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

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

• What do you mean with embed? Just an injection? Must it preserve the order? Apr 23, 2020 at 22:09
• Yes, it is an order preserving injection. Apr 23, 2020 at 22:23

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.