a function h : X × [0 , → X such that h ( x,
1) = x and h ( x,
0) is a constant map. A space is contractible if there is a contraction h : X × [0 , → X . We often write h t ( x ) for h ( x, t ).Recall that a tree T is a partially ordered set such that each point has a well ordered setof predecessors. We define the height of x ∈ T and of T , the α -th-level Lev α ( T ), the chains and antichains in T as usual, see for instance [12, Section II.5] or [15]. A subset E of T is rder-dense iff for any y ∈ T there is y ∈ E with x < y . Recall that all trees are endowed withthe order (also called interval) topology. We often abbreviate ‘closed and unbounded’ by club .A tree is rooted iff it has a unique minimal element called the root . A subtree S is a subset of T with order restricted to S . Notice that the induced topology on a subtree S is finer (sometimesstriclty) than the one given by the order restricted to S . Both topologies agree if S is closed in T . We assume that our trees are Hausdorff, that is, if x, y ∈ T are at a limit level and have thesame predecessors, then x = y . (This could be false for a subtree.) An ω -tree has countablelevels and height ω . A tree is Suslin if it has height ω and its chains and antichains are atmost countable. When x ∈ T and α is an ordinal, write T ( x ) = { y ∈ T : y ≥ x } , x ↾ α forthe unique predecessor of x at level α (if x is below the α -th level, x ↾ α = x ) and T ≤ α forthe subset of elements at level ≤ α . If E ⊂ T , set E ↓ = { x ∈ T : ∃ y ∈ E with x ≤ y } to beits downward closure. We say that the tree T is R -special iff there is a strictly increasing (notnecessarily continuous) function T → R . Recall that R -special ω -trees exist in ZFC .The road space R T of a tree T is obtained by joining consecutive points by an interval [0 , T as a subset of R T . The topology on the interior of theadded intervals is that of (0 , x ∈ T ⊂ R T , in order for R T to be (locally) connectedany open set containing x must contain a small portion of each interval emanating from x . Inorder for the space to be locally metrizable (and hence first countable), we take these portionsuniformly as follows. Denote by x [0 , y ⊂ R T the interval between the two consecutive points x, y ∈ T . If A ⊂ [0 , x A y is understood as the corresponding subset of x [0 , y . Forsingletons we usually write x a y instead of x { a } y . If x ∈ T , denote by s ( x ) the set of itsimmediate successors and set W x,n = [ y ∈ s ( x ) x [0 , /n ) y . If x ∈ s ( z ) with z ∈ T , a basis for theneighborhoods of x is given by { W x,n ∪ z (1 − /n, x : n ∈ ω } . A basic neighborhood of x ∈ T at a limit level is obtained by choosing some z ∈ T , z < x and n ∈ ω and taking the segment { y ∈ R T : z < y < x } union each W w,n for those w ∈ T with z < w < x . This makes R T locally embeddable in R (as seen by induction). The induced topology on T ⊂ R T is that of T , and R T is arc connected if and only if T is rooted. ω -compactsubsets of trees Notations: Given a tree T , if A, B ⊂ T and x ∈ T , A < B means that each member of A is < each member of B . We denote { x } < A by x < A and T ( x ) ∩ A by A ( x ). Notice thatLev α ( A ↓ ) = Lev α ( T ) ∩ A ↓ . Lemma 2.1.
Let T be a tree of height ω and S ⊂ T be uncountable and endowed with thesubspace topology. Then the following hold.(a) The subspace topology on S ↓ always agrees with the topology given by the induced order. If S is closed in T , then the subspace topology agrees with the topology given by the induced orderon S .(b) If S is ω -compact, then it intersects a stationary subset of levels of S ↓ (and of T ).(c) An antichain is closed discrete in T , and a closed discrete subset of T is an at most countableunion of antichains.(d) If S is ω -compact then so is S ↓ .(e) If S is ω -compact and does not contain an uncountable chain, then S ↓ is Suslin.(f ) If A ⊂ S ⊂ T , then A is a maximal antichain in S if and only if it is a maximal antichainin S ↓ .Proof. (a) and (f) are straightforward. For (b), if S avoids a club set of levels of S ↓ then taking ne point between consecutive avoided levels (when available) yields an uncountable discretesubset which is closed in S ↓ (and thus in T ). Item (c) is proved e.g. in [15, Thm 4.11], and(d)–(e) follow immediately from it. Lemma 2.2.
