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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.

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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.

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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
@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.

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