Let $\mathfrak{g}$ be a finite dimensional solvable Lie algebra. Given that we can take subspaces $\mathfrak{a}_{i}$ of $\mathfrak{g}$ such that $\mathrm{dim}(\mathfrak{a}_{i}/\mathfrak{a}_{i+1})=1$ and $\mathfrak{g}=\mathfrak{a}_{0}\supseteq \mathfrak{a}_{1} \supseteq \cdots \supseteq \mathfrak{a}_{n}=0$, why is it that for all $i$ there is some $j$ such that $\mathfrak{g}^{j}\supseteq \mathfrak{a}_{i} \supseteq \mathfrak{a}_{i+1} \supseteq \mathfrak{g}^{j+1}$? Here $\mathfrak{g}^{j} = [\mathfrak{g}^{j-1},\mathfrak{g}^{j-1}]$ and $\mathfrak{g}^{0} = \mathfrak{g}$. For instance, if $\mathfrak{g}$ is $n$-dimensional, and $\mathfrak{a}_{1}=\mathrm{span}\{v_{1},v_{2},\dots,v_{n-1} \}$, I don not see clearly why $\mathfrak{g}^{1}\subseteq \mathfrak{a}_{1}$. I would appreciate your help in making this a bit clear. Also I have not found results on the dimension of the ideals $\mathfrak{g}^{j}$.
I am following the proof of the following result in Lie groups: beyond an introduction.
An $n$-dimensional Lie algebra $\mathfrak{g}$ is solvable if and only if there exists a sequence of subalgebras $$ \mathfrak{g}=\mathfrak{a}_{0}\supseteq \mathfrak{a}_{1} \supseteq \cdots \supseteq \mathfrak{a}_{n}=0 $$ such that, for each $i$, $\mathfrak{a}_{i+1}$ is an ideal in $\mathfrak{a}_{i}$ and $\mathrm{dim}(\mathfrak{a}_{i}/\mathfrak{a}_{i+1})=1$