What does this have to do with philosophy? It's pure Math. — Alkis Piskas

No mathematician will ever tell you that discussing whether abstract objects, like mathematical objects, really exist in the world or are just contained within minds is a mathematical topic: this is what a substantive portion of this discussion led into, so this is more accurately construed as philosophy of mathematics informed by insights from mathematics (it's very difficult to do philosophy of any x without pretty extensive detail into x itself).

The point that you seem to be missing is that it's a simply a matter of definition -and definition alone- that powersets don't contain "all subsets" of the "original set". The original set IS the "set of all that exists". To conclude that the original set does not exist is nonsensical. It is borne of a failure of conceptual understanding on your part. — ThinkOfOne

Powersets do contain all subsets of their original set, this is a well-proven theorem by Cantor that any elementary introduction to set theory should teach you. The set of all exists, by its very definition, includes the cardinality of a set strictly larger than it is, and is therefore incoherent/a contradiction (the proof of this is a very trivial exercise: Suppose E, then there exists P(E), P(E) is cardinally larger than E, therefore there exists x's that are members of P(E) and not E and thus E isn't E).

This does not mean the things that exist, like my keyboard or this screen, do not actually exist, but rather they cannot all be collected into set. If your standards of "conceptual understanding" is mathematical inconsistency, this is a problem on your end.

I address something like that here. The set of both metals weighs 3g but none of its members weigh 3g. It doesn't then follow that we should treat the existence of this set as being additional to the existence of each of its members, else the total weight of things which exist would be 6g, which is false in this example. — Michael

This is a category error namely in that sets never weigh anything (you're confusing mereological complexes with sets, a confusion many participants of this discussion took on and continued to presume despite my clarification of the distinction in an earlier post right here) Sets lack the physical properties of mass or weight, even if all their members have this property (hint: they're distinct from their members).

What you're looking for is either 1. plural quantification (in plural logic) or 2. mereological composition - both of these are distinct formalisms that are not isomorphic to our set theoretic apparatus. In plural quantification, you are quantifying over nothing over and above the x's, and not any notion of collection they form treated as a single. In mereological composition, you're quantifying over parts bearing a particular relation and arrangement, often proper-parthood, to some whole, and in that venue it's under dispute whether the wholes are identical to just the collection of their parts (this is not at all isomorphic to whether sets are identical to their members, whose answer is a trivial no)

That said, mereological wholes may have physical properties: for instance, an apple is a complex composed of its atoms, and it has weigh, though I'm not sure if weigh would also be possessed by its super subatomic particles, I suppose that debate would be fleshed out more there

Plural logic would be the most natural formalism of asking about, say, the weigh of a collection which all have physical properties, because the plural quantifier quantifiers over nothing aside those x's in the collection itself and would simply be a summation of their properties. When comparing these collective relations, the plural ≺ would be the "less committal", the mereological P being in the middle and the set theoretic ∈ being the most committal.

Apples can be members of sets but are never themselves sets. There are no instances at all where a physical object like an apple is a set, especially if it's just because it has parts.

The reason you and some of the others are getting mixed up here on this non-issue is that you're employing the inappropriate apparatus (set theory instead of mereology) to think of this whole-parts identity problem, but you're thinking about it with presumptions that are already at fault, like supposing of complex objects as literally being sets or that they themselves have physical properties like weight which are both incoherent.

This is a category error namely in that sets never weigh anything — Kuro

I addressed that when I first brought up this example. If you’re saying that the set exists, in addition to its members, then you’re presumably saying that the set exists as some abstract thing, à la Platonism. That’s a view I take issue with.

I should add that I wasn't using the word "set" in the mathematical sense here. I moved beyond maths and was considering physical collections in response to litewave's comments. A collection of two coins has a weight (the sum of each coin). My argument was that even though a collection of two coins is conceptually distinct from each of its individual coins it is wrong to say that three ontologically distinct things exist (the one coin, the other coin, and the collection).

When I meet a married couple I don’t meet a married couple and the husband and the wife. Meeting the married couple is meeting the husband and the wife, and vice versa. The married couple isn’t an entity that’s additional to the husband and the wife, even though there are things we can say about the married couple that we can’t say about the husband or the wife individually.

If you try to say that the married couple and the husband and the wife all exist, and so 3 things exist, you’re counting the husband and the wife twice (or rather, 1.5 times each). — Michael

I've read your other (later) post prior to this one, but coincidentally it indulges in the same error that I pointed out in my response to your later post: you default to the idea that all sorts of plural quantification are identical with the sets corresponding to the plurality being quantified.

