This is a draft. Knowing me, it is probably nonsensical and filled with errors. I'm trying to enumerate some very large countable ordinal assuming of course that math isn't broken and I can't enumerate $\omega_1$ itself. ... And no, jic, I don't think math is broken. Why? Do you? :spin: :giggle:

I'll take a moment to step back from this and at some point return to see what corrections, comments, etc., I care to make. Feel free to look at it funny in the meantime.

For $\alpha \geq 2$, let $t(\alpha)$ equal a doublet of variables $(a,b)$ if $\alpha = 2$, a triplet of variables $(a,b,c)$ if $\alpha = 3$, a quadrulplet of variables $(a,b,c,d)$ if $\alpha = 4$, and so on, for any ordinal $\alpha$.

Let each element of $(\phi_{\alpha})_{2 \leq \alpha < \omega_1}$ be

1) $\phi_{\alpha} : \omega_1 \setminus \{0\} \rightarrow \{ t(\alpha) : a,b,c,\dots \in t(\alpha) \implies a,b,c,\dots < \omega_1 \}$ is bijective,

2) $a,b,c,\dots \leq \kappa$ for each $a,b,c,\dots \in \phi_{\alpha}(\kappa)$, and

3) $\zeta < \alpha \implies min\{ \phi_{\zeta}^{-1}(b) : \exists k \in b \text{ where } k \geq \phi_{\zeta}^{-1}(b)\} < min\{ \phi_{\alpha}^{-1}(b) : \exists k \in b \text{ where } k \geq \phi_{\alpha}^{-1}(b)\}$.

$$f(x) = \phi_{t^{-1}(x)}^{-1}(x)$$

$$k(\alpha) = \{ x : f(x) = \alpha \}$$

$$h(\alpha) = min\{ t^{-1}(x) : x \in k(\alpha) \text{ and } \forall y \in x(y < \alpha) \}$$

For any set of ordinals, $A$, let $g(A)$ be the set of all ordered doublets, triplets, quadruplets, and so on, that can be comprised from the elements of $\omega_1 \cap A$:

$$g(A) = \{ t(\alpha) : t(\alpha) \setminus (A \cap \omega_1) = \emptyset \text{ and } 2 \leq \alpha < \omega_1 \}$$

Define a transfinite sequence $T = t_1, t_2, t_3, \dots$ over $\omega_1$ iterations where:

a) Let $A = \{ t_i \in T : i < n \}$. E.g., $A = \{0, 1, 2 \}$ on the first iteration.

b) Let $B = \{ f(x) : x \in (g(A) \setminus \{ x \in g(A) : t^{-1}(x) > h(f(x))\} ) \}$. Using the previous elements of the sequence $A$, this step creates a set $B$ of all the new ordinals implied by letting function $f$ range over $g(A)$. The restriction of $g(A)$ to $A \cap \omega_1$ ensures that $B$ remains countable for this particular $T$ sequence.

c) Let $C = B \setminus A$ and let $c_1, c_2, c_3, \dots$ be an enumeration of $C$ that is also well ordered if $|C| \neq \aleph_0$. This step removes any redundant elements from $B$ before potentially well ordering them so that we can add them to $T$.

d) If $|C| \neq \aleph_0$, then set $t_n = c_1, t_{n+1} = c_2, t_{n+2} = c_3, \dots$.

e) If $|C| = \aleph_0$, then let $Tâ€™ = tâ€™_1, tâ€™_2, tâ€™_3, \dots$, be a subsequence of the remaining undefined elements of $T$ and set $tâ€™_1 = c_1, tâ€™_3 = c_2, tâ€™_5 = c_3, \dots$.

I'll take a moment to step back from this and at some point return to see what corrections, comments, etc., I care to make. Feel free to look at it funny in the meantime.