Let T be a tree of height ω and S ⊂ T be uncountable and ω -compact in thesubspace topology. Then the following hold.(a) S ↓ is the disjoint union of a countable set, a Suslin tree and at most countably many copiesof ω . In particular, S ↓ is an ω -tree.(b) There is α ∈ ω such that | S ( x ) | = | S ↓ ( x ) | = ℵ when x is above level α .(c) If C ⊂ S ↓ is uncountable, there is x ∈ S such that S ↓ ( x ) ⊂ C ↓ and | S ( x ) | = | S ↓ ( x ) | = ℵ .(d) C ↓ is ω -compact for any C ⊂ S ↓ .(e) If C ⊂ S ↓ is club in S ↓ , then { γ : Lev γ ( C ↓ ) ⊂ C } is club in ω .(f ) If C ⊂ S ↓ is club in S ↓ , then S ∩ C intersects a stationary subset of levels.(g) If E n ⊂ S ↓ are closed and order-dense in S ↓ for n ∈ ω , then ∩ n ∈ ω E n is also closed andorder-dense.(h) Let E, F ⊂ S ⊂ T be closed in S . If | E ∩ F | ≤ ℵ , then | E ↓ ∩ F ↓ | ≤ ℵ .(i) Let f : S → Y be continuous where Y is a metrizable space. Then there is β ∈ ω such that f ( S ( x )) is a singleton whenever x is above level β in T . In particular, f has a countable image.Proof. We will use several times (without acknowledging it explicitly) the fact that an at mostcountable intersection of club subsets of ω is club.(a) If S contains an uncountable branch B , then there is some minimal x B ∈ B such that S ( x B ) is linearly ordered. Indeed, otherwise S contains an uncountable antichain (take pointsbranching away from B above each height) and hence an uncountable closed discrete subset byLemma 2.1 (c). Since the minimal elements of { x B : S ( x B ) ⊂ S is an uncountable branch } is an antichain, S contains at most countably many disjoint branches. Notice that a maximalbranch in S ↓ contains an unbounded branch of S . Removing the branches above each x B in S ↓ ,what remains is either countable or a Suslin tree by Lemma 2.1 (e).(b) follows immediately from (a) and the equivalent statement for Suslin trees.(c) If C ∩ B is uncountable for some branch B ⊂ S ↓ , we are over. If not, by (a) and (b) wecan assume that S ↓ is a Suslin tree such that | S ( x ) | = ℵ for each x ∈ S ↓ . Then the result iswell known (see, for instance, the claim in Theorem 2.1 in [7]).(d) is immediate by (a) since an uncountable (downward closed) subset of a Suslin tree is aSuslin tree.(e) By (d), C ↓ is ω -compact. Fix α given by (b) and some β > α . Since each uncountablemaximal branch B of C ↓ is a copy of ω , C contains a club set of levels of B . By (a) we mayassume that C ↓ is Suslin. By Lemma 2.1 (f), for each n ∈ ω we may find a countable antichain A n ⊂ C which is maximal in C ↓ and such that A n +1 > A n and each member of A n is aboveheight β . Let γ = sup n ∈ ω sup { height( x ) : x ∈ A n } . By construction Lev γ ( C ↓ ) is the set of limitpoints of ∪ n ∈ ω A n , hence since C is closed Lev γ ( C ↓ ) ⊂ C . This shows that { γ : Lev γ ( C ↓ ) ⊂ C } is unbounded in ω , and closedness is obvious.(f) If C ∩ B is unbounded for some maximal uncountable branch B ⊂ S ↓ , then it is home-omorphic to a club subset of ω . By Lemma 2.1 (b) S ∩ B ∩ C is thus stationary. If C ∩ B isbounded for each uncountable branch of S ↓ , we may assume by (a) that S ↓ is a Suslin tree inwhich C is unbounded. By (c) it follows that S is unbounded in C ↓ as well. By (e) and Lemma2.1 (b), S intersects C on a stationary set of levels.