noetherian induction 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 $X = \varnothing$?
 A: 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 $y<x$ 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.
A: $\mathscr P$ 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!
A: 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 $X \setminus Y$.