**Define $t(\alpha)$ for any ordinal $\alpha$:**For $\alpha \geq 2$, let $t(\alpha)$ equal a doublet of variables $(a,b)$ if $\alpha = 2$, a triplet of variables $(a,b,c)$ if $\alpha = 3$, a quadrulplet of variables $(a,b,c,d)$ if $\alpha = 4$, and so on, for any ordinal $\alpha$.

**Define $(\phi_{\alpha})_{2 \leq \alpha < \omega_1}$:**Let each element of $(\phi_{\alpha})_{2 \leq \alpha < \omega_1}$ be

*almost regressive*such that:1) $\phi_{\alpha} : \omega_1 \setminus \{0\} \rightarrow \{ t(\alpha) : a,b,c,\dots \in t(\alpha) \implies a,b,c,\dots < \omega_1 \}$ is bijective,

2) $a,b,c,\dots \leq \kappa$ for each $a,b,c,\dots \in \phi_{\alpha}(\kappa)$, and

3) $\zeta < \alpha \implies min\{ \phi_{\zeta}^{-1}(b) : \exists k \in b \text{ where } k \geq \phi_{\zeta}^{-1}(b)\} < min\{ \phi_{\alpha}^{-1}(b) : \exists k \in b \text{ where } k \geq \phi_{\alpha}^{-1}(b)\}$.

**Define function $f$:**$$f(x) = \phi_{t^{-1}(x)}^{-1}(x)$$

**Define $k(\alpha)$ for any ordinal $\alpha$:**$$k(\alpha) = \{ x : f(x) = \alpha \}$$

**Define $h(\alpha)$ for any ordinal $\alpha$:**$$h(\alpha) = min\{ t^{-1}(x) : x \in k(\alpha) \text{ and } \forall y \in x(y < \alpha) \}$$

**Define function $g$:**For any set of ordinals, $A$, let $g(A)$ be the set of all ordered doublets, triplets, quadruplets, and so on, that can be comprised from the elements of $\omega_1 \cap A$:

$$g(A) = \{ t(\alpha) : t(\alpha) \setminus (A \cap \omega_1) = \emptyset \text{ and } 2 \leq \alpha < \omega_1 \}$$

**Define the sequence $T$:**Define a transfinite sequence $T = t_1, t_2, t_3, \dots$ over $\omega_1$ iterations where:

**Step 1)**Let $t_1 = 0, t_2 = 1, t_3 = 2$, and iteration counter $m = 1$.**Step 2)**Each $t_n$, where $n \geq 4$, is defined by the previous elements of the sequence. Starting with $n = 4$:a) Let $A = \{ t_i \in T : i < n \}$. E.g., $A = \{0, 1, 2 \}$ on the first iteration.

b) Let $B = \{ f(x) : x \in (g(A) \setminus \{ x \in g(A) : t^{-1}(x) > h(f(x))\} ) \}$. Using the previous elements of the sequence $A$, this step creates a set $B$ of all the new ordinals implied by letting function $f$ range over $g(A)$. The restriction of $g(A)$ to $A \cap \omega_1$ ensures that $B$ remains countable for this particular $T$ sequence.

c) Let $C = B \setminus A$ and let $c_1, c_2, c_3, \dots$ be an enumeration of $C$ that is also well ordered if $|C| \neq \aleph_0$. This step removes any redundant elements from $B$ before potentially well ordering them so that we can add them to $T$.

d) If $|C| \neq \aleph_0$, then set $t_n = c_1, t_{n+1} = c_2, t_{n+2} = c_3, \dots$.

e) If $|C| = \aleph_0$, then let $Tâ€™ = tâ€™_1, tâ€™_2, tâ€™_3, \dots$, be a subsequence of the remaining undefined elements of $T$ and set $tâ€™_1 = c_1, tâ€™_3 = c_2, tâ€™_5 = c_3, \dots$.

**Step 3)**Let $j$ be the first ordinal such that $t_j$ is undefined. Set $n = j$, increase the iteration counter by letting $m = m + 1$, and then repeat step 2.
Last edited: