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- Summary
- Assistance is requested. An understanding of Fodor's Lemma, club sets, and normal functions is required.

The following assertion quoted from the paper below seems as though it couldn’t be true. It is the issue that I would like some help addressing please:

“The restriction of ##g(A)## to ##A \cap \omega_1## ensures that ##B## remains countable for this particular ##T## sequence.”

...

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##. Note that this notation is used to avoid confusion as ##a=b## does not imply ##(a,b) = (b,a) = (a) = (b)##, whereas it does imply ##\{a,b\} = \{b,a\} = \{a\} = \{b\}## in general.

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 (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##.

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 \{ 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.

d) 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##.

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

f) If ##|C| = \aleph_0##, then let ##T’ = t’_1, t’_2, t’_3, \dots##, be a subsequence of the remaining undefined elements of ##T## and set ##t’_1 = c_1, t’_3 = c_2, t’_5 = c_3, \dots##.

“The restriction of ##g(A)## to ##A \cap \omega_1## ensures that ##B## remains countable for this particular ##T## sequence.”

...

**Define ##t(\alpha)## for any ordinal ##\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##. Note that this notation is used to avoid confusion as ##a=b## does not imply ##(a,b) = (b,a) = (a) = (b)##, whereas it does imply ##\{a,b\} = \{b,a\} = \{a\} = \{b\}## in general.

**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 (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##.

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 \{ 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.

d) 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##.

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

f) If ##|C| = \aleph_0##, then let ##T’ = t’_1, t’_2, t’_3, \dots##, be a subsequence of the remaining undefined elements of ##T## and set ##t’_1 = c_1, t’_3 = c_2, t’_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.
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