## results on the bound of some nonincreasing sequence? [closed]

Hello, everyone. I want to know whether there are nontrivial results of the following type:\

Given an operation $F$, then for any set $A \neq \emptyset$ that satisfies the property: $\exists K(A)>0$,s.t.,$A\supseteq F(A)\supseteq F\circ F(A)\supseteq \cdots \supseteq \underbrace{F\circ \cdots \circ F}_{K(A)}(A)=\emptyset$,then we can find a precise upper bound of $K(A)$ by using the information of $F,A$.

For example,consider A is the variety corresponding to a finite set of differential polynomials$f_{1},\cdots,f_{n}$ such that the differential ideal it generated is the unit ideal,(here A is obtained by regarding the differential polynomials as ordinary ones in polynomial ring) now take $F$ as the operation that putting $\delta f_{1},\cdots,\delta f_{n}$ to the set $f_{1},\cdots,f_{n}$ (here $\delta$ denotes the derivation operation),we obtain a set $A_{1}$ and then taking its variety,we obtain $F(A)$,continue,putting ${\delta}^{2} f_{1},\cdots,{\delta}^{2} f_{n}$ to set $A_{1}$,we obtain $A_{2}$,then $F\circ F(A)$,$\cdots,$. we know that the set sequence is nonincreasing and will become $\emptyset$ in some step.

Since I want to know how to attack this type of problem, any known results linked to this type may be illuminating.

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What is a "super bound"? – Ricky Demer Dec 24 2010 at 16:21
"upper bound". That much is clear, but the question is so vague that one can hardly say anything meaningful here. – fedja Dec 24 2010 at 16:50
Voting to close - too vague. Do you have any examples of such $F$ and $A$ so maybe we can get an idea of the kind of thing you're talking about? – Gerry Myerson Dec 24 2010 at 20:37
Sorry for the first version, I have editted it by adding an example in differential algebra. – Jiang Dec 25 2010 at 4:25
Your example does not seem to fit your criteria, since you end up with zero instead of the empty set. More generally, you need to ask a more focused question. Right now, it seems like almost any argument involving a finiteness condition can be rephrased as an argument roughly of the form you wrote, as long as the notion of "operation" is interpreted suitably liberally. – S. Carnahan Jan 2 2011 at 21:50