# Another Opus Here

#### Micrm@ss

For purposes of proving $\phi(\omega_1)$ true, select $\omega_1$ as the order type for $S$:
$$S = \{1\}_1, \{2, 3\}_2, \{4,5,6\}_3, \dots, \{\omega, \omega +1, \omega +2, \dots \}_{\omega}, \{\omega \cdot 2, \omega \cdot 2 +1, \omega \cdot 2 + 2, \dots, \omega \cdot 3\}_{\omega + 1}, \dots, \text{ (inclusive of all ordinals in } \omega_1 \text{)}$$
Is this countable? If it is, why?
If it is not, then why is the following true:

Then we have an uncountable set, $\omega_1$, being equal to a countable union of countable sets.

#### AplanisTophet

Is this countable? If it is, why?
The order type of the $S$ I've selected for proving $\phi(\omega_1)$ true is $\omega_1$, which is uncountable by definition. I'll further note that $\bigcup S$ would also be uncountable by requirements #2 and #3.

If it is not, then why is the following true:
Statement #2 comes with the assumption that there are no elements of $S$ with a cardinality equal to that of $\omega_1$. If each element of $S$ is therefore countable and the cardinality of each element of $S$ equals the cardinality of its ordinal position $i$ within $S$ (that is, with the particular $S$ I've given for proving $\phi(\omega_1)$ true, you'll find that for each $s_i \in S$ we have $|s_i| = |i|$), then the ordinal position $i$ of each $s_i \in S$ is also countable. We then have a countable union of countable sets by definition. Naturally, statement #2 leads to a contradiction.

#### Micrm@ss

The order type of the $S$ I've selected for proving $\phi(\omega_1)$ true is $\omega_1$, which is uncountable by definition. I'll further note that $\bigcup S$ would also be uncountable by requirements #2 and #3.

Statement #2 comes with the assumption that there are no elements of $S$ with a cardinality equal to that of $\omega_1$. If each element of $S$ is therefore countable and the cardinality of each element of $S$ equals the cardinality of its ordinal position $i$ within $S$ (that is, with the particular $S$ I've given for proving $\phi(\omega_1)$ true, you'll find that for each $s_i \in S$ we have $|s_i| = |i|$), then the ordinal position $i$ of each $s_i \in S$ is also countable. We then have a countable union of countable sets by definition. Naturally, statement #2 leads to a contradiction.
You don't have a countable union. You said yourself S is uncountable. You have an uncountabe union of countable sets. In this case, this turns out to be uncountable. There is no contradiction.

#### AplanisTophet

You don't have a countable union. You said yourself S is uncountable. You have an uncountabe union of countable sets. In this case, this turns out to be uncountable. There is no contradiction.
Statement #1 assumes what you assert, that $S$ is an uncountable union of countable sets. It leads to a contradiction too. That is how I arrived at statement #2 in the first place.

Yes, of course $\omega_1$ is not a countable union of countable sets, but when you assume statement #2 it is. There is a contradiction, because if there weren't, it would imply that $\omega_1$ is a countable union of countable sets which we both agree it cannot be.

#### Micrm@ss

So statement 2 says there is no element $s\in S$ such that $|s| = |\omega_1|$. Ok. I agree that this is a true statement. I agree every element in $S$ is countable.

So what is the countable set that you take the union of?

#### AplanisTophet

So statement 2 says there is no element $s\in S$ such that $|s| = |\omega_1|$. Ok. I agree that this is a true statement. I agree every element in $S$ is countable.

So what is the countable set that you take the union of?
For each $s_i \in S$, we have $|s_i| = |i|$. That is due my choice of $S$. Assume statement #2 and $\omega_1$ is a countable union of countable sets by definition because of requirement #2.

#### Micrm@ss

For each $s_i \in S$, we have $|s_i| = |i|$. That is due my choice of $S$. Assume statement #2 and $\omega_1$ is a countable union of countable sets by definition because of requirement #2.
No, I don't see it. Take your specific form of $S$:
$$\{1\}, \{2,3\}, \{4,5,6\}, ..., \{\omega, \omega + 1 , ...\}, \{\omega+\omega, \omega + \omega + 1,...\},...$$
Can you show me the bijection between this set and $\mathbb{N}$, assuming statement number 2.

#### AplanisTophet

No, I don't see it. Take your specific form of $S$:
$$\{1\}, \{2,3\}, \{4,5,6\}, ..., \{\omega, \omega + 1 , ...\}, \{\omega+\omega, \omega + \omega + 1,...\},...$$
Can you show me the bijection between this set and $\mathbb{N}$, assuming statement number 2.
That is not my $S$. It is missing the $\omega \cdot 3$. The ordinal indexes $i$ are also part of the proof. Statement #2 says there is no $s_i$ with a cardinality greater than that of $\mathbb{N}$. The cardinality of each $s_i$ is equal to the cardinality of its ordinal index $i$. Therefore, statement #2 also says each index $i$ is countable. There is a choice function across the indexes.

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#### Micrm@ss

That is not my $S$. It is missing the $\omega \cdot 3$. The ordinal indexes $i$ are also part of the proof. Statement #2 says there is no $s_i$ with a cardinality greater than that of $\mathbb{N}$. The cardinality of each $s_i$ is equal to the cardinality of its ordinal index $i$. Therefore, statement #2 also says each index $i$ is countable. There is a choice function across the indexes.
I agree with all that you wrote. Now please provide an explicit bijection between $\mathbb{N}$ and $S$.