Short proof of a theorem of Juhasz

arXiv:1102.4564v2 [math.GN] 24 Feb 2011 SHORT PROOF OF A THEOREM OF JUH´ASZ SANTI SPADARO Abstract. We give a simple proof of the increasing strengthening of Arhangel’skii’s Theorem. 1. Introduction The pair (X, τ) denotes a Hausdorfftopological space. In chapter 6 of his book [6], Istv´an Juh´asz proves the following theorem. Theorem 1.1. ([6], 6.11) Suppose X = S α<λ Xα, where Xα ⊂Xβ whenever α < β and t(Xα) · ψ(Xα) · L(Xα) ≤κ for every α < λ. Then |X| ≤2κ. Where ψ(X), t(X) and L(X) are respectively the pseudocharacter, the tightness and the Lindel¨of number of X (see [3] or [6]). This may be considered an increasing strengthening (in the sense of Juh´asz and Szentmikl´ossy [8]) of the following theorem of Arhangel’skii and Shapirovskii. Theorem 1.2. (Arhangel’skii and Shapirovskii, [9]) |X| ≤2ψ(X)·t(X)·L(X) Juh´asz’s proof of Theorem 1.1 is five pages long. We offer a simpler and shorter proof with some help from the technique of elementary submodels. The background on this technique needed to read this paper is quite basic and can be found in the first few sections of [2]. Recall the definition of a free sequence, which already had a crucial role in the original proof [1] of Arhangel’skii’s famous theorem saying that the cardinality of a first-countable Lindel¨of space never exceeds the continuum (a special case of Theorem 1.2). Definition 1.3. A set {xα : α < κ} is called a free sequence of length κ if for every β < κ we have {xα : α < β} ∩{xα : α ≥β} = ∅. As usual, we let F(X) = sup{|F| : F ⊂X is a free sequence }. It is known that F(X) ≤L(X) · t(X). Indeed, assume that L(X) · t(X) ≤κ. If X contained a free 2000 Mathematics Subject Classification. 54A25. Key words and phrases. Arhangel’skii Theorem, increasing union, free sequence, elementary submodel. Supported by the Center for Advanced Studies in Mathematics at Ben Gurion University. 1 2 SANTI SPADARO sequence F of size κ+ then by L(X) ≤κ, F would have a complete accumulation point. But no complete accumulation point of a free sequence can lie in the closure of an initial segment of it. And this contradicts t(X) ≤κ. 2. The main proof Before proving Theorem 1.1 we need a very simple lemma. Define Φ(X) = sup{L(X\ {x}) : x ∈X}. Lemma 2.1. Φ(X) = L(X) · ψ(X). Proof. Since any open cover of X \ {x} can be extended to a cover of X by the addition of a single open set, we have L(X) ≤Φ(X). If for some x ∈X we have L(X \ {x}) ≤κ then for every y ̸= x select Uy such that x /∈Uy. Then U = {Uy : y ̸= x} covers X \ {x} and hence we can find a subcover V having cardinality ≤κ. Then T{X \ U : U ∈U} = {x}, which proves that ψ(x, X) ≤κ. So, taking suprema we have that ψ(X) ≤Φ(X), and hence ψ(X) · L(X) ≤Φ(X). To prove the other direction, fix x ∈X suppose that L(X)·ψ(X) = κ. Let U be an open collection such that |U| ≤κ and T U = {x}. Then X \{x} = S{X \U : U ∈U} and L(X \ U) ≤κ for every U ∈U. Thus L(X \ {x}) ≤κ. □ Proof of Theorem 1.1. If λ ≤2κ then we are done by Theorem 1.2, so we can assume that λ = (2κ)+. Let M be an elementary submodel of H(µ), where µ is a large enough regular cardinal, such that [M]κ ⊂M, |M| = 2κ, {κ, τ, λ} ⊂M and {Xα : α < λ} ∈ M. Call a set C ⊂X bounded if |C| ≤2κ. Claim 1. If C ∈[X ∩M]≤κ then C is bounded. Proof of Claim 1. The proof of this claim is an extension of the proof of Subclaim 2 of Example 2.2 in [2]. We will achieve Claim 1 if we can prove that C ⊂X ∩M. So, suppose that this is not true and choose p ∈C \ M. For every θ < λ such that p ∈Xθ we can use ψ(Xθ) ≤κ to find open sets {Uθ α : α < κ} such that Xθ \ {p} = S α<κ Xθ \ Uθ α. By L(Xθ \ Uθ α) ≤κ we can find relative open sets {V θ αβ : β < κ} in Xθ covering Xθ \ Uθ α such that p /∈V θ αβ for every α, β < κ. Then we have C∩(Xθ\{p}) = S α,β<κ Cθ αβ∩Xθ for every θ < λ, where Cθ αβ = V θ αβ∩C. Note now that by κ-closedness of M, Cθ αβ ∈M for every α, β and θ. Moreover, since p /∈M (∀θ ∈λ ∩M)(C ∩Xθ ∩M = [ α,β<κ Cθ αβ ∩Xθ ∩M) So M |= (∀θ < λ)(C ∩Xθ = [ α,β<κ Cθ αβ ∩Xθ) SHORT PROOF OF A THEOREM OF JUH´ASZ 3 which implies H(µ) |= (∀θ < λ)(C ∩Xθ = [ α,β<κ Cθ αβ ∩Xθ) which is a contradiction because p ∈Xθ for some θ < λ. △ Now we claim that X ⊂M. Suppose not and choose p ∈X \ M. Claim 2. The collection U = {U ∈M ∩τ : p /∈U} is an open cover of X ∩M. Proof of Claim 2. Fix x ∈X ∩M and let V = {V ∈τ : x /∈V }. Note that V ∈M and V covers X \{x}. Suppose you have constructed subcollections {Vα : α < β} of V such that Vα ∈M, |Vα| ≤κ for every α < β and a set {xα : α < β} ⊂X∩M such that ClX\{x}({xα : α < γ}) ⊂S S α<γ Vα for every γ < β. By Claim 1, ClX\{x}({xα : α < β}) is bounded and hence there is λβ < λ such that ClX\{x}({xα : α < β}) ⊂Xλβ. By Lemma 2.1, L(ClX\{x}({xα : α < β})) ≤κ, so there is a subcollection Vβ of V such that |Vβ| ≤κ and ClX\{x}({xα : α < β}) ⊂S Vβ. If S α≤β Uα does not cover X \ {x} then we can fix a point xβ ∈((X \ {x}) ∩M) \ S α≤β Uα. If the induction doesn’t stop at an ordinal below κ+ then F = {xα : α < κ+} is a free sequence of length κ+ in X \ {x}. Since F is bounded we can

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