(g) Closedness is immediate, and order-density follows immediately by (a) and the fact thatthe result holds for Suslin trees and ω .(h) Let E, F be the closures in S ↓ of E, F . If E ∩ F is unbounded in S ↓ , by (f) S ∩ E ∩ F =( S ∩ E ) ∩ ( S ∩ F ) = E ∩ F is unbounded, a contradiction. Hence E ∩ F is bounded and thusdisjoint above some level α . It follows that E and F cannot be both unbounded in the sameuncountable branch. It is well known (see e.g. [18, Thm 6.18] or [7, Thm 2.1]) that if A, B are isjoint closed sets in a Suslin tree, then A ↓ ∩ B ↓ is at most countable. Together with (a), thisshows that | E ↓ ∩ F ↓ | ≤ ℵ .(i) Our proof is a slight adaptation of Stepr¯ans topological proof in [17] that a real valued mapwith domain a Suslin tree has a countable image. Denote the distance in Y by dist( · , · ). By (b)me may assume that S ( x ) is uncountable for each x . Set D ( ǫ ) = { x ∈ S : diam( f ( S ( x ))) ≤ ǫ } ,where diam stands for diameter, that is, the supremum of the distances between two points ina set. Assume for now that D ( ǫ ) is order-dense in S (and thus in S ↓ ) when ǫ >
0. Let D ( ǫ )denote the closure of D ( ǫ ) in S ↓ . By (g), D = ∩ n ∈ ω,n> D (1 /n ) is closed and order dense in S ↓ , hence by (f) S ∩ D intersects stationary many levels above each x ∈ S (since | S ( x ) | = ℵ ).Moreover, D is upward closed in S . Denote by A the minimal elements of D ∩ S . Then A is anantichain of S ↓ , let β be the supremum of the heights of its members. For each x ∈ S abovelevel β , the diameter of f (cid:0) S ( x ) (cid:1) is 0, hence f (cid:0) S ( x ) (cid:1) = f ( { x } ) and the lemma is proved.To finish, we now prove that D ( ǫ ) is order-dense in S . Suppose that it is not the case and let x ∈ S be such that diam (cid:0) f ( S ( y )) (cid:1) > ǫ for each y > x , y ∈ S . We build inductively antichains A αn ( n ∈ ω, α ∈ ω ) such that the following hold. • A αn ⊂ S is maximal above x , that is, in S ↓ ( x ), • A αn +1 > A αn > A βm for each n, m ∈ ω and α > β , • If u ∈ A αn , v ∈ A αn +1 , then dist( f ( u ) , f ( v )) ≥ ǫ/ A αn is defined. Set E = { z ∈ S : z > A αn and dist( f ( z ) , f ( u )) ≥ ǫ/
4, where u is the member of A αn below z } . It is enough to see that E is order-dense in S ↓ ( x ), since then we may put its minimal elementsin A αn +1 . Let thus w ∈ S ↓ ( x ), w > u ∈ A αn . Up to going further up, we may assume that w ∈ S .If dist( f ( w ) , f ( u )) ≥ ǫ/
4, then w ∈ E . If not, then dist( f ( w ) , f ( u )) < ǫ/
4. Choose v ∈ S ( w )such that dist( f ( w ) , f ( v )) > ǫ/ (cid:0) f ( S ( y )) (cid:1) > ǫ for each y > x ). Then dist( f ( v ) , f ( u )) ≥ dist( f ( v ) , f ( w )) − dist( f ( w ) , f ( u )) > ǫ/ , and v ∈ E . If A γn is chosen for each n ∈ ω and each γ < α , set A α to be an antichain in S ,maximal in S ↓ ( x ), whose members are all > ∪ n ∈ ω,γ<α A γn . This defines A αn for each n ∈ ω, α ∈ ω .Set β ( α ) to be sup { height( y ) : y ∈ ∪ n ∈ ω A αn } , let C be the closure in ω of { β ( α ) : α ∈ ω } ,and C ′ be its derived set (that is its limit points). By construction, if y ∈ S, y > x and theheight of y in S ↓ is in C ′ , then f is not continuous at y as they is a sequence of points in S converging to y whose images are ≥ ǫ/ S ( x )is ω -compact), there must be such an y , a contradiction. This shows that D ( ǫ ) is order-denseand concludes the proof. Proof.
