() @ Appl. Gen. Topol. 16, no. 1(2015), 75-80doi:10.4995/agt.2015.3244 c© AGT, UPV, 2015 When is a space Menger at infinity? Leandro F. Aurichi a and Angelo Bella b a Instituto de Ciências Matemáticas e de Computação, Universidade de São Paulo, São Carlos, SP, Brazil (aurichi@icmc.usp.br) b Dipartimento di Matematica, Città Universitaria, Catania, Italy (bella@dmi.unict.it) Abstract We try to characterize those Tychonoff spaces X such that βX \ X has the Menger property. 2010 MSC: 54F65; 54D40; 54D20. Keywords: Menger; remainder. 1. Introduction A space X is Menger (or has the Menger property) if for any sequence of open coverings {Un : n < ω} one may pick finite sets Vn ⊆ Un in such a way that⋃ {Vn : n < ω} is a covering. This equivals to say that X satisfies the selection principle Sfin(O, O). It is easy to see the following chain of implications: σ-compact −→ Menger −→ Lindelöf An important result of Hurewicz [4] states that a space X is Menger if and only if player 1 does not have a winning strategy in the associated game Gfin(O, O) played on X. This highlights the game-theoretic nature of the Menger property, see [7] for more. Henriksen and Isbell ([3]) proposed the following: Definition 1.1. A Tychonoff space X is Lindelöf at infinity if βX \ X is Lindelöf. They discovered a very elegant duality in the following: Received 26 August 2014 – Accepted 2 January 2015 http://dx.doi.org/10.4995/agt.2015.3244 L. F. Aurichi and A. Bella Proposition 1.2 ([3]). A Tychonoff space is Lindelöf at infinity if and only if it is of countable type. A space X is of countable type provided that every compact set can be included in a compact set of countable character in X. A much easier and well-known fact is: Proposition 1.3. A Tychonoff space is Čech-complete if and only if it is σ- compact at infinity. These two propositions suggest the following: Question 1.4. When is a Tychonoff space Menger at infinity? Before beginning our discussion here, it is useful to note these well known facts: Proposition 1.5. The Menger property is invariant by perfect maps. Corollary 1.6. X is Menger at infinity if, and only if, for any Y compactifi- cation of X, Y \ X is Menger. Fremlin and Miller [6] proved the existence of a Menger subspace X of the unit interval [0, 1] which is not σ-compact. The space X can be taken nowhere locally compact and so Y = [0, 1] \ X is dense in [0, 1]. Since the Menger property is invariant under perfect mappings, we see that βY \ Y is still Menger. Therefore, a space can be Menger at infinity and not σ-compact at infinity. Another example of this kind, stronger but not second countable, is Example 3.1 in the last section. On the other hand, the irrational line shows that a space can be Lindelöf at infinity and not Menger at infinity. Consequently, the property M characterizing a space to be Menger at infin- ity strictly lies between countable type and Čech-complete. Of course, taking into account the formal definition of the Menger property, we cannot expect to have an answer to Question 1.4 as elegant as Henriksen- Isbell’s result. 2. A characterization Definition 2.1. Let K ⊂ X. We say that a family F is a closed net at K if each F ∈ F is a closed set such that K ⊂ F and for every open A such that K ⊂ A, there is an F ∈ F such that F ⊂ A. Lemma 2.2. Let X be a T1 space. If (Fn)n∈ω is a closed net at K, for K ⊂ X compact, then K = ⋂ n∈ω Fn. Proof. Simply note that for each x /∈ K, there is an open set V such that K ⊂ V and x /∈ V . � Lemma 2.3. Let Y be a regular space and let X be a dense subspace of Y . Let K ⊂ X be a compact subset. If (Fn)n∈ω is a closed net at K in X, then (Fn Y )n∈ω is a closed net at K in Y . c© AGT, UPV, 2015 Appl. Gen. Topol. 16, no. 1 76 When is a space Menger at infinity? Proof. In the following, all the closures are taken in Y . Let A be an open set in Y such that K ⊂ A. By the compactness of K and the regularity of Y , there is an open set B such that K ⊂ B ⊂ B ⊂ A. Thus, there is an n ∈ ω such that K ⊂ Fn ⊂ B ∩ X. Note that K ⊂ F n ⊂ B ⊂ A. � Lemma 2.4. Let X be a compact Hausdorff space. If K = ⋂ n∈ω Fn, where (Fn)n∈ω is a decreasing sequence of closed sets, then (Fn)n∈ω is a closed net at K. Proof. If not, then there is an open set V such that K ⊂ V and, for every n ∈ ω, Fn \ V 6= ∅. By compactness, there is an x ∈ ⋂ n∈ω Fn \ V = K \ V . Contradiction with the fact that K ⊂ V . � Theorem 2.5. Let X be a Tychonoff space. X is Menger at infinity if, and only if, X is of countable type and for every sequence (Kn)n∈ω of compact subsets of X, if (F np )p∈ω is a decreasing closed net at Kn for each n, then there is an f : ω −→ ω such that K = ⋂ n∈ω F n f(n) is compact and ( ⋂ k≤n F k f(k) )n∈ω is a closed net for K. Proof. In the following, every closure is taken in βX. Suppose that X is Menger at infinity. By Lemma 1.2 X is of countable type. Let (F np )p,n∈ω be as in the statement. Note that, by Lemma 2.3 and Lemma 2.2, ⋂ p∈ω F np = ⋂ p∈ω F np for each