**Disclaimer:**I have a sense of humor and a purpose, so Iâ€™m going to try and enumerate $\omega_1$ itself. Clearly that is a crankish proposition, so this disclaimer is meant to clarify what I mean to accomplish and acknowledge that there should obviously be an error if I conclude that I have enumerated $\omega_1$. Do not assume that I actually think that I have done so, as I am not delusional (nor should you be). Given that clarification, I conclude that I have enumerated $\omega_1$!! Yay!!!! In all seriousness, the functions that I am using can enumerate any countable ordinal, so that is my primary reason for analyzing them. I also show an example of Solovay (regarding partitioning a regular uncountable cardinal, in this case $\omega_1$, into $\omega_1$ pairwise disjoint stationary sets).

**If I ask one thing of any reader,**it is to determine whether the set $C$ is countable or uncountable. I believe it is neither. The set $C$ is defined as a local variable under the definition of the sequence $T$ (assume the non-transfinite option) below. It may be the source of my error.

When the time comes, note that I believe:

1) $C$ should be uncountable because there are $\omega_1$ functions $\phi_{\alpha}$ and each may lead to countably many elements being added to $C$.

2) We have $sup(C) \in \omega_1$ on any given iteration because $sup(C)$ must be less than $sup(C)$ on some future iteration, therefore $C$ cannot be unbounded in $\omega_1$.

I have thought about this for a long time and am stuck. Any guidance is appreciated.

**Where $a,b$ are ordinals:**

1) $a=b$ does not imply $(a,b) = (a)$, nor does it imply $(a,b) = (b)$.

2) $a \neq b$ does not imply $(a,b) = (b,a)$.

This is done to differentiate between standard set operations and the properties I wish to assign to ordered tuplets of ordinals that are necessary to carry out my work.

**Define $t(\alpha)$ and $t^{-1}(\alpha)$ for any ordinal $\alpha \geq 2$:**

Let $t(\alpha)$ equal a doublet of ordinals $(a,b)$ if $\alpha = 2$, a triplet of ordinals $(a,b,c)$ if $\alpha = 3$, a quadruplet of ordinals $(a,b,c,d)$ if $\alpha = 4$, and so on, for any ordinal $\alpha$.

Similarly, let $t^{-1}(x)$ equal $2$ for any doublet of ordinals $x$, $3$ for any triplet of ordinals $x$, $4$ for any quadruplet of ordinals $x$, and so on, as determined by the order type of $x$.

**Use of set builder notation:**

Consider the set of all doublets of ordinals such that each element of each doublet is a member of $\omega_1$. It will become helpful to use set-builder notation in the following manner to define such a set:

$$\{t(2) : a,b \in t(2) \implies a,b \in \omega_1 \} = \{ (a,b) : a,b \in \omega_1 \} = \{(0,0),(0,1),(1,0),(a \in \omega_1,b \in \omega_1),\dots\}$$

**Define the set $P$:**

$$P = \{ t(\omega) : a,b,c,\dots \in t(\omega) \implies a,b,c,\dots \, \text{is a computable sequence} \, \text{and } a,b,c, \dots \in \{0,1\} \} \equiv \, \text{the set of computable binary sequences}$$

**Define the functions $(r_{\alpha})_{\omega \leq \alpha < \omega_1}$:**

Let $r_{\alpha} : \alpha \rightarrow \omega$ be bijective.

**Define 'Likewise Computable':**

Let $t(\alpha) = ((\beta + a)_0, (\beta + b)_1, (\beta + c)_2, \dots)$ be likewise computable for any ordinal $\alpha$ if and only if $\omega \leq \alpha < \omega_1$, $\beta \in \omega_1$, and an $\omega$-type ordering by index $( (\gamma)_{r_{\alpha}^{-1}(0)}, (\zeta)_{r_{\alpha}^{-1}(1)}, (\mu)_{r_{\alpha}^{-1}(2)},\dots )$ of the $\alpha$-type ordering $( a_{r_{\alpha}(0)}, b_{r_{\alpha}(1)}, c_{r_{\alpha}(2)},\dots )$ is an element of $P$.

