In this context, by “monstrosity” I meant some technically-a-set that couldn’t be constructed. Like the a = {a} example, only I suspect that you can create an example of that type that would technically satisfy the Axiom of Foundation. I am probably wrong here, but I need to think about this more until I see how exactly I am wrong. (That will probably take weeks.)
But that “every connected directed graph is a picture of a set” is the kind of perspective I have in mind. Being given the entire structure of sets, interconnected, at once. Now the question is whether the structure can be designed in a way that technically satisfies the Axiom of Foundation, while somehow is not construable, e.g. because it is infinite in both directions. So no set contains itself, directly nor indirectly, it’s just a line of sets infinite in both directions, like the integers, where each set contains an infinite amount of “previous” sets; and probably also something else, to make the Axiom of Foundation happy. -- Chances are, I am just talking complete nonsense here.
One formulation of the axiom of foundation says that there’s no “infinite descending chain” of sets. That is, if you have a set a with an element b with an element c with …, then this cannot go on for ever. So at least some versions of your “infinite in both directions” are precisely ruled out by Foundation.
The version from Wikipedia seems okay with an infinite descending chain of sets, as long as each of them would also contain e.g. an empty set. In combination with other axioms, though… well, that’s where my knowledge of set theory becomes insufficient.
Okay, here is one of the questions I couldn’t answer by self-study; let me use this opportunity to ask you: The Wikipedia page on Axiom schema of replacement says:
Suppose P is a definable binary relation (which may be a proper class) such that...
What exactly does “definable” mean in this context?
I am asking precisely because I want to play a “this is not a set in my model of set theory” card with regards to the set of all sets in the infinite descending chain, and I am not sure whether there is or isn’t a loophole I could use.
I think you may be misunderstanding the definition in the Wikipedia article, which is understandable because the formulation there (which is the usual one) is rather hard to get one’s head around.
So, this version of the axiom of foundation says: every nonempty set has an element disjoint from it. What does that have to do with infinite descending chains of sets? Well, suppose you have a containing b containing c, etc.; consider the set A = {a,b,c,...}. I claim that every element of this set fails to be disjoint from A. Consider, e.g., f. This contains g, and so does A, so f and A are not disjoint.
The usual way of formulating the replacement schema goes like this. Suppose you’ve got a formula P in the language of set theory, which has free variables x, y, A, and some finite number of w1,...,wn. (The idea is that this is going to specify a function on A, parameterized by w1,...,wn, with P meaning that the function maps x to y.) Replacement says that for each such formula the following holds: “For all A, w1, …, wn, if for all x in A there is a unique y such that P, then there is a set Y such that for all x in A there is a y in Y such that P”. In other words: if P describes a function on the set A, there’s an actual set Y containing the values of that function.
What if I am an asshole mathematician, and I insist that there is no set {a,b,c...}?
I mean, I know that it exists, you know that it exists, but I propose a model of set theory where individual sets a, b, c… exist, but there is no set {a, b, c...}. Can you prove there is one?
Wikipedia—if I understand it correctly—assumes that there is a function 0->a, 1->b, 2->c… and uses axiom of replacement. Following my passive-aggressive strategy, I insist that there is no such function in my model either. (We see the function from outside, but there is no in-universe set {{0, a}, {1, b}, {2, c}… }.) Can you construct one, if you are only given the sets in the chain with no “metadata”?
Specifically: V(0) = an infinite descending chain {a, b, c...} such that a ∋ b ∋ c...; V(1) = everything you can directly build from V(0) using ZF axioms (all subsets, pairs, unions, replacements...); and so on, as usual.
This is probably my Dunning–Kruger moment, but I’d appreciate if you’d play along. Does this attempt to create a set universe contradict the ZF axioms somewhere (where exactly?), or is this a model containing an infinitely descending chain where the axiom of foundation is still technically true.
(Meta: I probably have no further questions. Except perhaps if I wouldn’t understand some part of your answer. This is the most complicated thing I was able to think about set theory so far.)
EDIT:
Tried to do my homework, here is the part where I got stuck:
Imagine that you have a countably infinite amount of ur-elements. Is it possible to have a universe that would be a model of ZF with these ur-elements, but wouldn’t contain a “set of all ur-elements” as a set?
(For example, because it would only contain sets that contain, however indirectly, only a finite number of the ur-elements. That is, the set of ur-elements is countable from our perspective, but not necessarily in-universe. As Wikipedia says: “there is no absolute notion of countability”.)
