Why is the minimal uncountable well-ordered set $S_{\Omega}$ unique?

Question

In section 10 of Topology by Munkres, the minimal uncountable well-ordered set $S_{\Omega}$ is introduced. Furthermore, it is remarked that,

Note that $S_{\Omega}$ is an uncountable well-ordered set every section of which is countable. Its order type is in fact uniquely determined by this condition.

However, how to justify its uniqueness?

Answer 1

This stems from the fact that given two well-ordered sets $X, Y$ exactly one of the following holds (see this previous question):

1. $X$ is order-isomorphic to $Y$;
2. There is a (unique) $a \in X$ such that $\{ x \in X : x < a \}$ is order-isomorphic to $Y$; or
3. There is a (unique) $b \in B$ such that $\{ y \in Y : y < b \}$ is order-isomorphic to $X$.

So suppose $X$ is another well-ordered set which is not order-isomorphic to $S_\Omega$. Then either

• there is a (unique) $b \in S_\Omega$ such that $\{ y \in S_\Omega : y < b \}$ is order-isomorphic to $X$, and since $\{ y \in S_\Omega : y < b \}$ is countable, it follows that $X$ is also countable.

• there is a (unique) $a \in X$ such that $\{ x \in X : x < a \}$ is order-isomorphic to $S_\Omega$. But then $\{ x \in X : x <_X a \}$ is itself an initial section of $X$ which is uncountable.

So any well-ordered set which is not order-isomorphic to $S_\Omega$ is either countable, or has an uncountable proper initial section.

Answer 2

Suppose that $\alpha$ and $\beta$ are two uncountable ordinals. Suppose further that for every $\gamma<\alpha$, $\gamma$ is countable. Likewise, suppose that for very $\eta<\beta$, $\eta$ is countable. You want to show that $\alpha=\beta$.

Suppose that $\alpha<\beta$. Then $\eta=\alpha$ is uncountable, and satisfies $\eta<\beta$. This is a contradiction. Thus $\alpha\not<\beta$. Likewise $\beta\not<\alpha$.