This thread deals with ordinal numbers:

Ordinal Number -- from Wolfram MathWorld

The standard less-than ordinal comparison, $<$, implies that $\alpha < \beta$ if and only if $\alpha$ is order isomorphic to an initial segment of $\beta$. The class of ordinal numbers are well ordered by the relation.

Define the relation $\prec$ such that, for any two sets of ordinal numbers $A$ and $B$, we have:

$$A \prec B \iff \forall b \in B(\forall a \in A( a < b))$$

Define $\phi$ as a boolean function where $\phi(\gamma)$ is true if and only if $\gamma$ is an ordinal number that can be partitioned into at least one arbitrary sequence of sets, $S$, such that three requirements are met:

1) $S = s_1, s_2, s_3, \dots$ (that is, $S$ is an infinite sequence),

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

3) given $s_i, s_j \in S$, the following relation holds true: $s_i \prec s_j \iff i < j$.

For example, $\phi(\omega)$ is true because the first limit ordinal, $\omega$, can be partitioned into the sequence of finite ordinals meeting the above criteria:

$$S = \{1\}, \{2\}, \{3\}, \dots$$

Other possible sequences exist as well that suffice to prove $\phi(\omega)$ true. E.g., the following two sequences $Sâ€™$ and $Sâ€™â€™$ also suffice (and there are infinitely many more as well):

$$Sâ€™ = \{1,2,3\}, \{4, 5, 6\}, \{7,8,9\}, \dots, \text{ and}$$

$$Sâ€™â€™ = \{1 \}, \{2,3\}, \{4,5.6\}, \dots$$

Let it be emphasized that there are no restrictions on the cardinality of any of the elements of an arbitrary sequence $S$ that suffices to prove $\phi(\gamma)$ true for a given ordinal $\gamma$. As a result, even though $S$ is restricted to being a standard (countably infinite) sequence, the cardinality of $\bigcup S$ could potentially be uncountable.

Let it also be emphasized that $\phi(\gamma)$ will not hold true for any ordinal $\gamma$ that is not a limit ordinal. This is because $\gamma$ will have a greatest element that must also be in one and only one element of $S$ due requirement #2. That one element of $S$ containing the greatest element of $\gamma$ would also have to be the last element of $S$ due requirement #3 and in contradiction of requirement #1.

Conjecture #1: $\phi(\gamma)$ will hold true for any limit ordinal.

By definition, the smallest uncountable ordinal, $\omega_1$, is a limit ordinal that is the set of all countable ordinals.

Given that $\omega_1$ is a limit ordinal, we know that it has no immediate predecessor under the standard ordering. That is, if $y < \omega_1$, we know there also exists some ordinal $x$ where $y < x < \omega_1$ holds true.

What happens if we assume the existence of a set $S$ that suffices to prove $\phi(\omega_1)$ true?

In that case, one of two statements must be true and the other false:

1) There exists an element $s_j \in S$ such that $|s_j| = |\omega_1|$ or

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

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

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

Assume statement #2 is true. Then we have an uncountable set, $\omega_1$, being equal to a countable union of countable sets. Such a theorem would be consistent with the negation of the axiom of choice (at least one model exists where the real numbers are a countable union of countable sets given the negation of the axiom choice), noting the axiom of (countable) choice is required to show that a countable union of countable sets is itself countable. See https://en.wikipedia.org/wiki/Axiom_of_choice

If my conjecture is true, then either statement #1 or statement #2 must be true if my logic is correct. Also, if the axiom of choice is adopted, $\omega_1$ would be countable by statement #2, contradicting the definition of $\omega_1$.

Maybe uncountable ordinals simply donâ€™t exist as sets and instead need to be characterized as proper classes? See Burali-Forti Paradox -- from Wolfram MathWorld, showing that the class of ordinals is a proper class.

If no uncountable ordinals exist as sets, then the Continuum Hypothesis is resolved.

I wonder what implications this has for GÃ¶delâ€™s constructible universe, which is built on the hierarchy of the ordinals?

Ok. I assume I didnâ€™t break math or anything. Any help?

