How about sharing? The ordinals are interesting. What did we learn?

Those familiar with the thread up until now may find this post redundant save for my attempt to rewrite restriction #1.

We find $\phi(\gamma)$ is true for an ordinal $\gamma$ if and only if we have a suitable $S$ where, again, $S$ is a sequence of sets of ordinals that must adhere to three restrictions (the first loosened per above):

1) $S$ = an infinite ordinal sequence of sets, where â€œinfinite ordinal sequenceâ€ implies the sequence can be either a traditional sequence of order type $\omega$ or of any ordinal greater than it, so long as the order type has no greatest element under the standard $<$ ordering (let $s_i$ denote an element of $S$ where $i$ is an ordinal number representing the position of $s_i$ within $S$),

2) $\{ s_i : s_i \in S \}$ is a partition of $\gamma$, and,

3) where $s_i, s_j \in S$, we have $s_i \prec s_j \iff i < j$ (the relation $\prec$ is defined in the original post).

For purposes of proving $\phi(\omega_1)$ true, select $\omega_1$ as the order type for $S$:

$$S = \{1\}_1, \{2, 3\}_2, \{4,5,6\}_3, \dots, \{\omega, \omega +1, \omega +2, \dots \}_{\omega}, \{\omega \cdot 2, \omega \cdot 2 +1, \omega \cdot 2 + 2, \dots, \omega \cdot 3\}_{\omega + 1}, \dots, \text{ (inclusive of all ordinals in } \omega_1 \text{)}$$

Either statement #1 or statement #2 below is true:

1) There exists an element $s_i \in S$ such that $|s_i| = |\omega_1|$.

2) There does not exist an element $s_i \in S$ such that $|s_i| = |\omega_1|$

Assume statement #1 is true. Then the least upper bound of $s_i$ is both uncountable and less than $\omega_1$, resulting in a contradiction:

$$(\text{sup }s_i < \omega_1 \text{ and } |\text{sup }s_i| = |\omega_1|) \implies \omega_1 \text{ is not the smallest uncountable ordinal.}$$

Assume statement #2 is true. Then we have an uncountable set, $\omega_1$, being equal to a countable union of countable sets. Assume the axiom of choice and statement #2 implies that $\omega_1$ is countable, contradicting the definition of $\omega_1$.

With the axiom of choice, there would be no smallest uncountable ordinal (and because there is no smallest, none at all), so the (general) continuum hypothesis would be solved.

If the axiom of choice is negated, it may be consistent that $\omega_1$ is uncountable despite being a countable union of countable sets.

Sound good?