0
$\begingroup$

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?

Thanks in advance.

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 $\emptyset\notin X$.

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

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.

$\endgroup$
1
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Asaf Karagila Feb 16 '18 at 12:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.