Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $A$ be a nonempty subset of $\omega$, the set of natural numbers. If $\bigcup A=A$ then $A=\omega$.

I have proved '$\bigcup n^+ = n$ for any $n\in\omega$' and '$\bigcup\omega = \omega$' I think this question should be proved with the above results..

Help please

share|improve this question
1  
You should really learn some LaTeX. –  Asaf Karagila Apr 25 '12 at 8:09
1  
You may also want to read why and how to accept an answer. –  Asaf Karagila Apr 25 '12 at 8:18

1 Answer 1

up vote 5 down vote accepted

Lemma: If $A$ is a set of ordinals, then $\bigcup A$ is an ordinal.

The proof appears here.

Suppose that $A\subseteq\omega$ then $\bigcup A$ is an ordinal. If $A=\bigcup A$ then $A$ was an ordinal, that is either $n\in\omega$ or $\omega$ itself.

However you know that for $n\neq 0$, $\bigcup n\neq n$, so you are only left with the option $A=\omega$.


To prove this directly in this case you can argue that if $n\in A$ then $n\subseteq\bigcup A$. From this follows that if $A$ is non-empty and $\bigcup A=A$ for every $k\in\omega$ such there is $n\in A$ and $k<n$ we have $k\in\bigcup A$, and therefore $k\in A$. So if $A$ is unbounded $A=\omega$; if $A$ is bounded then $A=n$ but we know that $\bigcup n\neq n$ which is a contradiction.

share|improve this answer
    
Its good but i want to prove this by showing that a set satisfying above condition is a successor set. –  Katlus Apr 25 '12 at 13:41
    
I have proved that empty set is an element of A. Can you help me how to show that if n is an element of A, successor of n is an element of A? –  Katlus Apr 25 '12 at 13:44

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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