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Let $C$ be an uncountable set.

Can we construct a set $A \subseteq C^2$ such that it has a cocountable number of cocountable horizontal fibers, and a cocountable number of countable vertical fibers?

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2 Answers 2

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Let $C = \omega_1$ and $<$ be the ordinal ordering of $C$. Since $C = \omega_1$, it is the least uncountable ordinal (cardinal). Hence for all $\eta \in \omega_1$, $\{\alpha < \eta\}$ is countable. Define $A \subset C^2$ as follows:

$A = \{(\alpha,\beta) \in C^2 : \alpha < \beta\}$.

Then for any fixed $\alpha$, $\{\beta : (\alpha, \beta) \in C^2\} = \{\beta : \beta > \alpha\}$ which is uncountable. However for any fixed $\beta$, $\{\alpha : (\alpha, \beta) \in C^2\} = \{\alpha : \alpha < \beta\}$ which is countable.

This $A$ is the desired set.

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Since removing a countable set from C does not change its cardinality (at least in ZFC), it is equivalent to ask whether there is A such that ALL the horizontal fibers are cocountable and all the vertical fibers are countable.

The answer is yes if $C=\aleph_1$ because we can let A be a well-ordering of C. Apparently, for $C=\mathbb{R}$, such an $A$ exists if and only if the continuum hypothesis ($|\mathbb{R}|=\aleph_1$) holds: see page 10 here.

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