Because if that is impossible, then if you take the infinite descending chain a, b, c..., the universe would also have to contain the set {a, b, c...}, which contradicts the Axiom of Foundation. So, what I tried, is really impossible.
On the other hand, if it is possible, then let’s take the model of ZF that contains the ur-elements a, b, c..., but does not contain the set {a, b, c...}, and now replace the ur-elements with sets from the infinitely descending chain. I believe that what you would get after replacement, would satisfy all the ZF axioms. (I could try to prove it, but if the answer to the question above is “no”, it would probably be a waste of time.)
My other comment was written before I saw your edit. Here are some remarks on the edited bit.
ZF can’t have urelements, if that means objects that aren’t sets but can be elements of sets; everything in ZF is a set. Nor if it means things other than “the empty set” that have no elements; the axiom of extensionality means that anything with no elements is the empty set.
It’s possible that you mean something else by “urelements”, but rather than trying to guess what I’ll let you tell me.
Yes, by “urelements” I meant “elements that are not sets”. However, this was just a different way to express the question that if we treat a, b, c… as black boxes, i.e. ignoring the question of what is inside them; whether from the facts that a is a set, b is a set, c is a set… inevitably follows that {a, b, c...} is also a set in given model of set theory.
For example, using the axiom of Pair, we can prove that {a, b} or {g, x} are also sets. Using also the axiom of Union, we can prove that any finite collections, such as {a, b, g, h, x} are sets. But using Pair and Union alone, we cannot prove that about any infinite collection.
The axiom of Infinity only proves the existence of one specific infinite set: ω.
My guess is that using all axioms together, we can only prove existence of sets that involve a finite number of a, b, c...
This needs to be put a bit more precisely, though, because by the very fact of them being an infinitely descending chain, including one of them means involving an infinite amount of them; for example {{}, c} is simultaneously {{}, {d}}, which is also {{}, {{e}}}, etc. But suppose that we stop expanding the expression whenever we hit one of these chain-sets. Thus, the set {{}, c} would involve cexplicitly, but d, e… only implicitly.
Now we can rephrase my guess, that using all axioms together, we can only prove existence of sets that involve explicitly a finite number of a, b, c… That means, {a, b, c...} is not among them.
Axioms of Existence, Infinity, Pair, and Powerset do not increase the number of a, b, c… used explicitly. Axioms of Union, Comprehension can only “unpack” these sets one level deeper. -- Union applied to c would only prove the existence of d. Comprehension applied to c would result in either empty set or d. Each of them applied to a structure that explicitly contains a finite number of a, b, c… would result in a structure that also explicitly contains a finite, albeit maybe twice larger, number of a, b, c...
Axiom of Foundation should be okay with such structures, because even if you have a set containing finitely many of a, b, c..., just choose the one furthest down the chain. For example, for set {{{}}, a, c, e}, choose e = {f}, and the intersection is empty.
How to construct the universe: Take a ZF universe, and for each set in that universe also create all possible sets that replace some of the empty sets in the structure by sets a, b, c… under condition that during each replacement you only use a finite number of a, b, c… (but with each of the selected finite few, you can replace an arbitrary, finite or infinite, number of the empty sets in the original structure). My hypothesis is that this, together with the rules a = {b}, b = {c}… is also a ZF universe, and it is one containing an infinite descending chain of sets (a, b, c...).
You can only ever prove something about specific objects about which nothing else is known if there are only finitely many of them. Because, in the situation you describe, the only way we can say anything about them is by naming them explicitly, and any proof has only finitely many symbols in it.
But e.g. the following could be true in some system: if you have a set that (“really”) begins an infinite descending chain, you can prove the existence of a set containing all the elements in some infinite descending chain, even though you couldn’t do it if all you had were names for those elements and no information about the relationships between them.
I don’t think I understand your prescription for a Strange Universe. What do you mean by “all possible sets that replace some of the empty sets in the structure”? In an actual ZF universe there is only one empty set.
What do you mean by “all possible sets that replace some of the empty sets in the structure”? In an actual ZF universe there is only one empty set.
I meant empty sets in the, uhm, description/graph of the set. For example, in the description of the set {{}, {{}}}, there are two instances of empty set: here {{}, {{}}}, and here {{}, {{}}}.
Visually, if you would draw the set as a tree—the topmost node represents the set itself, each node has below it nodes representing the elements (if two nodes have the same set for a subset, it would be drawn on the diagram below each node separately; we are making a tree structure that only splits downwards, never joins) -- then every set in standard model of ZF is a tree, with some nodes having infinitely many nodes directly below them, but each individual path downwards is finite. And each path downwards ends with the empty set = a node that does not have any nodes below it.
