I want to prove the following problem
Let $Y$ be an open or a closed subscheme of a scheme $X$, with the natural inclusion $i:Y \hookrightarrow X $. Prove that $Y$ is separated over $X$.
I don't understand what the role of "open or close subscheme" is here? Does it represent "open or closed immersion".
If that's the case, then the problem can be reduced to what I mentioned in my title.
(This "separated" means: Let $f:X\to Y$ is a morphism between scheme $X$ and scheme $Y$. If the diagonal morphism $\Delta(X):X \to X\times_Y X$ is closed, then we call $f$ is separated)
Please give me some hints, thank you so much!
Geometrically, this follows from the fact that the intersection of an open/closed subscheme with itself is the subscheme itself again. In a general category with the necessary limits, the following holds: a morphism $g\colon Y\to X$ is a monomorphism iff the commutative square $$\require{AMScd} \begin{CD} Y @>{\mathrm{id}}>> Y\\ @V{\mathrm{id}}VV @VV{g}V\\ Y @>{g}>> X \end{CD}$$ is a pullback diagram in this category. (Edit: This is the case iff the diagonal $\Delta\colon Y\to Y\times_X Y$ is an isomorphism.) In particular, any monomorphism of schemes is separated, so you're done.