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...
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...