Let $A\subseteq\mathbb{R}$ with $|A|=\omega_1$.

Prove $A$ does not embed into $\mathbb{Q}\times\omega_1$.

$\mathbb{Q}\times\omega_1$ is ordered with the lexicographic ordering.

I know that $A$ does not embed into $\omega_1$ since otherwise $A$ and $\omega_1$ are isomorphic and $\omega_1$ embeds into $A$. Then we can find a unique rational for each element in $\omega_1$, this is an uncountable amount of unique rationals, which is a contradiction.

I don't know how to extend this proof that $A$ does not embed into $\omega_1$ into a proof that $A$ does not embed into $\mathbb{Q}\times\omega_1$.

  • $\begingroup$ What do you mean with embed? Just an injection? Must it preserve the order? $\endgroup$
    – J. De Ro
    Apr 23, 2020 at 22:09
  • $\begingroup$ Yes, it is an order preserving injection. $\endgroup$
    – James Gwin
    Apr 23, 2020 at 22:23

1 Answer 1


HINT: Use the fact that $\Bbb Q$ is countable to show that if $f:A\to\Bbb Q\times\omega_1$ were an embedding, there would be a $q\in\Bbb Q$ such that $f[A]\cap(\{q\}\times\omega_1)$ was uncountable.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .