Regular semisimple elements form a dense open subset

The set ${\mathfrak{h}}_{\mathrm{r}\mathrm{e}\mathrm{g}}$ from The regular root-hyperplane arrangement in a Cartan subalgebra is the complement of finitely many hyperplanes, so it is dense and Zariski open in $\mathfrak{h}$. For $h\in {\mathfrak{h}}_{\mathrm{r}\mathrm{e}\mathrm{g}}$, The centralizer of a Cartan element from its vanishing roots gives $C_{\mathfrak{g}}(h)=\mathfrak{h}$; moreover $h$ is semisimple because the Cartan subalgebra acts diagonalizably in the root decomposition.

Let $G_{\mathrm{a}\mathrm{d}}$ be the connected adjoint group and consider $\mathrm{\Phi }:G_{\mathrm{a}\mathrm{d}}\times {\mathfrak{h}}_{\mathrm{r}\mathrm{e}\mathrm{g}}\to \mathfrak{g},(g,h)\mapsto \mathrm{Ad}(g)h\mathrm{.}$ At $(1,h)$ its differential is $(x,k)\mapsto [x,h]+k$. The root decomposition and the inequalities $\alpha (h)\mathrm{\ne }0$ give $[\mathfrak{g},h]={⨁}_{\alpha \in \mathrm{\Phi }}{\mathfrak{g}}_{\alpha }$, so this differential is surjective. Therefore the image $U$ is open. The same surjectivity makes the algebraic map $\mathrm{\Phi }$ dominant; its constructible image contains a nonempty Zariski-open subset of $\mathfrak{g}$, so $U$ is dense. Every point of $U$ has centralizer dimension $\dim \mathfrak{h}$ by step 1.1. Since the regular locus is dense by Regular elements form a connected dense open subset, it meets the nonempty open set $U$; at a point of that intersection the centralizer dimension is both $\mathrm{rank}\mathfrak{g}$ and $\dim \mathfrak{h}$. Hence $\mathrm{rank}\mathfrak{g}=\dim \mathfrak{h}$.

The equality $\mathrm{rank}\mathfrak{g}=\dim \mathfrak{h}$ from step 2.1 now shows that every point of $U$ is regular semisimple. Conversely, let $x$ be regular semisimple. By [A1], $x$ lies in a Cartan subalgebra; by Cartan subalgebras are conjugate in a complex semisimple Lie algebra, conjugate that Cartan to the fixed $\mathfrak{h}$. The conjugate of $x$ is regular, so its centralizer has dimension $\dim \mathfrak{h}$; the centralizer formula then forces it to avoid every root hyperplane. Hence it lies in ${\mathfrak{h}}_{\mathrm{r}\mathrm{e}\mathrm{g}}$, and $x\in U$. Thus $U$ is exactly the regular semisimple locus and is dense and open.