TITLE: Subspace of Noetherian space still Noetherian
QUESTION [7 upvotes]: I see this everywhere, but cannot find a proof of it. Is it just this easy; it feels wrong.
Suppose $X$ is noet and $Y\subset X$ with subspace topology. If $Y$ has a descending chain of closed sets $Y_1 \supset Y_2 \supset Y_3 \supset \ldots$, where each $Y_i$ is formed by intersection some closed set in $X$ with $Y$. That is $(X_1\cap Y)\supset (X_2\cap Y) \supset \ldots$ but I can't seem to finish this because it is not as if I can just remove the intersections with $Y$ right?

REPLY [6 votes]: This is (Bourbaki, Alg. comm., chap. II, §4, n° 2, prop. 8).  Take a decreasing chain $(Y_n)$ of subsets of $Y$, closed in $Y$; note $Y_n = \overline{Y_n} \cap Y$, and the closures $\overline{Y_n}$ form a decreasing sequence of closed subsets of $X$.  As this sequence stabilizes, so does $(Y_n)$.