Order type of subsets of $\mathbb{R}$

Question

I need to show that there is no subsets of $\mathbb{R}$ of order type $\omega_1$.

This is a very tricky one, because I was first trying to prove that there are no subsets with cardinality $\omega_1$ by showing that every uncountable subset of $\mathbb{R}$ has cardinality of $P(\omega)$ and thus not $\omega_1$, but I that was not true as there are subsets with cardinality $\omega_1$. The trick for me is to work with order type instead of cardinality.

Answer 1

Hint: Use the fact that between any two real numbers (in particular, between any two real numbers in your subset) you can find a rational number.

A full proof is hidden below.

Suppose $S\subset\mathbb{R}$ and $f:\omega_1\to S$ is an order-isomorphism. For each $\alpha<\omega_1$, choose a rational number $q_\alpha$ between $f(\alpha)$ and $f(\alpha+1)$. This gives a strictly order-preserving map (in particular, an injection) $\alpha\mapsto q_\alpha$ from $\omega_1$ to $\mathbb{Q}$, which is impossible.