Are there such a complete metric space $X$ of weight $k<\mathfrak{c}$ ($w(X)=k$) and a family $\{F_{\alpha}: \alpha<k\}$ of closed subsets of $X$ that $k<|X\setminus \bigcup F_{\alpha}|<\mathfrak{c}$ holds ?
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4$\begingroup$ Not if the Continuum Hypothesis holds. The $k=\aleph_0$ and there is no cardinality between $\aleph_0$ and $\mathfrak{c}$. $\endgroup$– KP HartCommented Apr 4, 2022 at 16:30
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1$\begingroup$ @KPHart, note that if $\kappa=\aleph_0$ then the answer is no, regardless of the size of the continuum, since then $X\setminus \bigcup F_\alpha$ is completely metrizable and therefore it is either countable or has size continuum. $\endgroup$– Ramiro de la VegaCommented Apr 4, 2022 at 16:34
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1 Answer
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As K.P. and Ramiro both point out in the comments, it follows from $\mathsf{CH}$ that the answer is no. I claim that it is also consistent that the answer is yes.
It is consistent that $\mathfrak{c} > \aleph_2$ and that there is a partition $\mathcal P$ of the Cantor space $2^\omega$ into $\aleph_2$ closed sets, such that all but $\aleph_1$ members of $\mathcal P$ are singletons. (See Theorem 3.11 in this paper for a proof.)
Let $Y$ be any completely metrizable space of weight $\kappa = \aleph_1$, and let $X$ be the disjoint sum of $Y$ and the Cantor space. Let $F_0 = Y$, and let $\{F_\alpha :\, 1 \leq \alpha < \omega_1 \}$ be an enumeration of the members of $\mathcal P$ that are not singletons.
answered Apr 4, 2022 at 18:08
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$\begingroup$ Will, Thanks! Great answer! $\endgroup$ Commented Apr 5, 2022 at 3:13