Let $G$ be a group and $H$ be an ascendant subgroup of $G$. Suppose that $X \leq H \leq K$, where $K$ has certain properties. I want to show that $K = N_G(X)H$. I have shown that $N_G(X)H \leq K$
Now there exists a chain of subgroups: $$H = H_0 \leq H_1 \leq \ldots \leq H_\alpha \leq H_{\alpha + 1}\leq \ldots \leq H_\tau = G$$
Let $x\in K$ and let $\alpha \leq \tau$ be the least ordinal such that $x \in N_G(X)H_\alpha$. How can I deduce that $\alpha$ is not a limit ordinal?
The author just states that 'Clearly, $\alpha$ is not a limit ordinal'
Probably in this context the chain of subgroups is assumed to satisfy $H_\alpha=\bigcup_{\beta<\alpha}H_\beta$ whenever $\alpha$ is a limit ordinal. That is, you get no new elements in $H_\alpha$ for limit ordinals $\alpha$; you just take the union of $H_\beta$ for $\beta<\alpha$. It follows that $N_G(X)H_\alpha=\bigcup_{\beta<\alpha}N_G(X)H_\beta$ if $\alpha$ is a limit ordinal. So if $x\in N_G(X)H_\alpha$, there is some $\beta<\alpha$ such that $x\in N_G(X)H_\beta$. That is, $\alpha$ is not the least such ordinal.