# If $X\subset\{\alpha:\alpha\mbox{ is an ordinal} \}$, prove that $\bigcap X\in X$

What I want to prove is Theorem 121 of the appendix of Kelley's book General topology:

If $X\subset\mathcal O$ (here $\mathcal O$ denotes the class of all ordinals) is non-empty, then $\bigcap X\in X$. Moreover, it is the least element of $X$.

My attempt:

Since $X\subset\mathcal O$, the elements of $\bigcap X$ are ordinals and thus $\bigcap X$ is a transitive set. By a previous result, I know that for two ordinals $\alpha$ and $\beta$, $\alpha\in\beta$ or $\beta\in\alpha$ or $\alpha=\beta$, and only one is true. So I think this proves $\bigcap X$ is trichotomic. Hence, $\bigcap X$ is an ordinal.

But I need to proof that $\bigcap X\in X$, and I don't know how. I thought consider $\bigcap X\in\mathcal O \setminus X$ and try to get a contradiction. But I didn't kno how.

I also thought: if $\alpha\in\bigcap X$, then $\alpha\in \kappa$, for some $\kappa \in X$. So $\alpha\subset\bigcap X$ and $\alpha\subset\kappa$ and also:

$$\alpha\subset \kappa\cap\left(\bigcap X \right) .$$

But any of this aideas convence myself $\bigcap X\in X$.

Any ideas for that?

The second part would be easier:

If there is $\kappa\in X$ such that $\kappa\in\bigcap X$, then

$$\kappa\in\alpha \qquad \forall \alpha \in X.$$

And $X$ hasn't got any element disjoint with itself.

Is that proof right?

Remark. We are considering only proper inclusions. If $X=\mathcal O$, then

$$\bigcap X=\emptyset ,$$

and the theorem follows eaisly.

By the same reason, we should consider $X\subset \mathcal O$ such that $\emptyset\notin X$.

• And if $X = \varnothing$, then $\cap X = \;$ ? – GEdgar Feb 14 '18 at 1:07
• No. The theorem doesn't allow X to be empty. – Dog_69 Feb 14 '18 at 10:08
• Please, other hints??? :( – Dog_69 Feb 14 '18 at 23:39

There is a simpler proof, if you already laid some ground work to it. Let me write down the main steps, and I will leave you to fill in the details and missing parts.

1. For two ordinals, $\alpha\in\beta$ if and only if $\alpha\subsetneq\beta$.
2. The ordinals are linearly ordered by $\in$, therefore they are also linearly ordered by $\subseteq$.
3. If $X$ is a non-empty class of ordinals, then there is a minimum element not only in $\in$, but also in $\subseteq$.

From the last point, the conclusion should follow immediately, since $\bigcap X$ is the intersection of all the members of $X$. If there is a minimum element in $\subseteq$, say $\alpha$, then $\bigcap X\subseteq\alpha$ by definition of $\bigcap$, but vice versa follows by the fact that for all $\beta\in X$, $\alpha\subseteq\beta$ holds as well.

The proof of that $\bigcap X$ is the least element is in some way the proof of that it belongs to $X$. But I'm traying to rewrite the proof.

Since $X$ is non-empty, we know it has a least element; calle it $\alpha_0$. On the other hand, since the elements of $X$ are ordinals, we also know that $\bigcap X$ is transitive. And is easy to prove that is in fact an ordinal.

Finally, I'm going to prove that $\bigcap X=\alpha_0$. For that, we only have to consider the definition:

$$\bigcap X = \{ z : z\in x \forall x\in X \}.$$

But the above condition is equivalent to belong to $\alpha _0$ and thus

$$\bigcap X = \{ z : z\in \alpha_0 \} =\alpha_0.$$

• This proof is fine, I've posted an outline of the same argument from a slightly different perspective. I hope that will be illuminating for future references and future readers. – Asaf Karagila Feb 16 '18 at 12:57