Let T be a rooted R -special tree with root r and let f : T → R be a strictly increasingfunction. By replacing f ( x ) by sup y 1) = x and then define h such that x travels downwards in R T , starting to moveexactly at time t = f ( x ) and reaching y < x ( y ∈ T ) exactly at time t = f ( y ). Since f isstrictly increasing, there is time available to cross the interval between consecutive points. Inless readable formulas, let x at level α be given. If f ( x ) ≤ t , we set h ( x, t ) = x . If β < α and f ( x ↾ β ) ≤ t ≤ f ( x ↾ β + 1), we set h ( x, t ) to be x ↾ β k x ↾ β +1 where k = t − f ( x ↾ β ) f ( x ↾ β + 1) − f ( x ↾ β ) . inally, if t ≤ f ( r ), we set h ( x, t ) = r . It should be clear that h is continuous (on T × [0 , h ( x, 0) = r for all x ∈ T . Notice that h has the property that at time f ( x ), all of T ( x )is squished onto x . Actually, each point on the tree starts to move exactly when all the pointsabove it reach it, all at the same time. This enables to extend easily the map to all of R T , apoint in the segment x [0 , y does not move until y reaches it, and then it follows it until theend. This gives the required contraction.Notice that if one sets j t ( x ) = h ( x, − t ) for t ∈ [0 , j t ( x ) = x when t < j t ( x ) = r when t > 1, then j is actually a flow, that is j t ◦ j s = j t + s . Our proof relies on simple consequences of the properties given in Lemma 2.2, especially (i).When S is a subset of a tree T , the height of a point of S is to be understood as its height in T . Recall that S ( x ) = T ( x ) ∩ S . Lemma 3.1. Let X be a space containing a closed metrizable G δ subset B ⊂ X , S be anuncountable ω -compact subset of a tree of height ω and f : S → X be continuous. Then thereis α ∈ ω such that either f ( S ( x )) ∩ B = ∅ or f ( S ( x )) is a singleton whenever x is at height ≥ α .Proof. By Lemma 2.2 (b), above some level each S ( x ) is uncountable, we assume for simplicitythat this holds for each x ∈ S . Let U n be open sets such that ∩ n ∈ ω U n = B . By Lemma 2.2 (h),there is β ∈ ω such that for each n we have (cid:16) (cid:0) f − ( B )) (cid:1) ↓ ∩ (cid:0) f − ( X − U n ) (cid:1) ↓ (cid:17) − T ≤ β = ∅ . It follows that if x ∈ S is at level above β , either f ( S ( x )) ⊂ B or f ( S ( x )) ∩ B = ∅ . Let E = { x ∈ S : Lev( x ) ≥ β and f ( S ( x )) ⊂ B } . Then E is an upward closed subspace of S , inparticular it is an ω -compact subspace of T . By Lemma 2.2 (i), there is α ≥ β such that f isconstant on S ( x ) whenever x ∈ E is at level ≥ α . This proves the lemma. Corollary 3.2. Let X be a space and U ⊂ X be open such that U is contained in a metrizableopen V ⊂ X . Let S be an uncountable ω -compact subset of a tree of height ω . Let h : S × [0 , → X be continuous. Then there is α ∈ ω such that for each t ∈ [0 , and each x ∈ S above level α , either h − t ( U ) ∩ S ( x ) = ∅ , or h t is constant on S ( x ) . Again, h t stands for h ( · , t ). Proof. Let { t n : n ∈ ω } be a countable dense subset of [0 , B = U . Since B is closed inthe metrizable subset V , it is a G δ . The previous lemma shows that there is some α such thatwhen x is above level α , either h − t n ( B ) ∩ S ( x ) = ∅ or h t n is constant on S ( x ) for each n ∈ ω .The result follows by continuity. Proof of Theorem 1.3. Let h : S × [0 , → X be continuous such that h ( x ) = u ∈ X and h : S → X has uncountable image. We recall that we can assume