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I'm confused by the Noetherian Induction exercise in Hartshorne.

Let $X$ be a Noetherian topological space, and let $\mathscr{P}$ be a property of closed subsets of $X$. Assume that for any closed subset $Y$ of $X$, if $\mathscr{P}$ holds for every proper closed subset of $Y$, then $\mathscr{P}$ holds for $Y$. (In particular, $\mathscr{P}$ must hold for the empty set.) Then $\mathscr{P}$ holds for $X$.

Using Zorn's Lemma, we can say that the set $\Sigma$ of closed subsets of $Y$ where $\mathscr{P}$ holds has a maximal element, but how do we know that the maximal element is $X$? I want to say that we can take the union of all closed subsets in $\Sigma$ and that should be $X$ but why is the union of all closed subsets in $\Sigma$ equal to $X$?

Thanks!

nekodesu
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  • I think you need to exchange $Y$ by $X$ in your question. For $Y$ already know this by assumption. – red_trumpet Aug 26 '18 at 08:41
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    For the empty set we know, that $\mathscr{P}$ hold for every proper subset, because there are no proper subsets. Thus $\mathscr{P}$ holds for the empty set as well, by assumption. – red_trumpet Aug 26 '18 at 08:42
  • In my Hartshorne's book this is Exercise II.3.16 .. I'm not sure if I have to correct this post. Are there multiple versions of this book? – quantum Mar 27 '20 at 06:50

1 Answers1

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Consider the set $S$ of closed subsets that do not satisfy $\mathscr{P}$.

Assume that $S$ is not empty. Then you can find an infinite descending chain in $S$ as follows:

for any closed subset $Y$ of $X$, if $\mathscr{P}$ holds for every proper closed subset of $Y$, then $\mathscr{P}$ holds for $Y$

can be rewritten by contrapositive

for any element $Y$ of $S$, there is a proper closed subset of $Y$ that also lies in $S$

But this contradicts $X$ being Noetherian.

Arnaud Mortier
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