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Let $A$ be a nonempty subset of $\omega$, the set of natural numbers. I want to prove this statement:

If $\bigcup A=A$ then $n\in A \implies n^+\in A$.

Help...

Rudy the Reindeer
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Katlus
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    When you're viewing the natural numbers as the sets of smaller natural numbers, it is good style to explicitly call them "finite ordinals" or something like that. Otherwise, readers are likely to think that you're only considering the arithmetic properies of the numbers (rather than one specific set-theoretic implementation of them). – hmakholm left over Monica Apr 25 '12 at 15:03
  • Also closely related: this question by the OP. – Asaf Karagila Apr 25 '12 at 16:18
  • @asaf: I read your proof. You showed that A is fully ordered by membership assuming that w is well ordered by membership, but is there another way to prove that UA is an element of w or is w, not assuming that w is well ordered? – Katlus Apr 25 '12 at 16:43
  • The book im studying has not even defined what is ordinal.. Thus i wrote 'element of w or w' rather than ordinal.. – Katlus Apr 25 '12 at 16:45
  • Katlus, I have no way of knowing what the book have taught you or haven't taught you. At least tell us what book you are studying from, so people knowing it might have a better way of helping you. This extends to my next point, you have not supplied any definition of $\omega$ or otherwise a natural number. There are several which are equivalent, I cannot know which one you were given as the basic definition and which one will come later. – Asaf Karagila Apr 25 '12 at 17:30
  • Charles C.Pinter. This book defines w as the intersection of all the successor sets. – Katlus Apr 25 '12 at 17:58

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$\bigcup A\subseteq A$ says that $A$ is transitive and is therefore an ordinal. Now if $A$ were a successor ordinal $\alpha+1 = \alpha\cup\{\alpha\}$, then $\alpha\in A$ but $\bigcup A = (\bigcup\alpha)\cup\alpha \not\ni \alpha$. Thus $A$ must be either $0$ or $\omega$.

Or more directly: Assume $n\in A$. Then $n\in\bigcup A$, that is, there exits $y$ such that $n\in y \in A$. Then $n^+ \le y$. In the case $n^+=y$ we have $n\in A$ directly. Otherwise $n^+\in y\in A$ so $n^+\in\bigcup A$.

hmakholm left over Monica
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  • Thanks! But i want to show that A=w by showing that A is a successor set. Your proof is really good but it shows that A=w first, then show that A is successor set. Can you help me how to show that A is a successor set directly? By the way, I have proved that empty set is an element of A. – Katlus Apr 25 '12 at 15:25
  • See edit. ${}{}{}$ – hmakholm left over Monica Apr 25 '12 at 15:34
  • This problem is at right after introducing elementary properties of natural numbers while showing that 'w is well ordered by membership' is at 3 chapters later. – Katlus Apr 25 '12 at 16:26
  • This is first time im studying set theory.. It feels weird to study further and come back to the problem i couldnt prove and prove it with the concept not in that chapter... – Katlus Apr 25 '12 at 16:34
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Oh yes, this can be pretty confusing! Here is how I think of it:

$\bigcup A = A$ really means $\bigcup A \subseteq A$ and $\bigcup A \supseteq A$.

Now assume $a \in A$. From $A \subseteq \bigcup A$, we get $a \in \bigcup A$, which, looking at the definition of $\bigcup A$, means that $a$ is an element of some element of $A$, say $b$. That is, $a \in b \in A$. But $b$ is a natural number, and since $a \in b$, $b$ is a $strictly$ $bigger$ natural than $a$. If $b = a^+$, we have $a^+ \in A$, and we are done. Otherwise, $a^+$ is an element of $b$. But then from $\bigcup A \subseteq A$, we get $a^+ \in A$ again.

And looking at the proof, we see that $A$ must have actually been $\omega$ all along!

lesnikow
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