# Infinite Chain of Implication Statements - Can it have a Conclusion?

First, does an infinite string of implication statements have a conclusion? If so, is there a such thing as a "closure" of such a beast, giving a conclusion?

I am also trying to determine if the following is valid. I begin with uncountable set S, not necessarily well-ordered. I show $\exists$ a nonempty $A \subsetneq S$ such that property P(A) is true, and that this implies P(B) is true for some nonempty $B \subsetneq A$, and this implies P(C) is true for some nonempty $C \subsetneq B$, and so on. What are the necessary and sufficient conditions for which this process would "terminate" with a conclusion? If none, can I modify this general approach to make it conclude (akin to some kind of logical "closure")?

• Your question seem to be too broad; furthermore you need to make the distinction between well-ordered and well-orderable. Assuming the axiom of choice we can always well-order a set. May 3, 2013 at 1:59
• I only mention it may not be well-ordered because I rely, in no way, on any ordering, as only the existence of these sets is proven. Anyway, the third paragraph is the most important to me and the most specific. I just would like to know if the decreasing sequence of subsets "ends" at some subset. May 3, 2013 at 2:02
• I was talking on the third paragraph being too broad. Transfinite recursion is equivalent to the schema of replacement. If you don't have $\sf ZF$ you can't really do transfinite recursion. If you have the axiom of choice then you can prove the existence of well-ordered, and otherwise it would be very wise to state that as well, because so many things are ill-behaved when the axiom of choice goes missing. May 3, 2013 at 2:04
• Consider $A_k=\{n\in\Bbb N\mid k<n\}$, this is a decreasing chain of subsets of $\Bbb N$ without an "end". May 3, 2013 at 2:05
• That is the plan. Yes. May 3, 2013 at 2:38