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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?

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Why do you need to remove the intersection, its still a chain in $X$ – Sean Ballentine Feb 25 '13 at 23:06
True, but if $Y$ is not closed then it is not a descending chain of closed sets. – Steven-Owen Feb 25 '13 at 23:09
Can you take the closures? – Sean Ballentine Feb 25 '13 at 23:13
No, the def of Noetherian Space is a DCC on closed sets – Steven-Owen Feb 25 '13 at 23:13
You could use (prove?) the equivalent characterization of a Noetherian space as hereditarily compact. Then this is trivial. – Shawn Henry Feb 25 '13 at 23:16
up vote 5 down vote accepted

If $Y_1\supseteq Y_2\supseteq \cdots$ is a descending sequence of closed sets in $Y$, let $Y_i=X_i\cap Y$, where $X_i$ is closed in $X$. Then the sequence $X'_i:=X_1\cap\cdots \cap X_i$ gives the same intersections with $Y$ as $X_i$ does and is a descending sequence of closed sets in $X$. Therefore, $X'_i$ must become eventually constant, so $Y_i$ must as well.

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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 $(Y_n)$.

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