I'm stuck on the proof of the following propsition in Humphreys Linear Algebraic Groups:
Let $G$ be an algebraic group acting morphically on a variety $X$. Then each orbit is a smooth, locally closed subset of $X$, whose boundary is a union of orbits of strictly lower dimension.
Proof: Let $Y$ be an orbit in $X$. As the continuous image of a closed set, $Y$ is constructible, and hence contains an open dense subset of $\overline{Y}$. But $G$ acts transitively on $Y$ (and hence stablizes $\overline{Y}$), so $Y$ is smooth and contains a neighborhood in $\overline{Y}$ of each of its points, i.e. $Y$ is open in $\overline{Y}$. Therefore, $\overline Y \setminus Y$ is closed and of strictly lower dimension than $\overline{Y}$. Being $G$-stable, this boundary is just a union of other orbits.
Two questions: first, why is $Y$ is open in $\overline{Y}$? I don't understand why each point in $Y$ admits an open neighborhood in $\overline{Y}$. Second, why is $\overline{Y} \setminus Y$ of strictly lower dimension than $\overline{Y}$?
Let $x\in X$ with $Y=G\cdot x$. There is a subset $U\subseteq Y$, which is open and dense in $\bar{Y}$, so you can write the orbit as union of open sets
$$G\cdot x=\bigcup_{g\in G} g\cdot U, $$ where $g\cdot U$ is open as isomorphic image of an open set.
$\overline{G\cdot x}\setminus G\cdot x$ is a closed subset and every irreducible component of it is a proper subset of a irreducible component of $\overline{G\cdot x}$, because $G\cdot x$ is dense in $\overline{G\cdot x}$. So by definition of krull dimension it has stricly lower dimension.