n ∈ ω. Thus, for each n ∈ ω, (V n p )p∈ω, where V np = βX \ F n p , is an increasing covering for βX \ X. Since βX \ X is Menger, there is an f : ω −→ ω such that βX \ X ⊂ ⋃ n∈ω V n f(n) . Note that K = ⋂ n∈ω F n f(n) is compact and it is a subset of X. By Lemma 2.4, ( ⋂ k≤n F k f(k) )n∈ω is a closed net at K in βX, therefore, ( ⋂ k≤n F k f(k) )n∈ω is a closed net at K in X. Conversely, for each n ∈ ω, let Wn be an open covering for βX \ X. We may suppose that each W ∈ Wn is open in βX. By regularity, we can take a refinement Vn of Wn such that, for every x ∈ βX \ X, there is a V ∈ Vn such that x ∈ V ⊂ V ⊂ WV for some WV ∈ Wn. Since X is of countable type, By Lemma 1.2 we may suppose that each Vn is countable. Fix an enumeration for each Vn = (V n k )k∈ω. Define A n k = βX \ ( ⋃ j≤k V nj ). Note that each Kn = ⋂ k∈ω An k is compact and a subset of X. By Lemma 2.4, (An k )k∈ω is a closed net at Kn. Thus, (A n k ∩ X)k∈ω is a closed net at Kn in X. Therefore, there is f : ω −→ ω such that K = ⋂ n∈ω (An f(n) ∩ X) is compact and ( ⋂ k≤f(n) A k f(k) ∩ X)n∈ω is a closed net at K. So, by Lemma 2.3, K = ⋂ n∈ω (An f(n) ∩ X). Since ⋂ n∈ω (An f(n) ∩ X) = ⋂ n∈ω An f(n) and by the fact that K ⊂ X, it follows that βX \X ⊂ ⋃ n∈ω βX \An f(n) ⊂ ⋃ n∈ω Int( ⋃ j≤f(n) V n j ) ⊂⋃ n∈ω ⋃ j≤f(n) WV nj . Therefore, letting Un = {WV n j : j ≤ f(n)} ⊂ Wn, we see that the collection ⋃ n∈ω Un covers βX \ X, and we are done. � Property M given in the above theorem does not look very nice and we wonder whether there is a simpler way to describe it, at least in some special cases. c© AGT, UPV, 2015 Appl. Gen. Topol. 16, no. 1 77 L. F. Aurichi and A. Bella Recall that a metrizable space is always of countable type. Moreover, a metrizable space is complete if and only if it is σ-compact at infinity. Therefore, we could hope for a “nicer” M in this case. Question 2.6. What kind of weak completeness characterizes those metrizable spaces which are Menger at infinity? Proposition 2.7. Let X be a Tychonoff space. If X is Menger at infinity then for every sequence (Kn)n∈ω of compact sets, there is a sequence (Qn)n∈ω of compact sets such that: (1) each Kn ⊂ Qn; (2) each Qn has a countable base at X; (3) for every sequence (Bn k )n,k∈ω such that, for every n ∈ ω, (B n k )k∈ω is a decreasing base at Kn, then there is a function f : ω −→ ω such that K = ⋂ n∈ω Bn f(n) is compact and ( ⋂ k≤n Bk f(k) )n∈ω is a closed net at K. Proof. Suppose X is Menger at infinity. Let (Kn)n∈ω be a sequence of com- pact sets. Since X is Menger at infinity, X is Lindelöf at infinity. Thus, by Proposition 1.2, for each Kn, there is a compact Qn ⊃ Kn such that Qn has a countable base. Now, let (Bn k )k,n be as in 3. Since each Qn is compact and X is regular, each (Bn k )k∈ω is a decreasing closed net at Qn. Thus, by Theorem 2.5, there is an f : ω −→ ω as we need. � To some extent, the Menger property is closer to σ- compactness rather than to Lindelöfness. Since a Čech- complete space has the Baire property, we may ask: Question 2.8. Is it true that a space Menger at infinity has the Baire property? We thank M. Sakai for calling our attention to the above question. He also noticed a partial answer to it: Theorem 2.9 (Sakai). Let X be a first countable Tychonoff space. If X is Menger at infinity, then X is hereditarily Baire. Proof. According to a result of Debs [2], a regular first countable space is hereditarily Baire if and only if it contains no closed copy of the space of rationals Q. To finish, it suffices to observe that Q is not Menger at infinity. � We end this section presenting a selection principle that at first glance could be related with the Menger at infinity property. Definition 2.10. We say that a family U of open sets of X is an almost covering for X if X \ ⋃ U is compact. We call A the family of all almost coverings for X. Note that the property “being Menger at infinity” looks like something as Sfin(A, A), but for a narrow class of A. We will see that the “narrow” part is important. Proposition 2.11. If X satisfies Sfin(A, A), then X is Menger. c© AGT, UPV, 2015 Appl. Gen. Topol. 16, no. 1 78 When is a space Menger at infinity? Proof. Let (Un)n∈ω be a sequence of coverings of X. By definition, for each n ∈ ω, there is a finite Un ⊂ Un, such that K = X \ ⋃ n∈ω ⋃ Un is compact. Therefore, there is a finite W ⊂ Un such that K ⊂ ⋃ W . Thus, X = W ∪⋃ n∈ω ⋃ Un. � Example 2.12. The space of the irrationals is an example of a space that is Menger at infinity but does not satisfy Sfin(A, A) (by the Proposition 2.11). Example 2.13. The one-point Lindelöfication of a discrete space of cardinality ℵ1 is an example of a Menger space which does not satisfy Sfin(A, A). Example 2.14. ω is an example of a space that satisfies Sfin(A, A), but it is not compact. Proof. Let (Vn)n∈ω be a sequence of almost coverings for ω. Therefore, for each n, Fn = ω \ ⋃ Vn is finite. For each n, let Vn ⊂ Vn be a finite subset such that Fn+1 \Fn ⊂ ⋃ Vn and min(ω \ ⋃ k