Ordinal Number -- from Wolfram MathWorld

The standard less-than ordinal comparison, $<$, implies that $\alpha < \beta$ if and only if $\alpha$ is order isomorphic to an initial segment of $\beta$. The class of ordinal numbers are well ordered by the relation.

Define the relation $\prec$ such that, for any two sets of ordinal numbers $A$ and $B$, we have:

$$A \prec B \iff \forall b \in B(\forall a \in A( a < b))$$

Define $\phi$ as a boolean function where $\phi(\gamma)$ is true if and only if $\gamma$ is an ordinal number that can be partitioned into at least one arbitrary sequence of sets, $S$, such that three requirements are met:

1) $S = s_1, s_2, s_3, \dots$ (that is, $S$ is an infinite sequence),

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

3) given $s_i, s_j \in S$, the following relation holds true: $s_i \prec s_j \iff i < j$.

For example, $\phi(\omega)$ is true because the first limit ordinal, $\omega$, can be partitioned into the sequence of finite ordinals meeting the above criteria:

$$S = \{1\}, \{2\}, \{3\}, \dots$$

Other possible sequences exist as well that suffice to prove $\phi(\omega)$ true. E.g., the following two sequences $Sâ€™$ and $Sâ€™â€™$ also suffice (and there are infinitely many more as well):

$$Sâ€™ = \{1,2,3\}, \{4, 5, 6\}, \{7,8,9\}, \dots, \text{ and}$$

$$Sâ€™â€™ = \{1 \}, \{2,3\}, \{4,5.6\}, \dots$$

Let it be emphasized that there are no restrictions on the cardinality of any of the elements of an arbitrary sequence $S$ that suffices to prove $\phi(\gamma)$ true for a given ordinal $\gamma$. As a result, even though $S$ is restricted to being a standard (countably infinite) sequence, the cardinality of $\bigcup S$ could potentially be uncountable.

Let it also be emphasized that $\phi(\gamma)$ will not hold true for any ordinal $\gamma$ that is not a limit ordinal. This is because $\gamma$ will have a greatest element that must also be in one and only one element of $S$ due requirement #2. That one element of $S$ containing the greatest element of $\gamma$ would also have to be the last element of $S$ due requirement #3 and in contradiction of requirement #1.

Conjecture #1: $\phi(\gamma)$ will hold true for any limit ordinal.

By definition, the smallest uncountable ordinal, $\omega_1$, is a limit ordinal that is the set of all countable ordinals.

Given that $\omega_1$ is a limit ordinal, we know that it has no immediate predecessor under the standard ordering. That is, if $y < \omega_1$, we know there also exists some ordinal $x$ where $y < x < \omega_1$ holds true.

What happens if we assume the existence of a set $S$ that suffices to prove $\phi(\omega_1)$ true?

In that case, one of two statements must be true and the other false:

1) There exists an element $s_j \in S$ such that $|s_j| = |\omega_1|$ or

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

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

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

Assume statement #2 is true. Then we have an uncountable set, $\omega_1$, being equal to a countable union of countable sets. Such a theorem would be consistent with the negation of the axiom of choice (at least one model exists where the real numbers are a countable union of countable sets given the negation of the axiom choice), noting the axiom of (countable) choice is required to show that a countable union of countable sets is itself countable. See https://en.wikipedia.org/wiki/Axiom_of_choice

If my conjecture is true, then either statement #1 or statement #2 must be true if my logic is correct. Also, if the axiom of choice is adopted, $\omega_1$ would be countable by statement #2, contradicting the definition of $\omega_1$.

Maybe uncountable ordinals simply donâ€™t exist as sets and instead need to be characterized as proper classes? See Burali-Forti Paradox -- from Wolfram MathWorld, showing that the class of ordinals is a proper class.

If no uncountable ordinals exist as sets, then the Continuum Hypothesis is resolved.

I wonder what implications this has for GÃ¶delâ€™s constructible universe, which is built on the hierarchy of the ordinals?

Ok. I assume I didnâ€™t break math or anything. Any help?

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