There is a statement on page 149 of James E. Humphreys' Linear Algebraic Groups:
Let $G$ be an algebraic group. $S \subseteq G$ is an arbitrary torus contained in a Borel subgroup $B$. $\mathscr{B}^S$ denotes the the set of Borel subgroups of $G$ containing $S$. Then $\mathscr{B}^S$ can be identified with a closed subgroup of $G/B$.
I don't know why is the set closed.
Let Y = \cup_{B' \in \mathscr{B}^S} B', and $\phi: G \rightarrow G/B$ be the canonical map. Then $\mathscr{B}^S$ can be identified with $\phi(Y)$. As $|\mathscr{B}^S|$ may be finite or infinite, I don't even know whether $Y$ is closed, then why is $\phi(Y)$ closed?
Thank you very much.