Proving the existence of the least limit ordinal In Jech, he defines $\omega$ to be the least limit ordinal and guarantees its existence by the axiom of infinity. I wanted to see this, so he directs us to this problem 

If a set $X$ is inductive, then $X \cap \textit{Ord}$ is inductive. The set $N = \cap\{X : X \text{ is inductive}\}$ is the least limit ordinal $\omega \neq 0$.

How would I start with this problem. Do I first need to show that $N$ itself is an ordinal? 
 A: $N$ is a set by axiom of infinity, which states: there exists an inductive set. From $N\cap Ord=N$, we see that $\bigcup N$ is an ordinal. So if we show $N=\bigcup N$, then we get that $N$ is a limit ordinal.
$N\subset \bigcup N$ can be shown from the fact: $x\in N\implies x\cup\{x\}\in N$. To show $\bigcup N\subset N$, first note that $x\in N$ implies $x$ is either $0$ or a successor ordinal because every non-empty limit ordinal is an inductive set.
If we show that $N$ is transitive, then $\bigcup N\subset N$ is clear. Assume there is $x\in N$ such that there exists $y\in x$ with $y\notin N$. Since elements of $N$ are ordinals, we may let $\alpha\in N$ be the least ordinal with the above property. Then $\alpha$ is clearly non-empty, so $\alpha$ is a successor ordinal, say $\alpha=\beta+1=\beta\cup\{\beta\}$. Say $y\in\alpha$ is such that $y\notin N$. Then by the choice of $\alpha$, we see that $y=\beta$.
If $\beta=0$, then this is clearly a contradiction, so we are done. Otherwise, let $M:=\{\theta\in N:\theta<\beta\}$ which is a non-empty set and $0\in M$. Moreover, for $x\in M$, we know $x+1\in N$. If $x+1<\beta$ for all $x\in M$, then $x+1\in M$, so that $M$ is an inductive set, so we reach a contradiction since then $M$ is a strict subset of $N$. If $x+1=\beta$ for some $x\in M$, then $\beta\in N$ so we reach a contradiction.
