Old exam question about well ordering on $\mathbb{R}$

Question

I am not sure how to start tackling this question and would love a hint:

Let $<$ the regular order relation on $\mathbb{R}$, and $<_w$ well ordering on $\mathbb{R}$. We define a coloring function $h:[\mathbb{R}]^2 \rightarrow 2$ like so: $$ h(\{x,y\}) = \left\{\begin{matrix} 1 &((x < y) \wedge (x <_w y)) \vee ((y < x) \wedge(y<_wx)) \\ 0 & otherwise \end{matrix}\right. $$ Prove that there isn't an homogeneous $H \subseteq \mathbb{R}$ such that $|H| = \aleph_1$.

I'm not sure if I'm "allowed" or need to assume AC here. Looks like the question is related to combinatoric properties of large cardinals (the result would be that $\mathbb{R}$ is not weakly compact, if I'm not mistaking), but non of the theorems we saw seems relevant.

Answer 1

The question is not related to large cardinals. Rather it is closer to Ramsey theory, and its various degrees of failure in infinite sets.

Suppose that there was such homogeneous $H$ of size $\aleph_1$.

If $H$ is homogeneous and its color is $1$ then there is an order embedding of an uncountable ordinal into $\Bbb R$, which is impossible.

If $H$ is homogeneous and its color is $0$ then there is a sequence $h_n\in H$ which is $<$-increasing. By homogeneity this means that whenever $n<k$, $h_k<_w h_n$. This is a decreasing sequence in the well-order, which is impossible.

To fill in the details in the above you proof you need to know that every well-ordered subset of $\Bbb R$ is countable (that is, well-ordered by the usual ordering). It's not hard to prove, and there are various proofs (especially if we are allowed to use the axiom of choice, then we can appeal to all sort of topological proofs).

Answer 2

Hint: Suppose there is such a homogeneous set, and use it to construct a subset of $\mathbb{R}$ whose order-type under $<$ is $\omega_1$. This must be a contradiction because, by selecting rational points between points in this sequence, you get a strictly increasing $\omega_1$-sequence of rationals.