2002
DOI: 10.1016/s0166-8641(01)00220-6
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κ-normality and products of ordinals

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Cited by 16 publications
(19 citation statements)
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“…Nagami [17] proved that if every finite subproduct of X = i∈ω X i has covering dimension n and X is normal, then X has also covering dimension n. With the result of [8], these results yield: Lemma 1.9 [10,17,8]. Let α i be an ordinal and…”
Section: Strong Zero-dimensionality and κ -Normalitymentioning
confidence: 96%
See 2 more Smart Citations
“…Nagami [17] proved that if every finite subproduct of X = i∈ω X i has covering dimension n and X is normal, then X has also covering dimension n. With the result of [8], these results yield: Lemma 1.9 [10,17,8]. Let α i be an ordinal and…”
Section: Strong Zero-dimensionality and κ -Normalitymentioning
confidence: 96%
“…Note that the proof does not use elementary submodel techniques. Lemma 1.8 [10]. Let α i be an ordinal and Y i = {β < α i : cf β ω} for each i ∈ ω.…”
Section: Strong Zero-dimensionality and κ -Normalitymentioning
confidence: 99%
See 1 more Smart Citation
“…The converse is not always true. The space u\ x ui\ + 1 is mildly normal, see [2] and [3], but not almost normal because the closed subset A = u>i x {wi} is disjoint from the regularly closed subset B = {(A, A) : a < LO\} and they cannot be separated by two disjoint open subsets, see [1].…”
Section: A Subset a Of A Topological Space X Is Called Regularly Closmentioning
confidence: 99%
“…For instance, ω 1 × (ω 1 + 1) is mildly normal but not normal. Moreover, using elementary submodels, it is proved in [4] that products of arbitrary many ordinals are mildly normal. In [6], it is proved that for A, B ⊆ ω 1 , A × B is normal if and only if A or B is non-stationary or A ∩ B is stationary in ω 1 .…”
Section: Introductionmentioning
confidence: 99%