By “replacing some empty sets” I meant replacing some of those nodes at the bottom with the sets from the infinite chain a, b, c… For example, from the set {{}, {{}}}, you would achieve {a, {{}}} by replacing the “first” empty set; {{}, {a}} by replacing the “second” empty set; and {a, {a}} by replacing both empty sets. You would do this for all sets from the chain, even combinations like {a, {b}}.
The only restriction is that when the graph of the set contains an infinite number of empty-set-nodes, for example in {{}, {{}}, {{{}}}, {{{{}}}}… }, you can replace either finite or infinite number of them, but you may only use a finite number of different sets from the descending chain. So for example, you could infinitely many “a”s and infinitely many “b”s; or perhaps two “a”s and infinitely many “b”s; or just two “a”s and three “b”s; but you cannot use infinitely many different letters. So it is allowed to create {{}, {a}, {{{}}}, {{{b}}}… }, or {a, {a}, {{a}}, {{{a}}}… } (with infinitely many “a”s), or {a, {b}, {{a}}, {{{b}}}… } (with infinitely many “a”s and “b”s), but not {a, {b}, {{c}}, {{{d}}}… } (with infinitely many different letters used).
I think that the class of sets created this way satisfies the ZF axioms.
EDIT: I will try to send you an e-mail during this weekend, because this definitely makes more sense with pictures, at least in my head. Thank you for your patience so far!
Ah, OK, I think I now understand your intended construction. I’m trying to figure out whether it satisfies the ZF axioms, but right now it’s past my bedtime. One thing that definitely isn’t true is the following stronger version of “if Foundation holds in the original model then it also holds in your universe”: Write x->y to mean that set y in your universe is obtained from set x in some other model of ZF by “replacing some empty sets with things from a,b,...”). Then (I repeat: this is a thing that isn’t true): If x has an element w disjoint from x, and x->y, and when x->y w turns into z, then z is disjoint from y. So if Foundation is true in your universe it’s not for the very most obvious reason. (Counterexample to that stronger claim: let x = {{{}}, {{},t}} for some choice of t, let w = {{},t}; let y = {{b}, {a,t}} so that w → {a,t}; then although w is disjoint from x it isn’t true that z is disjoint from y, because both contain a={b}.
I’ll return to this tomorrow, if I find the time, and think some more about whether your universe is guaranteed to be a model of ZF...
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
In this context, by “monstrosity” I meant some technically-a-set that couldn’t be constructed. Like the a = {a} example, only I suspect that you can create an example of that type that would technically satisfy the Axiom of Foundation. I am probably wrong here, but I need to think about this more until I see how exactly I am wrong. (That will probably take weeks.)
But that “every connected directed graph is a picture of a set” is the kind of perspective I have in mind. Being given the entire structure of sets, interconnected, at once. Now the question is whether the structure can be designed in a way that technically satisfies the Axiom of Foundation, while somehow is not construable, e.g. because it is infinite in both directions. So no set contains itself, directly nor indirectly, it’s just a line of sets infinite in both directions, like the integers, where each set contains an infinite amount of “previous” sets; and probably also something else, to make the Axiom of Foundation happy. -- Chances are, I am just talking complete nonsense here.
One formulation of the axiom of foundation says that there’s no “infinite descending chain” of sets. That is, if you have a set a with an element b with an element c with …, then this cannot go on for ever. So at least some versions of your “infinite in both directions” are precisely ruled out by Foundation.
The version from Wikipedia seems okay with an infinite descending chain of sets, as long as each of them would also contain e.g. an empty set. In combination with other axioms, though… well, that’s where my knowledge of set theory becomes insufficient.
Okay, here is one of the questions I couldn’t answer by self-study; let me use this opportunity to ask you: The Wikipedia page on Axiom schema of replacement says:
What exactly does “definable” mean in this context?
I am asking precisely because I want to play a “this is not a set in my model of set theory” card with regards to the set of all sets in the infinite descending chain, and I am not sure whether there is or isn’t a loophole I could use.
I think you may be misunderstanding the definition in the Wikipedia article, which is understandable because the formulation there (which is the usual one) is rather hard to get one’s head around.
So, this version of the axiom of foundation says: every nonempty set has an element disjoint from it. What does that have to do with infinite descending chains of sets? Well, suppose you have a containing b containing c, etc.; consider the set A = {a,b,c,...}. I claim that every element of this set fails to be disjoint from A. Consider, e.g., f. This contains g, and so does A, so f and A are not disjoint.
