Let $A\subseteq\mathbb{R}$ with $|A|=\omega_1$. Prove $A$ does not embed into $\mathbb{Q}\times\omega_1$.

Question

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 $\mathbb{Q}\times\omega_1$.

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.