**Define the functions $(\phi_{\alpha})_{2 \leq \alpha < \omega_1}$:**

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

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

2) $$\alpha \geq \omega \implies \phi_{\alpha} : \omega_1 \setminus \{0\} \rightarrow \{ t(\alpha) : a,b,c,\dots \in t(\alpha) \implies a,b,c,\dots \in \omega_1 \, \text{and } t(\alpha) \, \text{is likewise computable}\} \text{ is bijective},$$

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

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

**Define function $f$:**

$$f(x) = \begin{cases}

\phi_{t^{-1}(x)}^{-1}(x) & \text{if, given } x \, \text{has order type } \alpha, 2 \leq \alpha < \omega \, \text{or } x \, \text{is likewise computable} \\

\text{empty string} & \text{otherwise} \\

\end{cases}$$

**Define $k(\alpha)$ for any ordinal $\alpha \in \omega_1$:**

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

**Define function $h$:**

$$h(\alpha) = \begin{cases}

min\{ t^{-1}(x) : x \in k(\alpha) \text{ and } \forall y \in x(y < \alpha) \} & \text{if } \alpha \in \omega_1 \\

1 & \text{otherwise} \\

\end{cases}$$

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 (potentially transfinite) sequence $T = t_1, t_2, t_3, \dots$ over $\omega_1$ iterations where:

a) If $m$ is a countable limit ordinal and $T$ is of order type $\omega$, free up space in $T$ by first letting $(s_n)_{n \in \mathbb{N}}$ be defined for odd indexes and undefined for even indexes: $s_{n \cdot 2 - 1} = t_{n}$. Then, set $t_1 = s_1, t_2 = s_2, t_3 = s_3, \dots$. Finally, if $m = \omega$, set $t_j$ undefined for any index $j > i$ where $t_j = t_i$ and $i,j < \omega$.

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

c) Let $B = \{ f(x) : x \in (g(A) \setminus \{ y \in g(A) : t^{-1}(y) \neq h(f(y))\} ) \}$. 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.

d) Let $C = B \setminus A$ and let $c_1, c_2, c_3, \dots$ be a (potentially transfinite) 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 $T$.

e) If $|C| < \aleph_0$ or if a transfinite $T$ is desired, set $t_n = c_1, t_{n+1} = c_2, t_{n+2} = c_3, \dots$ and then proceed to step 3. If a transfinite $T$ is not desired, proceed to sub-step f.

f) Let $Tâ€™ = tâ€™_1, tâ€™_2, tâ€™_3, \dots$ be a subsequence of the remaining undefined elements of $T$ and set $tâ€™_1 = c_1, tâ€™_3 = c_2, tâ€™_5 = c_3, \dots$.

$$h(\alpha) = \begin{cases}

min\{ t^{-1}(x) : x \in k(\alpha) \text{ and } \forall y \in x(y < \alpha) \} & \text{if } \alpha \in \omega_1 \\

1 & \text{otherwise} \\

\end{cases}$$

**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 (potentially 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) If $m$ is a countable limit ordinal and $T$ is of order type $\omega$, free up space in $T$ by first letting $(s_n)_{n \in \mathbb{N}}$ be defined for odd indexes and undefined for even indexes: $s_{n \cdot 2 - 1} = t_{n}$. Then, set $t_1 = s_1, t_2 = s_2, t_3 = s_3, \dots$. Finally, if $m = \omega$, set $t_j$ undefined for any index $j > i$ where $t_j = t_i$ and $i,j < \omega$.

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

c) Let $B = \{ f(x) : x \in (g(A) \setminus \{ y \in g(A) : t^{-1}(y) \neq h(f(y))\} ) \}$. 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.

d) Let $C = B \setminus A$ and let $c_1, c_2, c_3, \dots$ be a (potentially transfinite) 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 $T$.

e) If $|C| < \aleph_0$ or if a transfinite $T$ is desired, set $t_n = c_1, t_{n+1} = c_2, t_{n+2} = c_3, \dots$ and then proceed to step 3. If a transfinite $T$ is not desired, proceed to sub-step f.

f) 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. If $j>n$, set $n = j$. Increase the iteration counter by letting $m = m + 1$, and then repeat step 2.