that T ⊃ S has height ω .The set of x ∈ S such that S ( x ) has uncountable image under h is uncountable and downwardclosed, hence by Lemma 2.2 (c) there is x such that the image of S ( z ) under h is uncountablefor each z ≥ x . Up to replacing S ↓ by S ↓ ( x ) we assume that this holds for all z ∈ S ↓ . For x ∈ S , set τ ( x ) = sup { t : h t is constant on S ( x ) } . Then τ is an increasing map S → R , however τ is a priori neither continuous nor strictlyincreasing. We will show that there is a closed unbounded C ⊂ ω such that τ is strictlyincreasing on the subspace of members of S at levels belonging to C . Such a subspace is ncountable (and thus unbounded) in S by Lemma 2.2 (f). Hence as a partially ordered space S contains an R -special tree and thus (at least) an uncountable antichain (see for instance [15,Thm 4.29]), a contradiction with the fact that S is ω -compact.It is enough to show that for each x ∈ S , there is α such that τ ( y ) > τ ( x ) whenever y > x is atlevel ≥ α . Indeed, since the levels of S ↓ are countable by Lemma 2.2 (a) and τ is increasing, asimple induction provides C . Let x ∈ S be fixed. By continuity, h τ ( x ) is constant on S ( x ) withvalue u = h τ ( x ) ( x ) and thus τ ( x ) < h ( S ( x )) is uncountable. (While it is not needed,notice that h restricted to the subspace S ( x ) × [ τ ( x ) , 1] ’contracts’ all of S ( x ) to the point u .)Since X is locally metrizable, we may choose an open U ∋ u such that B = U is contained inan open metrizable set. Let α be given by Corollary 3.2. We may assume that α > height( x ).Assume that there is y at level above α such that for each t > τ ( x ), h t is not constant on S ( y ).By definition of α , this implies that h t ( S ( y )) ∩ U = ∅ and in particular that h t ( y ) U for each t > τ ( x ). But this contradicts the continuity of h since h τ ( x ) ( y ) = u ∈ U . Hence, h t is constanton S ( y ) for at least one t > τ ( x ) and thus τ ( y ) > τ ( x ). This finishes the proof. The proof is almost exactly the same as that of Theorem 1.3 once we have an equivalent ofCorollary 3.2 in this context. This is given by Corollary 3.5 below. Lemma 3.4 below plays therole of Lemma 2.2 (i). We first state the following easy fact. Lemma 3.3. Let S ⊂ ω be stationary, endowed with the subspace topology. Then an at mostcountable family of club subsets of S has a club intersection.Proof. A direct proof is quite easy, but notice that since S is an ω -compact subspace of thetree ω , the result is also a consequence of Lemma 2.2 (f)–(g). Lemma 3.4. Let S be a stationary subset of ω endowed with the subspace topology. If Y is regular, [ ℵ , ℵ ] -compact and has G δ points, then any continuous f : S → Y is eventuallyconstant, that is, there is α ∈ ω such that f ( β ) = f ( α ) for each β ≥ α , β ∈ S .Proof. We start by showing that there is some c ∈ Y such that f − ( { c } ) is club in S . If f ( S ) iscountable, then this is immediate. We thus assume that f ( S ) is uncountable. Since Y is [ ℵ , ℵ ]-compact, f ( S ) has a condensation point c ∈ Y . Since Y is regular and has G δ points, we maychoose open sets U n ∋ c , n ∈ ω , such that ∩ n ∈ ω U n = ∩ n ∈ ω U n = { c } . Since c is a condensationpoint, f − ( U n ) is club in S for each n , hence f − ( { c } ) = f − ( ∩ n ∈ ω U n ) = ∩ n ∈ ω f − ( U n ) is clubin S by Lemma 3.3.Now, since f − ( Y − U n ) is closed, it must be bounded, otherwise