In the following paragraphs, I will outline my proof for the claim:
If $X$ is a non-empty set of ordinals, then $\bigcup X = \text{sup}(X)$
I've taken note of the post here ($\bigcup X$ = $\sup(X)$ for a set $X$ of ordinals),but my question is NOT related to "How Do I Prove This".
I have included my proof of this statement for reference…but for the actual question, you can scroll to the bottom.
This proof amounts to demonstrating that $\bigcup X$ is the smallest ordinal $\alpha$ such that $\alpha \geq \xi$ for all $\xi \in X$.
At the end of my proof, I will clarify the specifics of my question.
As an important side note, my book (The Foundations of Mathematics By Kenneth Kunen) uses the following conventions:
When discussing ordinals, Greek letters (especially $\alpha, \beta, \gamma, \delta, \zeta, \eta, \xi, \mu$) "range over the ordinals"; that is, $\forall \alpha \varphi(\alpha)$ abbreviates $$ \forall x [x \text{ is an ordinal} \rightarrow \varphi(x)] $$ Also, $\alpha \lt \beta$ means (by definition) $\alpha \in \beta$, and $\alpha \leq \beta$ means $\alpha \in \beta \lor \alpha = \beta$
Also, the following definition will be used:
Let $R$ be a relation. $y \in X$ is $R$-maximal in $X$ iff $\neg \exists z (z \in X \land yRx))$
This proof is broken up into two parts:
Part $1$: Demonstrate that $\neg \exists \zeta (\zeta \lt \bigcup X \land \forall \alpha (\alpha \in X \rightarrow \alpha \leq \zeta ))$
Part $2$: Demonstrate that $\bigcup X$ satisfies the property that $\forall \alpha ( \alpha \in X \rightarrow \alpha \leq \bigcup X)$
The following lemmas are useful:
$\bigcup X$ is an ordinal
$\alpha \text{ is an ordinal } \land \delta \in \alpha \rightarrow \delta \text{ is an ordinal}$
Part $1$: Proof by Contradiction – show that $\exists \zeta (\zeta \lt \bigcup X \land \forall \alpha (\alpha \in X \rightarrow \alpha \leq \zeta ))$ leads to a contradiction.
By definition, $\bigcup X = \{ \alpha \in A \ | \ \exists \beta (\beta \in X \land \alpha \in \beta)\}$ where $A$ is the set that is posited to exist by my book's Union Axiom.
Let $\xi$ be an ordinal that satisfies the claim we want to produce a contradiction with.
Therefore, $\xi \in \bigcup X$. By definition of $\bigcup X$, this implies that $\exists \beta (\beta \in X \land \xi \in \beta)$. Call this object $\beta'$.
But $\xi \in \beta' \iff \xi \lt \beta'$ means that there is an ordinal in $X$ (i.e. $\beta'$) that is strictly greater than $\xi$…a contradiction. Therefore, $\neg \exists \zeta (\zeta \lt \bigcup X \land \forall \alpha (\alpha \in X \rightarrow \alpha \leq \zeta ))$
Part $2$:
Let $\alpha \in X$.
Show that either $\alpha \in \bigcup X$ or $\alpha = \bigcup X$.
There are two scenarios that are possible with respect to $\alpha$:
i) $\alpha$ is not $R$-maximal in $X$ $\iff \exists \beta ( \beta \in X \land \alpha \in \beta)$
ii) $\alpha$ is $R$-maximal in $X$ $\iff \neg \exists \beta (\beta \in X \land \alpha \in \beta)$
Scenario i: Prove that $\alpha \in \bigcup X$
By definition of $\bigcup X$, if $\exists \beta ( \beta \in X \land \alpha \in \beta)$ then $\alpha \in \bigcup X$.
Scenario ii: Prove that $\alpha = \bigcup X$ (i.e. prove that they are subsets of each other)
Let $\gamma \in \alpha$. By definition of $\bigcup X$, $\gamma \in \bigcup X$. Therefore, $\alpha \subseteq \bigcup X$.
Let $\gamma \in \bigcup X$. By definition of $\bigcup X$, $\exists \beta (\beta \in X \land \gamma \in \beta)$. Call this $\beta^*$. (i.e. $\gamma \in \beta^*$)
$\alpha$ is maximal and $\beta^* \in X$. By trichotomy of ordinals, $\beta^* \leq \alpha$.
Breaking this into two cases:
a) $\beta^* \lt \alpha \iff \beta^* \in \alpha$. Therefore, because $\beta^*$ is an ordinal, $\beta^* \subset \alpha$. But that means $\gamma \in \alpha$. Thus, $\bigcup X \subseteq \alpha$.
b) $\beta^*=\alpha$. Well, in such a case, it is immediately clear that $\gamma \in \alpha$. Thus, $\bigcup X \subseteq \alpha$.
I believe what I carried out above is correct and demonstrates that
If $X$ is a non-empty set of ordinals, then $\bigcup X = \text{sup}(X)$
My question is as follows:
Throughout this proof, it dawned on me that what I am effectively doing with my argument is creating a set that looks like this:
$\{\eta\in ON \ | \forall \alpha (\alpha \in X \rightarrow \alpha \leq \eta) \}$ (…where $ON$ is the set of all ordinals…)
and subsequently proving that $\bigcup X$ must be the minimal element of this set.
After doing a little digging, I found that $ON$ is not a set, and actually forms a Proper Class, which means that the notation "$\eta \in ON$" is invalid syntax in formal set theory.
How then should I describe the aforementioned set?
The only thing I can think to do is the following:
$\{\eta\in S(\bigcup X) \ | \forall \alpha (\alpha \in X \rightarrow \alpha \leq \eta) \}$
Where $S(\bigcup X)=\bigcup X \cup \{\bigcup X\}$
Is this the proper convention (or at least acceptable in and of itself?)
Best Answer
Your concern is unfounded: you are not looking at some set of ordinals and showing that $\bigcup X$ is the least element of that set. The axioms ensure the existence of the set $\bigcup X$, and you have presumably proved that $\bigcup X$ is an ordinal. Now you are simply showing $(1)$ that $X$ has no upper bound that is less than the ordinal $\bigcup X$, and $(2)$ that $\bigcup X$ is an upper bound for $X$. It follows immediately from $(1)$ and $(2)$ that $\bigcup X=\sup X$.
It’s true that once you’ve shown that $\bigcup X=\sup X$, you get as an immediate corollary that $\bigcup X=\min Y$ whenever $Y$ is a non-empty set of upper bounds for $X$, but the proof that $\bigcup X=\sup X$ does not make use of any such $Y$.