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.

I'd like to show that the class of all limit ordinals, LIM, is cofinal in the class of all ordinals ON.

I thought that I could do this by contradiction and assume that it wasn't cofinal in ON. Then there is an $\alpha \in$ ON such that for all $\beta \in $ LIM: $\beta < \alpha$, i.e. $\beta \in \alpha$ and hence LIM $\subset \alpha$.

This means that LIM is a set so we can consider $\bigcup$ LIM. Then we have $\beta < \bigcup$ LIM for all $\beta \in$ LIM by definition of $\bigcup$ LIM. So LIM $\subseteq \bigcup$ LIM.

$\bigcup$ LIM is either a limit or a successor ordinal.

If it is a limit ordinal then $\bigcup$ LIM $\in$ LIM and by transitivity of ON, $\bigcup$ LIM $\subseteq$ LIM, so $LIM = \bigcup$ LIM in this case which would mean that $\bigcup$ LIM $\in$ LIM $= \bigcup$ LIM which would be a contradiction because sets cannot contain themselves. Hence $\bigcup$ LIM cannot be a limit ordinal.

If $\bigcup$ LIM is a successor ordinal then $\bigcup$ LIM = $\beta \cup \{ \beta \}$ for some $\beta \in $ ON.

Now I'm not sure how to proceed from here. I'm also not sure about whether I'm on the right track with this proof because I think there should be a shorter way to prove this.

Thanks for your help.

share|improve this question

2 Answers 2

up vote 4 down vote accepted

Trivially, let $\beta$ be an ordinal, then $\beta+\omega$ is strictly larger than $\beta$, and it is a limit ordinal. Therefore the limit ordinals are cofinal in the class of ordinals.

Why is $\beta+\omega$ a limit ordinal? Well, recall that $\beta+\omega = \sup\{\beta+n\mid n<\omega\}$. Suppose it is not a limit ordinal then $\beta+\omega=\gamma+1$ therefore for some $n<\omega$ we have that $\beta+n+1=\beta+\omega$, which is a contradiction.


Just as well, and perhaps even nicer: given $\beta$ we can take $\beta^+=\min\{\gamma\mid\forall f\colon\beta\to\gamma, f\text{ not surjective}\}$, the "next initial ordinal" is also a limit ordinal by definition.

share|improve this answer
    
Is showing that every cofinal class of ON is proper also a one liner? –  Rudy the Reindeer Dec 29 '11 at 14:29
    
@Matt: Of course. Let $M\subseteq\mathrm{Ord}$. If it a set then $M\in V_\alpha$ for some $\alpha$ and thus bounded by $\alpha$; and if it not a set then for every $\alpha$ we have $M\cap\alpha\subsetneq M$, so there is $\beta\in M$ such that $\alpha<\beta$, therefore $M$ is cofinal. –  Asaf Karagila Dec 29 '11 at 15:01
    
Thank you, Asaf. –  Rudy the Reindeer Dec 29 '11 at 18:16
    
Can this proof be modified into not using the von Neumann universe? –  Rudy the Reindeer Dec 30 '11 at 8:09
1  
@Matt: It is, but I'm just talking about the rank of the set $M$, and a set of ordinals of rank $<\alpha$ is a subset of $\alpha$. This is the point, that $M$ is bounded by $\alpha$. –  Asaf Karagila Dec 30 '11 at 8:15

If $\mathbf{LIM}$ is not cofinal in $\mathbf{ON}$ then there is an ordinal $\alpha $ such that $\forall \lambda \in \mathbf{LIM}: \lambda \le \alpha$. This implies that $\mathbf{LIM}$ is a set so that we can construct the smallest ordinal exceeding all limit ordinals: $\bigcup \mathbf{LIM}$. $\bigcup \mathbf{LIM}$ is itself a limit ordinal (easy to show) but it is not in $\mathbf{LIM}$ which constitutes a contradiction.

share|improve this answer

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.