it intersects f − ( { c } ). Itfollows that f − ( Y − { c } ) = ∪ n ∈ ω f − ( Y − U n ) is bounded in S , say by α . Hence f is constanton S above α . Corollary 3.5. Let S be a stationary subset of ω endowed with the subspace topology. Let Y be a regular space with G δ points. Let U ⊂ Y be open such that U is [ ℵ , ℵ ] -compact. Let h : S × [0 , → Y be continuous. Then there is α ∈ ω such that for each t ∈ [0 , either h − t ( U ) ∩ [ α, ω ) ∩ S = ∅ , or h t is constant on [ α, ω ) ∩ S .Proof. Fix a countable dense subset { t n : n ∈ ω } ⊂ [0 , α such that for each n either [ α, ω ) ∩ S ∩ h − t n ( U ) = ∅ or h t n is constant above α . The resultfollows by continuity.The proof of Theorem 1.5 can now be done exactly along the same lines as that of Theorem1.3, we thus only give a sketch. Set τ = sup { t : h t is eventually constant } . Then h τ iseventually constant and τ < h has uncountable image. Fix α minimal in S such that h τ is constant above α , take an open U containing h τ ( α ) such that U is [ ℵ , ℵ ]-compact. ByCorollary 3.5 this contradicts the continuity of h . .4 Proof of Theorem 1.3 by forcing Chapter VII of [12] is a convenient reference for this subsection. Let h X, τ i be a topologicalspace in the ground model V . Given a forcing extension V [ G ] by a generic filter G , we denoteby τ ( G ) the topology for X in V [ G ] with base τ . Notice that in general, τ = τ ( G ) since newunions may appear in V [ G ]. A function in the ground model f : X → Y which is continuousremains continuous in V [ G ]. Also, being a 1-to-1 function is preserved. Some properties of X are preserved in any forcing extension, for instance metrizability and the separation axioms T i i ≤ (see, e.g., [4, Lemma 22]). We will force with a Suslin tree with reversed order. Wegather in the next lemma the well known properties of such forcings we are going to need. Lemma 3.6. Let T be a Suslin tree in the ground model V . Let V [ G ] be the extension by ageneric filter G when forcing over V with T with the reverse order. Then the following hold.(a) Cardinals and cofinalities are preserved between V and V [ G ] .(b) If S ⊂ ω is stationary in V , it remains so in V [ G ] .(c) If f, B ∈ V [ G ] and f : ω → B , then f ∈ V (“no new countable sets are added”). As said above, a classical reference for the proofs is [12]: (a) is Theorem VII.5.10, (b) isExercices VII.(H1)–(H2) and (c) is Theorem VII.8.4. Proof of 1.3. Again, we may assume that T has height ω . Let S ⊂ T and h : S × [0 , → X be as in the statement of the theorem, so h is constant and h has uncountable image. Pickone preimage in S for each point in the image of h , this defines an uncountable subset W of S . Up to replacing S ↓ by S ↓ ( x ) for some x , by Lemma 2.2 (c)–(d) we may assume that W ↓ ⊃ S and | S ↓ ( y ) | = ℵ for all y ∈ S . By Lemma 2.2 (a) S ↓ contains either a Suslin treeor an uncountable chain (or both). In the latter case Theorem 1.5 shows that h cannot becontinuous, we may thus assume that S ↓ is a rooted Suslin tree. We now force with S ↓ withreversed order and let G be a generic filter. Then ∪ G is a copy of ω V [ G ]1 which is equal to ω V since no new countable sets are added by Lemma 3.6 (c). Also, ∪ G meets every level of S ↓ since D α = { x ∈ S : height( x ) ≥ α } is order-dense for each α (and the enumeration is in the groundmodel). 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