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So I think I've misunderstood the principle of Noetherian induction as stated in the Hartshorne exercise II.3.16, or his statement is slightly incorrect. He says: "Let $X$ be a Noetherian topological space, and let $\mathscr{P}$ be a property of closed subset 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 $X$."

Why does $\mathscr{P}$ hold for the empty set? What if $\mathscr{P}$ is the property of being nonempty and $X = \varnothing$?

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    It just means that you can't use induction to prove that $X$ is nonempty (which is not surprising).2011-06-26

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$\mathscr P$ holds for the empty set $\varnothing$ because it holds for every proper closed subset of $\varnothing$.

You should not let the fact that there is no proper closed subset of $\varnothing$ confuse you!

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    And yet I did... "Vacuously true" things have confused me in the past.2011-06-26
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It is just as Mariano says.

The situation here is completely analogous to the fact that if one wishes to prove a statement is true for all natural numbers by strong induction (or more generally for a family of values indexed by a well-ordered set) then, logically speaking, one does not have to single out a "base case" $P(0)$, because the induction step allows us to deduce $P(0)$ from $P(n)$ for all $n < 0$, of which there are none.

Many other people have found this confusing. (I believe it came up in a MO question asked by Bjorn Poonen a while back, which is not to imply that he was confused by it!) In practice, this little logical filigree doesn't seem to save any time: you still have to know how to prove $P(0)$ assuming nothing, and the argument for this is usually rather different and often easier than the general induction step(s).

Added: Just to place the last card on the table, this "Noetherian induction" is really exploiting the fact that (by definition) a topological space is Noetherian iff its closed sets satisfy the Descending Chain Condition, which in order-theoretic terms is expressed by saying that the containment relation among closed subsets is a well-founded partial ordering (or sometimes "well-partial ordering"). A partial ordering $(X,\leq)$ is well-founded iff every nonempty subset has a minimal element and this shows that a subset $Y$ of $X$ with the property:

$\forall x \in X \ (\forall y \in X, y < x \implies y \in Y) \implies x \in Y$

must be all of $X$: if not, consider the least element of $X \setminus Y$.

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    From a more general perspective, one may view Noetherian topological spaces as topological analogs of the order-theoretic notion of a [well-quasi-order.](http://en.wikipedia.org/wiki/Well-quasi-ordering) For example see Jean Goubault-Larrecq, [On Noetherian Spaces.](http://ieeexplore.ieee.org/stamp/stamp.jsp?tp=&arnumber=4276588)2011-06-26
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I believe that the answer lies hidden within the depth of "vacuously true argument".

An argument of the form $\forall x\varphi$ is true if and only if there is no $x$ such that $\lnot\varphi(x)$.

An example I often used with my students was "If it I am drinking beer during the class then you are all elephants" (or something similar along these lines, usually including alcohol of some sort and a ridiculous entailment). It does not matter that I am talking to people, because I am now allowed to drink alcohol during class (as a teacher/TA anyway).

If we say that $P(x)$ holds for $x$ if for all $y holds $P(y)$. In this case $<$ is proper inclusion of closed subsets; since there are no proper subsets of the empty set, the argument $P(x)$ holds for $\varnothing$ vacuously, as above.

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    @Mariano, I can. But if I am out of beer in my office then there is no feasible source of alcohol around. Sadly it happens more than I wish for.2011-06-26