There is this theorem that states: If $\alpha$ is an ordinal, then there is a limit ordinal $\mu$ with $\alpha &lt; \mu$. The definition of limit ordinal is just &quot;an ordinal which is a limit&quot;, but a better reformulation of this definition is: Let $\lambda$ be a non-zero ordinal. $\lambda$ is a limit ordinal if and only if for all ordinal $\alpha$, if $\alpha&lt;\lambda$ then $\alpha^+&lt;\lambda$. ($\alpha^+$ is the successor ordinal of $\alpha$) I don't understand how that theorem makes sense, the book I'm reading gives the proof as an exercise (which I can't do) so I can't even see why it's true. If for example I have the ordinal $1=\{\emptyset\}$ what is a limit ordinal for this? Is it $\omega$ for example (or $2\omega$, $3\omega$ etc)? Is this what it means?
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: $\beta &lt; \alpha$, 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 $\beta &lt; \bigcup$ 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 = $\beta \cup \{&#160;\beta \}$ 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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