Monotone normality and nabla products

Pub Date : 2020-06-26 DOI:10.4064/FM926-10-2020
H. Barriga-Acosta, P. Gartside
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引用次数: 1

Abstract

Roitman's combinatorial principle $\Delta$ is equivalent to monotone normality of the nabla product, $\nabla (\omega +1)^\omega$. If $\{ X_n : n\in \omega\}$ is a family of metrizable spaces and $\nabla_n X_n$ is monotonically normal, then $\nabla_n X_n$ is hereditarily paracompact. Hence, if $\Delta$ holds then the box product $\square (\omega +1)^\omega$ is paracompact. Large fragments of $\Delta$ hold in $\mathsf{ZFC}$, yielding large subspaces of $\nabla (\omega+1)^\omega$ that are `really' monotonically normal. Countable nabla products of metrizable spaces which are respectively: arbitrary, of size $\le \mathfrak{c}$, or separable, are monotonically normal under respectively: $\mathfrak{b}=\mathfrak{d}$, $\mathfrak{d}=\mathfrak{c}$ or the Model Hypothesis. It is consistent and independent that $\nabla A(\omega_1)^\omega$ and $\nabla (\omega_1+1)^\omega$ are hereditarily normal (or hereditarily paracompact, or monotonically normal). In $\mathsf{ZFC}$ neither $\nabla A(\omega_2)^\omega$ nor $\nabla (\omega_2+1)^\omega$ is hereditarily normal.
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单调正态性与纳布拉积
Roitman的组合原理$\Delta$等价于nabla乘积的单调正态性$\nabla (\omega +1)^\omega$。如果$\{ X_n : n\in \omega\}$是一个可度量空间族,并且$\nabla_n X_n$是单调正规的,那么$\nabla_n X_n$是遗传的仿紧的。因此,如果$\Delta$成立,那么盒子产品$\square (\omega +1)^\omega$是准紧凑的。$\Delta$的大片段保留在$\mathsf{ZFC}$中,产生了$\nabla (\omega+1)^\omega$的大的子空间,这些子空间是“真正的”单调正态的。在分别为:$\mathfrak{b}=\mathfrak{d}$, $\mathfrak{d}=\mathfrak{c}$或模型假设下,任意,大小为$\le \mathfrak{c}$或可分的可度量空间的可数nabla积单调正态。它是一致的和独立的,$\nabla A(\omega_1)^\omega$和$\nabla (\omega_1+1)^\omega$是遗传正态(或遗传似紧,或单调正态)。在$\mathsf{ZFC}$中,$\nabla A(\omega_2)^\omega$和$\nabla (\omega_2+1)^\omega$都不是遗传正常的。
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