9
$\begingroup$

In ZFC, cardinality of set of linear orders over $\omega$ is $2^{\aleph_0}$. By the argument given by here, we can prove (without the choice) the number of linear orders over $\omega$ is at least $2^{\aleph_0}$. In addition, we can prove if $A$ is a set of countable linear order-types then $|A|\ge \aleph_1$ in ZF.

Therefore we can prove $|A|\ge 2^{\aleph_0}$ and $|A|\ge\aleph_1$ in ZF. If we assume the choice then we can prove $|A|\le 2^{\aleph_0}$. However, if we assume the AD (in fact, it is enough that assuming $\aleph_1$ and $2^{\aleph_0}$ are incomparable) then $|A|>2^{\aleph_0}$.

My question is : if we assume the AD (or, every subset of reals is Lebesgue measurable) then $|A|=2^{\aleph_0}+\aleph_1$? If not, there are known results about the cardinality of set of countable linear-order types?


The set of countable linear-order types are defined as follows : Let $C\subset \mathcal{P}( \omega\times \omega)$ be a set of all linear order over $\omega$. Define $\le_1\,\sim\, \le_2$ iff $(\omega,\le_1)$ and $(\omega,\le_2)$ is isomorphic (as linearly ordered set.) Then $\sim$ is a equivalence relation over $C$, and $C/\sim$ is a set of all 'linear order-types' over a countable set.

$\endgroup$
7
$\begingroup$

Nice question!

To get us started, a simple variant of the argument at that link gives us that there is an injection from $\omega_1^{<\omega_1}$ into $C/\sim$: Given such a sequence $(\alpha_\iota\mid \iota<\beta)$, consider the ordered sum $$\underbrace{\alpha_0+(\omega^*+\omega)+\alpha_1+(\omega^*+\omega)+\dots}_\beta,$$ where the underbrace is simply my poor notation to indicate that the sum continues transfinitely to include all $\alpha_\iota$ in the sequence.

The set $\omega_1^{<\omega_1}$ of (well-ordered) countable sequences of countable ordinals is much larger than $\mathfrak c+\omega_1$ (see for instance Woodin's paper on The cardinals below $|[\omega_1]^{<\omega_1}|$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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