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I want to prove that if $X\subset \mathcal O$ is a non-empty subclass of the class of ordinals numbers, then

$$ \bigcap X \subset X .$$

The author reasons as follows:

1.- Since $\mathcal O$ is transitive, $X$ is an ordinal.

2.- Thus, every element of $X$ is also a proper subset of $X$.

3.- And also every element $z$ of every element $x\in X$ will also belong to $X$ and hence

$$ \bigcap X \subset X . $$

I understand the proof except for step 1. I know that $\mathcal O$ is an ordinal and thus transitive. So if $\alpha\in \mathcal O$ then $\alpha\subset \mathcal O$. But the converse may not be true. I'm thinking for example in $X=\mathcal O\setminus \{\emptyset\}$. Then $\emptyset\in\{\emptyset\}\in X$, but $\emptyset\notin X$. So $\{\emptyset\}\in X$ but $\{\emptyset\}\not\subset X$.

What do you think? Is the author right? Or is my counterexample valid? And how can I prove that $\bigcap X\subset X$?

Thanks

Edit The theorem is all completely wrong. The statement is that if $X\subset \mathcal O$ is a non-empty subclass of the class of ordinals numbers, then

$$ \bigcap X \in X .$$

The author proves (wrongly) that $\bigcap X\subset X$, which I don't know if it is right.

My aplogies.

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    $\begingroup$ It should be $\bigcap X\in X$, and $\bigcap X$ is the ordinal, not $X$. $\endgroup$ – Asaf Karagila Feb 13 '18 at 12:18
  • $\begingroup$ Yes yes, there are a lot of mistakes. Probably I will ask a new question asking for help to prove the correct statement. $\endgroup$ – Dog_69 Feb 13 '18 at 12:21
  • $\begingroup$ I think it's time for a new source material. $\endgroup$ – Asaf Karagila Feb 13 '18 at 12:21
  • $\begingroup$ Humm... well, that book has 5 parts. The part of set theory follows almost literally the appendix of Kelley's book, but it proves many theorems that in Kelley's book are only stated. What I should do is to compare both books before to ask a question. Be sure that this will never happen again. $\endgroup$ – Dog_69 Feb 13 '18 at 12:30
  • $\begingroup$ I have already post the correct question: math.stackexchange.com/questions/2649722/… $\endgroup$ – Dog_69 Feb 14 '18 at 15:34
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Easier counterexample of (1): obviously, X =$\{0,2\}\subset\mathcal O$ but isn't an ordinal because $2 = \{0,1\}\in X$ but $2 = \{0,1\}\not\subset X$.

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