LIM is cofinal in ON 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.
 A: 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. 
A: 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.
