2009
DOI: 10.1090/s0002-9939-09-09887-6
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Menger subsets of the Sorgenfrey line

Abstract: Abstract. A space X is said to have the Menger property if for every sequence {U n : n ∈ ω} of open covers of X, there are finite subfamilies V n ⊂ U n (n ∈ ω) such that n∈ω V n is a cover of X. Let i : S → R be the identity map from the Sorgenfrey line onto the real line and let X S = i −1 (X) for X ⊂ R. Lelek noted in 1964 that for every Lusin set L in R, L S has the Menger property. In this paper we further investigate Menger subsets of the Sorgenfrey line. Among other things, we show: (1) If X S has the Me… Show more

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Cited by 5 publications
(3 citation statements)
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References 25 publications
(27 reference statements)
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“…Pick a Sierpiński set S ⊆ R and endow it with the Sorgenfrey line topology. By Corollary 3.6 of [28], S is Menger. Since S does not have measure zero, it cannot be Rothberger [27]; thus, as every compact subset of the Sorgenfrey line is countable, it follows from Corollary 3.9 that S does not satisfy S 1 (K, O).…”
Section: Example 34mentioning
confidence: 91%
“…Pick a Sierpiński set S ⊆ R and endow it with the Sorgenfrey line topology. By Corollary 3.6 of [28], S is Menger. Since S does not have measure zero, it cannot be Rothberger [27]; thus, as every compact subset of the Sorgenfrey line is countable, it follows from Corollary 3.9 that S does not satisfy S 1 (K, O).…”
Section: Example 34mentioning
confidence: 91%
“…Let S be the Sorgenfrey line and let R be the set of reals. If i : S → R is the identity map and X ⊂ R then by X S we denote i −1 (X) (see [20]). Lelek showed in [16] that for every Lusin set L in R, L S is Menger and, therefore, almost Menger, but if L satisfies that (L × L) ∩ {(x, y) : x + y = 0} is uncountable, then L S × L S is not Menger and since S × S is regular and every subspace of a regular space is regular, we have that the square of L S is regular and not Menger.…”
Section: Subspaces and Productsmentioning
confidence: 99%
“…Such is any Lusin set or, under MA, any uncountable set of cardinality less than continuum. These sets are Hurewicz [9], [13], and hence totally paracompact [6], see also [2], [12], [14], [15], [25]. G. Gruenhage gave a direct proof for Lusin subspaces that the base of all half-open intervals contains no coarse base.…”
Section: Introductionmentioning
confidence: 99%