Well, what is an infinite descending chain? It’s precisely a sequence a0,a1,… such that ∀jaj+1∈aj. Which means, indeed, a function with the required property. And then, yes, you can use Replacement to guarantee that there’s a set containing the aj.
Now, if I understand you correctly, what you’re asking is whether the following scenario is possible. We have a model M of the axioms of ZF. (So, out in the “real world” we have a set S providing the “sets” in M, and a relation R corresponding to set membership in M, so that R(x,y) means that x is an element of y in M.) In the model there’s a set A, and “really” (i.e., outside, not inside, the model) we have a sequence (i.e., a function from the natural numbers, but we write it with subscript notation) a0,a1,… of elements of S such that ∀jR(aj+1,aj). So with our god’s-eye perspective we can see that A “has” an infinite descending chain, but inside M there is no such thing; there is no element of S that (within the model) is a function from the model’s natural numbers such that, etc. And there is no element of S that contains-in-M precisely the elements of our descending chain.
The answer (which was not obvious to me, but it’s years since I actually studied this stuff) is that this is possible, and for very much the sort of reason you have in mind.
Suppose we have a model of ZF. It satisfies the axiom of foundation. And suppose, for reasons I’ll explain in a moment, that in the “outside world” the axiom of choice holds. Now we construct what’s called an ultrapower of our model. The details are a bit intricate; here goes. Say that a “filter on the natural numbers” is a set of sets of natural numbers such that (1) if X is in the filter then so is anything that contains X and (2) if X and Y are in the filter then so is their intersection. The idea is that a filter is a “notion of largeness”; for any filter we can call sets in the filter “big” and sets not in the filter “small” and this will be a somewhat-reasonable use of language. Now say that an ultrafilter on the natural numbers is a filter that can’t have any more sets added to it while remaining a filter, other than by making literally every set of natural numbers “big”. Example: {sets containing 17} is an ultrafilter. (Proof: exercise for the reader.) Ultrafilters of this “easy” kind -- {sets containing n} for some fixed n—are called principal ultrafilters. Are there any other ultrafilters? It’s not obvious. There is no way to construct one, but if the axiom of choice holds then indeed there are. OK, suppose we’ve got an ultrafilter. Now, consider all infinite sequences of elements of S and say that two such sequences a,b are the same if {j:aj=bj} is “big” (i.e., is an element of the ultrafilter). This set, modulo this equivalence relation, is what we call an ultrapower of S. We can extend this to the whole model M by saying what it means for one element of the ultrapower to be an element of another: a is an element of b iff aj∈bj for a “big” set of j. Now an amazing thing is true: any sentence in the language of set theory that’s true in M is also “true” of the ultrapower. In particular, the axiom of foundation is “true”. BUT it turns out that the natural numbers in M are strange in exactly the sort of way we need. Consider the sequence (0,1,2,3,...). This is bigger than (0,0,0,...) and bigger than (1,1,1,...) and bigger than (999,999,999,...), etc. (Because the corresponding per-sequence-element is true at all but a finite set of positions, and any finite set is “small”.) It behaves like an “infinitely large natural number”. We can subtract 1 from it, getting (0,0,1,2,3,...) which is smaller. We can subtract 1 from that, getting (0,0,0,1,2,3,...) which is smaller still. (There’s no particular reason other than convenience for using 0 for the elements we can’t actually subtract 1 from. Remember, any “small” change to the sequence is ignored, and in particular for any finite set of ’em it doesn’t matter what entries we use.) And so on. So we have an infinite descending sequence of natural numbers—infinite “from the outside”, that is. On the inside it’s no such thing; it can’t be, because “there is no infinite descending sequence of natural numbers” is a fact about M, readily expressed in the language of set theory, and therefore true in the ultrapower too. (So, in particular, that nice simple sequence of elements is not a set in the ultrapower.) And if we “implement” natural numbers in the usual way, “smaller than” for natural numbers is exactly the same relation as “element of”, so in fact this very thing is an externally-infinite descending sequence of sets.
Yes, you understood my question correctly… and I need to spend some time thinking about your answer. (Mostly because it’s past midnight here, so I am leaving my computer for now.)
