Given an infinite cardinal $\kappa$, is there a connected Hausdorff space $(X,\tau)$ with $|X|=\kappa$, and for every infinite cardinal $\lambda \leq \kappa$ there is an open set $U\in \tau$ with $|U| = \lambda$?
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asked May 10, 2018 at 7:53
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1$\begingroup$ If you take an ordinal $\alpha$ and $X=\alpha\times [0,1[$ with lexicographic order, you already get open subsets of all cardinals in $[2^{\aleph_0},\alpha]$. $\endgroup$– YCorCommented May 10, 2018 at 8:02
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1$\begingroup$ Taking the 1-point union with a countable connected space $Y$ with$y\in Y$ and $Y-\{y\}$ connected also yields a countable open subset. If CH holds this is enough. $\endgroup$– YCorCommented May 10, 2018 at 8:03
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1$\begingroup$ Without CH, I don't even know if in ZFC it holds that for every infinite cardinal there a Hausdorff connected space of the given cardinal (it's clear for $\ge 2^{\aleph_0}$). $\endgroup$– YCorCommented May 10, 2018 at 8:05
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1$\begingroup$ I think not as connected Hausdorff spaces are uncountable. Apart from that, a very long line can give many desired cardinalities above the bare minimum for a connected open set. Gerhard "Connected Sets Are Not Arbitrary" Paseman, 2018.05.10. $\endgroup$– Gerhard PasemanCommented May 10, 2018 at 8:14
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1 Answer
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Let $C$ be Bing's countable connected Hausdorff space. Specify some point $0 \in C$. Let $D(\alpha)$ be the discrete space of cardinality $\alpha$. Let $F(\alpha)$ be the product $C \times D(\alpha)$ with the set $\{0\}\times D(\alpha)$ treated as a point $r(\alpha) \in F(\alpha)$. It is easy to see $F(\alpha)$ is connected and Hausdorff and has an open set $(C-0) \times D(\alpha)$ of cardinality $\aleph_0 \times \alpha = \alpha$.
Now let $\displaystyle D = \bigsqcup_{\alpha< \kappa}F(\alpha)$ be the disjoint union and treat the set $\{ r(\alpha): \alpha < \kappa)\}$ as a point. All the images of $(C-0) \times D(\alpha)$ are open in the quotient space and $D$ itself is open with cardinalty $\kappa$.
Edit: Here's an easier one. Start with $C \times D(\kappa)$ and treat $\{0\} \times D(\kappa)$ as a point. Then the sets $(C-0) \times \{d\}$ are open and disjoint and countable. By taking unions you get open sets of all possible cardinalities.