The usual way of formulating the replacement schema goes like this. Suppose you’ve got a formula P in the language of set theory, which has free variables x, y, A, and some finite number of w1,...,wn. (The idea is that this is going to specify a function on A, parameterized by w1,...,wn, with P meaning that the function maps x to y.) Replacement says that for each such formula the following holds: “For all A, w1, …, wn, if for all x in A there is a unique y such that P, then there is a set Y such that for all x in A there is a y in Y such that P”. In other words: if P describes a function on the set A, there’s an actual set Y containing the values of that function.
What if I am an asshole mathematician, and I insist that there is no set {a,b,c...}?
I mean, I know that it exists, you know that it exists, but I propose a model of set theory where individual sets a, b, c… exist, but there is no set {a, b, c...}. Can you prove there is one?
Wikipedia—if I understand it correctly—assumes that there is a function 0->a, 1->b, 2->c… and uses axiom of replacement. Following my passive-aggressive strategy, I insist that there is no such function in my model either. (We see the function from outside, but there is no in-universe set {{0, a}, {1, b}, {2, c}… }.) Can you construct one, if you are only given the sets in the chain with no “metadata”?
Specifically: V(0) = an infinite descending chain {a, b, c...} such that a ∋ b ∋ c...; V(1) = everything you can directly build from V(0) using ZF axioms (all subsets, pairs, unions, replacements...); and so on, as usual.
This is probably my Dunning–Kruger moment, but I’d appreciate if you’d play along. Does this attempt to create a set universe contradict the ZF axioms somewhere (where exactly?), or is this a model containing an infinitely descending chain where the axiom of foundation is still technically true.
(Meta: I probably have no further questions. Except perhaps if I wouldn’t understand some part of your answer. This is the most complicated thing I was able to think about set theory so far.)
EDIT:
Tried to do my homework, here is the part where I got stuck:
Imagine that you have a countably infinite amount of ur-elements. Is it possible to have a universe that would be a model of ZF with these ur-elements, but wouldn’t contain a “set of all ur-elements” as a set?
(For example, because it would only contain sets that contain, however indirectly, only a finite number of the ur-elements. That is, the set of ur-elements is countable from our perspective, but not necessarily in-universe. As Wikipedia says: “there is no absolute notion of countability”.)
Because if that is impossible, then if you take the infinite descending chain a, b, c..., the universe would also have to contain the set {a, b, c...}, which contradicts the Axiom of Foundation. So, what I tried, is really impossible.
On the other hand, if it is possible, then let’s take the model of ZF that contains the ur-elements a, b, c..., but does not contain the set {a, b, c...}, and now replace the ur-elements with sets from the infinitely descending chain. I believe that what you would get after replacement, would satisfy all the ZF axioms. (I could try to prove it, but if the answer to the question above is “no”, it would probably be a waste of time.)
My other comment was written before I saw your edit. Here are some remarks on the edited bit.
ZF can’t have urelements, if that means objects that aren’t sets but can be elements of sets; everything in ZF is a set. Nor if it means things other than “the empty set” that have no elements; the axiom of extensionality means that anything with no elements is the empty set.
It’s possible that you mean something else by “urelements”, but rather than trying to guess what I’ll let you tell me.
Yes, by “urelements” I meant “elements that are not sets”. However, this was just a different way to express the question that if we treat a, b, c… as black boxes, i.e. ignoring the question of what is inside them; whether from the facts that a is a set, b is a set, c is a set… inevitably follows that {a, b, c...} is also a set in given model of set theory.
For example, using the axiom of Pair, we can prove that {a, b} or {g, x} are also sets. Using also the axiom of Union, we can prove that any finite collections, such as {a, b, g, h, x} are sets. But using Pair and Union alone, we cannot prove that about any infinite collection.
The axiom of Infinity only proves the existence of one specific infinite set: ω.
My guess is that using all axioms together, we can only prove existence of sets that involve a finite number of a, b, c...
This needs to be put a bit more precisely, though, because by the very fact of them being an infinitely descending chain, including one of them means involving an infinite amount of them; for example {{}, c} is simultaneously {{}, {d}}, which is also {{}, {{e}}}, etc. But suppose that we stop expanding the expression whenever we hit one of these chain-sets. Thus, the set {{}, c} would involve c explicitly, but d, e… only implicitly.
Now we can rephrase my guess, that using all axioms together, we can only prove existence of sets that involve explicitly a finite number of a, b, c… That means, {a, b, c...} is not among them.