Thank you! It was a pleasure to be understood—unlike when I e.g. post a question on Stack Exchange. :D
Well, what is an infinite descending chain? It’s precisely a sequence a0,a1,… such that ∀j aj+1∈aj. Which means, indeed, a function with the required property. And then, yes, you can use Replacement to guarantee that there’s a set containing the aj.
Now, if I understand you correctly, what you’re asking is whether the following scenario is possible. We have a model M of the axioms of ZF. (So, out in the “real world” we have a set S providing the “sets” in M, and a relation R corresponding to set membership in M, so that R(x,y) means that x is an element of y in M.) In the model there’s a set A, and “really” (i.e., outside, not inside, the model) we have a sequence (i.e., a function from the natural numbers, but we write it with subscript notation) a0,a1,… of elements of S such that ∀j R(aj+1,aj). So with our god’s-eye perspective we can see that A “has” an infinite descending chain, but inside M there is no such thing; there is no element of S that (within the model) is a function from the model’s natural numbers such that, etc. And there is no element of S that contains-in-M precisely the elements of our descending chain.
The answer (which was not obvious to me, but it’s years since I actually studied this stuff) is that this is possible, and for very much the sort of reason you have in mind.
Suppose we have a model of ZF. It satisfies the axiom of foundation. And suppose, for reasons I’ll explain in a moment, that in the “outside world” the axiom of choice holds. Now we construct what’s called an ultrapower of our model. The details are a bit intricate; here goes. Say that a “filter on the natural numbers” is a set of sets of natural numbers such that (1) if X is in the filter then so is anything that contains X and (2) if X and Y are in the filter then so is their intersection. The idea is that a filter is a “notion of largeness”; for any filter we can call sets in the filter “big” and sets not in the filter “small” and this will be a somewhat-reasonable use of language. Now say that an ultrafilter on the natural numbers is a filter that can’t have any more sets added to it while remaining a filter, other than by making literally every set of natural numbers “big”. Example: {sets containing 17} is an ultrafilter. (Proof: exercise for the reader.) Ultrafilters of this “easy” kind -- {sets containing n} for some fixed n—are called principal ultrafilters. Are there any other ultrafilters? It’s not obvious. There is no way to construct one, but if the axiom of choice holds then indeed there are. OK, suppose we’ve got an ultrafilter. Now, consider all infinite sequences of elements of S and say that two such sequences a,b are the same if {j:aj=bj} is “big” (i.e., is an element of the ultrafilter). This set, modulo this equivalence relation, is what we call an ultrapower of S. We can extend this to the whole model M by saying what it means for one element of the ultrapower to be an element of another: a is an element of b iff aj∈bj for a “big” set of j. Now an amazing thing is true: any sentence in the language of set theory that’s true in M is also “true” of the ultrapower. In particular, the axiom of foundation is “true”. BUT it turns out that the natural numbers in M are strange in exactly the sort of way we need. Consider the sequence (0,1,2,3,...). This is bigger than (0,0,0,...) and bigger than (1,1,1,...) and bigger than (999,999,999,...), etc. (Because the corresponding per-sequence-element is true at all but a finite set of positions, and any finite set is “small”.) It behaves like an “infinitely large natural number”. We can subtract 1 from it, getting (0,0,1,2,3,...) which is smaller. We can subtract 1 from that, getting (0,0,0,1,2,3,...) which is smaller still. (There’s no particular reason other than convenience for using 0 for the elements we can’t actually subtract 1 from. Remember, any “small” change to the sequence is ignored, and in particular for any finite set of ’em it doesn’t matter what entries we use.) And so on. So we have an infinite descending sequence of natural numbers—infinite “from the outside”, that is. On the inside it’s no such thing; it can’t be, because “there is no infinite descending sequence of natural numbers” is a fact about M, readily expressed in the language of set theory, and therefore true in the ultrapower too. (So, in particular, that nice simple sequence of elements is not a set in the ultrapower.) And if we “implement” natural numbers in the usual way, “smaller than” for natural numbers is exactly the same relation as “element of”, so in fact this very thing is an externally-infinite descending sequence of sets.
Yes, you understood my question correctly… and I need to spend some time thinking about your answer. (Mostly because it’s past midnight here, so I am leaving my computer for now.)
Thank you! It was a pleasure to be understood—unlike when I e.g. post a question on Stack Exchange. :D