Containment of Bruhat cells on flag variety This question was posted at MSE, but it did not receive any answer there.
Let $G$ be a connected semisimple algebraic group over $\mathbb C$, $X$ the flag variety of $G$, $B_0$ a Borel subgroup, $\mathbb O$ a $B_0$-orbit on $X$.
Question: Can we always find a Borel subgroup $B_x$ so that the open $B_x$-orbit $\mathbb O'$ contains $\mathbb O$?
The obvious attempt would be to choose a point $\bar x \in \mathbb O$, and then take $B_x$ to be a Borel in opposite relative position to $B_{\bar x}$. But not all choices work. For example when $\mathbb O$ is the open orbit, choices are pretty restricted, and I'm failing to find a construction that works in general.
Any help would be appreciated.
 A: I'd say that the relevant fact here is as follows. For two Borels $B_1$ and $B_2$ with a common maximal torus $T$ let $x_1$ be the unique $T$-fixed point in the open $B_1$-orbit. Then the $B_2$-orbit $B_2x_1$ lies in the open $B_1$-orbit. For instance, you can see this by proving that
(a) for every $x\in B_2x_1$ the orbit closure $\overline{Tx}$ contains $x_1$ and
(b) the open $B_1$-orbit consists precisely of those points $x$ for which $x_1\in\overline{Tx}$.
(These are fairly basic properties of Schubert decompositions.)
In your setting this means that you can choose any maximal torus $T\subset B_0$, let $x'$ be the unique $T$-fixed point in $\mathbb O$ and consider the unique Borel $B\supset T$ for which the orbit $Bx'$ is open. This $B$ will be your $B_x$. In fact, this $B$ is the opposite (with respect to $T$) Borel of the stabilizer of $x'$. So in terms of your last paragraph you can choose any $\bar x$ but you have to choose the opposite Borel correctly: it must intersect $B_0$ in a maximal torus that fixes $\bar x$.
