I'm stuck on the following problem:
Let $X$ be a Noetherian space, and let $Y \subseteq X$ have the property that for every irreducible closed set $Z \subseteq X$, $Y \cap Z$ contains an open dense subset of $\overline{Y \cap Z}$. Show that $Y$ is constructible.
It seems to me that without loss of generality, we may assume that $X$ is Noetherian and irreducible: if $X_1, ... , X_n$ are the irreducible components of $X$, then $Y \cap X_i$, considered as a subspace of $X_i$, should have the same property as $Y$ as a subspace of $X$. Once we show that $Y \cap X_i$, it will follow that $Y$ is also constructible as a finite union of constructible sets (since $X_i$ is closed in $X$, constructible subspaces of $X_i$ are obviously the same thing as constructible subspaces of $X$ which are contained in $X_i$)
The only hint I've gotten is "use Noetherian induction" which I don't know where to apply. Would anyone kindly provide another hint?
You need to prove that $Y$ is constructible, i.e. can be written as a finite union of locally closed sets. If you prove that $Y$ contains a nonempty open $U$, then you could replace $Y$ by $Y \setminus U$, which is closed in $Y$, and continue inductively (check that $Y \setminus U$ still satisfies the assumptions, etc).
(This type of argument is actually one of the most classical uses of Noetherian induction.)