Union of Ordinals is Least Upper Bound

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Theorem

Let $A \subset \operatorname{On}$.

That is, let $A$ be a class of ordinals (every member of $A$ is an ordinal).


Then $\bigcup A$, the union of $A$, is the least upper bound of $A$:

$\displaystyle \forall x \in A: x \le A$
$\displaystyle \forall y \in A: y \le x \implies \bigcup A \le x$


Proof

First, we must show that $\displaystyle \bigcup A$ is an upper bound.

Take any member $a \in A$.

Then by Subset of Union:

$\displaystyle a \subseteq \bigcup A$

By Ordering on Ordinal is Subset Relation:

$a \le A$

By generalizing for all $a \in A$:

$\displaystyle \forall x \in A: x \le \bigcup A$


Similarly, suppose now that $x$ is an upper bound of $A$.

We shall denote $<$ for ordering on the ordinal numbers.

By Ordering on Ordinal is Subset Relation and Transitive Set is Proper Subset of Ordinal iff Element of Ordinal, $<$ is the same as both $\in$ and $\subsetneq$.

Then:

\(\displaystyle z \in \bigcup A\) \(\implies\) \(\displaystyle \exists y: \left({z \in y \land y \in A}\right)\) Definition of union
\(\displaystyle \) \(\implies\) \(\displaystyle \exists y: \left({z \in y \land y < x}\right)\) by hypothesis (since $y \in A$, $y < x$)
\(\displaystyle \) \(\implies\) \(\displaystyle z \in x\) by transitivity of $\in$: see Alternative Definition of Ordinal

Thus, by definition of subset:

$\displaystyle \bigcup A \subseteq x$

Therefore:

$\displaystyle \forall y \in A: y \le x \implies \bigcup A \le x$

$\blacksquare$


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