Regular semisimple elements form a dense open subset

Under [A1, F1], put $\rho ={\min }_{x}n(x)=\dim \mathfrak{h}$. The locus $U=\mathrm{Ad}(G){\mathfrak{h}}_{\mathrm{r}\mathrm{e}\mathrm{g}}$ is nonempty Euclidean open by [F1]. The ordinary regular locus is dense by Regular elements form a connected dense open subset, so it meets $U$. At a point of the intersection, the centralizer dimension is both $\mathrm{rank}\mathfrak{g}$ and $\dim \mathfrak{h}$, by [F1] and The centralizer of a Cartan element from its vanishing roots. Hence $\mathrm{rank}\mathfrak{g}=\rho =\dim \mathfrak{h}$.

If $x\in {\mathfrak{g}}^{\mathrm{s}\mathrm{r}}$, [F1] makes $x$ semisimple, so its generalized zero eigenspace is its actual centralizer and has dimension $n(x)=\rho =\mathrm{rank}\mathfrak{g}$ by step 1.1. Thus $x$ is regular semisimple. Conversely, if $x$ is regular semisimple, its adjoint operator is diagonalizable, so $n(x)=\dim C_{\mathfrak{g}}(x)=\mathrm{rank}\mathfrak{g}=\rho$, and $x\in {\mathfrak{g}}^{\mathrm{s}\mathrm{r}}$. Therefore the regular semisimple locus equals the dense open set ${\mathfrak{g}}^{\mathrm{s}\mathrm{r}}$ of [F1].

For regular semisimple $x$, [F1] makes $C_{\mathfrak{g}}(x)$ a Cartan subalgebra, and $x$ lies in it. By [A1] conjugate that Cartan to the fixed $\mathfrak{h}$. The conjugate $h$ of $x$ has centralizer dimension $\dim \mathfrak{h}$ by steps 1.1 and 2.1; the root centralizer formula The centralizer of a Cartan element from its vanishing roots forces $\alpha (h)\mathrm{\ne }0$ for every root $\alpha$, so $h\in {\mathfrak{h}}_{\mathrm{r}\mathrm{e}\mathrm{g}}$ by The regular root-hyperplane arrangement in a Cartan subalgebra. This proves the second assertion.