How is the Continuum Hypothesis equivalent to the existence of a well-ordering on $\Bbb R$ whose bounded initial segments are countable?

Question

There exists an well-ordering $(<)$ on $\Bbb R$ such that the set $\{x \in \Bbb R\mid x < y \}$ is countable for every $y \in \Bbb R.$

How to prove that the above statement is equivalent to continuum hypothesis? Any help will be highly appreciated.

Thank you very much.

Answer 1

Let $\omega_1$ be the least uncountable ordinal. The continuum hypothesis states that $|\omega_1| = |\mathbb{R}|$.

The important observation to see the equivalence of your statement and the continuum hypothesis is that the only uncountable ordinal which has the property that every proper initial segment is countable is $\omega_1$. To see this, suppose $\alpha$ is an uncountable ordinal, so that $\omega_1 \leq \alpha$. If $\omega_1 < \alpha$, then $\omega_1$ is a proper initial segment of $\alpha$ which is uncountable, contrary to hypothesis. Thus, $\alpha = \omega_1$.

So now suppose there exists a well-ordering $\leq$ on $\mathbb{R}$ such that for each $y \in \mathbb{R}$ the set $\{x \in \mathbb{R} \mid x < y\}$ is countable. The set $\{x \in \mathbb{R} \mid x < y\}$ is exactly the proper initial segment of $(\mathbb{R},\leq)$ determined by $y$, and every proper initial segment of $(\mathbb{R},\leq)$ arises in this way. It follows from the above observation that the order-type (i.e., the unique ordinal which is order isomorphic to) of $(\mathbb{R},\leq)$ is $\omega_1$. In particular, there is a bijection between $\omega_1$ and $\mathbb{R}$ given by such an order isomorphism.

Conversely, given a bijection $f\colon\omega_1 \to \mathbb{R}$, defined a binary relation on $\mathbb{R}$ by declaring $f$ to be an order isomorphism. By construction, $\leq$ is a well-ordering of $\mathbb{R}$ with the additional property that every proper initial segment of $(\mathbb{R},\leq)$ is countable, i.e., for every $y \in \mathbb{R}$, the set $\{x \in \mathbb{R} \mid x < y\}$ is countable.