Axioms of Existence, Infinity, Pair, and Powerset do not increase the number of a, b, c… used explicitly. Axioms of Union, Comprehension can only “unpack” these sets one level deeper. -- Union applied to c would only prove the existence of d. Comprehension applied to c would result in either empty set or d. Each of them applied to a structure that explicitly contains a finite number of a, b, c… would result in a structure that also explicitly contains a finite, albeit maybe twice larger, number of a, b, c...
Axiom of Foundation should be okay with such structures, because even if you have a set containing finitely many of a, b, c..., just choose the one furthest down the chain. For example, for set {{{}}, a, c, e}, choose e = {f}, and the intersection is empty.
How to construct the universe: Take a ZF universe, and for each set in that universe also create all possible sets that replace some of the empty sets in the structure by sets a, b, c… under condition that during each replacement you only use a finite number of a, b, c… (but with each of the selected finite few, you can replace an arbitrary, finite or infinite, number of the empty sets in the original structure). My hypothesis is that this, together with the rules a = {b}, b = {c}… is also a ZF universe, and it is one containing an infinite descending chain of sets (a, b, c...).
You can only ever prove something about specific objects about which nothing else is known if there are only finitely many of them. Because, in the situation you describe, the only way we can say anything about them is by naming them explicitly, and any proof has only finitely many symbols in it.
But e.g. the following could be true in some system: if you have a set that (“really”) begins an infinite descending chain, you can prove the existence of a set containing all the elements in some infinite descending chain, even though you couldn’t do it if all you had were names for those elements and no information about the relationships between them.
I don’t think I understand your prescription for a Strange Universe. What do you mean by “all possible sets that replace some of the empty sets in the structure”? In an actual ZF universe there is only one empty set.
I meant empty sets in the, uhm, description/graph of the set. For example, in the description of the set {{}, {{}}}, there are two instances of empty set: here {{}, {{}}}, and here {{}, {{}}}.
Visually, if you would draw the set as a tree—the topmost node represents the set itself, each node has below it nodes representing the elements (if two nodes have the same set for a subset, it would be drawn on the diagram below each node separately; we are making a tree structure that only splits downwards, never joins) -- then every set in standard model of ZF is a tree, with some nodes having infinitely many nodes directly below them, but each individual path downwards is finite. And each path downwards ends with the empty set = a node that does not have any nodes below it.
By “replacing some empty sets” I meant replacing some of those nodes at the bottom with the sets from the infinite chain a, b, c… For example, from the set {{}, {{}}}, you would achieve {a, {{}}} by replacing the “first” empty set; {{}, {a}} by replacing the “second” empty set; and {a, {a}} by replacing both empty sets. You would do this for all sets from the chain, even combinations like {a, {b}}.
The only restriction is that when the graph of the set contains an infinite number of empty-set-nodes, for example in {{}, {{}}, {{{}}}, {{{{}}}}… }, you can replace either finite or infinite number of them, but you may only use a finite number of different sets from the descending chain. So for example, you could infinitely many “a”s and infinitely many “b”s; or perhaps two “a”s and infinitely many “b”s; or just two “a”s and three “b”s; but you cannot use infinitely many different letters. So it is allowed to create {{}, {a}, {{{}}}, {{{b}}}… }, or {a, {a}, {{a}}, {{{a}}}… } (with infinitely many “a”s), or {a, {b}, {{a}}, {{{b}}}… } (with infinitely many “a”s and “b”s), but not {a, {b}, {{c}}, {{{d}}}… } (with infinitely many different letters used).
I think that the class of sets created this way satisfies the ZF axioms.
EDIT: I will try to send you an e-mail during this weekend, because this definitely makes more sense with pictures, at least in my head. Thank you for your patience so far!
Ah, OK, I think I now understand your intended construction. I’m trying to figure out whether it satisfies the ZF axioms, but right now it’s past my bedtime. One thing that definitely isn’t true is the following stronger version of “if Foundation holds in the original model then it also holds in your universe”: Write x->y to mean that set y in your universe is obtained from set x in some other model of ZF by “replacing some empty sets with things from a,b,...”). Then (I repeat: this is a thing that isn’t true): If x has an element w disjoint from x, and x->y, and when x->y w turns into z, then z is disjoint from y. So if Foundation is true in your universe it’s not for the very most obvious reason. (Counterexample to that stronger claim: let x = {{{}}, {{},t}} for some choice of t, let w = {{},t}; let y = {{b}, {a,t}} so that w → {a,t}; then although w is disjoint from x it isn’t true that z is disjoint from y, because both contain a={b}.
I’ll return to this tomorrow, if I find the time, and think some more about whether your universe is guaranteed to be a model of ZF...
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