# Intersection of Ordinals is Ordinal

## Theorem

Let $A$ be a non-empty class of ordinals.

Then its intersection $\bigcap A$ is an ordinal.

## Proof

Let $\alpha = \bigcap A$.

It will be demonstrated that $\alpha$ is an ordinal according to Definition $2$:

$\alpha$ is an **ordinal** if and only if it fulfils the following conditions:

\((1)\) | $:$ | $\alpha$ is a transitive set | |||||||

\((2)\) | $:$ | the epsilon relation is connected on $\alpha$: | \(\ds \forall x, y \in \alpha: x \ne y \implies x \in y \lor y \in x \) | ||||||

\((3)\) | $:$ | $\alpha$ is well-founded. |

### Set

By Intersection of Non-Empty Class is Set, $\alpha$ is a set.

$\Box$

### Transitive

Let $n \in \alpha$.

Let $m \in n$.

Let $\beta \in A$ be arbitrary.

By the definition of intersection:

- $n \in \beta$

Since $\beta$ is an ordinal, it is transitive.

Thus:

- $m \in \beta$

As $\beta$ is arbitrary, this holds for all $\beta \in A$.

Thus:

- $m \in \alpha$

Thus $\alpha$ is transitive.

$\Box$

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### $\in$-connected

Let $x, y \in \alpha$ be arbitrary such that $x \ne y$.

Since $A$ is non-empty, it has an element $\beta$.

By the definition of intersection:

- $x, y \in \beta$

Since $\beta$ is an ordinal, it is $\in$-connected.

Thus $x \in y$ or $y \in x$.

As $x$ and $y$ are arbitrary, this holds for all $x, y \in \alpha$.

That is, $\alpha$ is $\in$-connected.

$\Box$

### Well-founded

Let $\beta$ be a non-empty subset of $\alpha$.

Let $\gamma \in A$ (such exists because $A$ is non-empty).

By Intersection is Largest Subset:

- $\beta \subseteq \gamma$

Since $\gamma$ is an ordinal it is well-founded.

Thus $\beta$ has an element $x$ such that $x \cap \beta = \O$.

As this holds for all such $\beta$, $\alpha$ is well-founded.

$\blacksquare$

## Sources

- 2010: Raymond M. Smullyan and Melvin Fitting:
*Set Theory and the Continuum Problem*(revised ed.) ... (previous) ... (next): Chapter $5$: Ordinal Numbers: $\S 2$ Ordinals and transitivity: Exercise $2.2$