Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 3 down vote accepted

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.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.