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 $X \setminus Y$.