The set of the married couple {husband, wife} is not identical to the plurality of the husband and wife, nor is the arbitrarily infinitely many sets including the husband and wife (i.e. {{husband}, wife} identical with each other nor that first plurality.

Physical things, be they collections or singular, are never themselves sets: they just can join sets. By talking about a set of objects where that object is a member of the set, you've not counted the object again. It isn't the case that the wife exists infinitely many times because there are infinitely many sets that contain the wife.

OK, as you like. :smile:

OK.

The point that you seem to be missing is that it's a simply a matter of definition -and definition alone- that powersets don't contain "all subsets" of the "original set". The original set IS the "set of all that exists". To conclude that the original set does not exist is nonsensical. It is borne of a failure of conceptual understanding on your part.
— ThinkOfOne

Powersets do contain all subsets of their original set, this is a well-proven theorem by Cantor that any elementary introduction to set theory should teach you. The set of all exists, by its very definition, includes the cardinality of a set strictly larger than it is, and is therefore incoherent/a contradiction (the proof of this is a very trivial exercise: Suppose E, then there exists P(E), P(E) is cardinally larger than E, therefore there exists x's that are members of P(E) and not E and thus E isn't E).

This does not mean the things that exist, like my keyboard or this screen, do not actually exist, but rather they cannot all be collected into set. If your standards of "conceptual understanding" is mathematical inconsistency, this is a problem on your end. — Kuro

You seem to have missed the point. There is a distinction that needs to be made between the definition of a SET and the definition of a POWERSET. They are not one and the same.

There is a SET of "all that exists" and there is POWERSET of "all that exists".

Perhaps a thought experiment will help.
Let's say that there is a universe with a SET called Tegwar.
Tegwar contains two members {x, y}.
The POWERSET of Tegwar contains "all subsets, empty set and the original set itself".
If the only things that exist in this universe are x and y:
Does Tegwar contain "all that exists" in the universe?
Does the POWERSET of Tegwar contain "all that exists" + all subsets + empty set?

As I posted earlier:

All that needs to be decided is whether or not to allow "all subsets" of "all that exists" to be members of "all that exists".

Either way:
1. There is a set of "all that exists"
2. There is a powerset for "all that exists"

But either way, both both the SET and the POWERSET exist. Whether the SET is called Tegwar or "all that exists", the result is the same.

I'll add that if "all subsets" are allowed to be members, then the set is infinite.
If "all subsets" are not allowed to be members, then the set is finite.

But the cardinality of P(E) can only be greater than E's if there exists elements in P(E) that are not members of E — Kuro

This, I think, shows a more fundamental problem. You appear to equivocate. When you say that there exist elements in P(E) that are not members of E you're actually just saying that there are members of P(E) that are not members of E. But that something is a member of a set isn't that it exists. For example, Santa doesn't exist and so isn't a member of E, but it is a member of the set {Santa}.

If E is {John, Jane} then P(E) is {{}, {John}, {Jane}, {John, Jane}}. No member of P(E) is a member of E and so no member of P(E) exists. Therefore, that P(E) has a greater cardinality than E doesn't entail that E is not the set of all that exists. Whether you consider E or P(E), the only things which exist are John and Jane.

The one issue I see with this line of reasoning is that if sets exist then if something is a member of P(E) then it must be a member of E, which given the fact that P(E) has a greater cardinality than E entails a contradiction. So do sets exist, and if so, in what sense? Platonism?

You seem to have missed the point. There is a distinction that needs to be made between the definition of a SET and the definition of a POWERSET. They are not one and the same. — ThinkOfOne

There's no point I missed: no where in any of my entries I equivocated sets with powersets. Tegwar does not contain all that exists in that universe, namely because the set cannot contain either itself or the powerset (and its powerset, ad infinitum). The notion of a set of all that exists is not possible. Similarly, you cannot exhaust all that exists in any universe obeying set-theoretic principles: some universe where all that exists is some ball in space denoted by a letter is a figment of the imagination.

But that something is a member of a set isn't that it exists. For example, Santa doesn't exist and so isn't a member of E, but it is a member of the set {Santa}. — Michael

Sets do not have meontological members, because set-membership itself is a relation requiring that there are two relata of the set and the given member, yet the necessary condition can't be satisfied when one of the relata quite literally isn't there. Since Santa does not exist, {Santa} as a set doesn't exist in the real world (though there are possible, hypothetical universes out there where Santa does exist, and thus the singleton exists as well).

There actually is one set with no members, but it is a unique set. This is called the empty set, uniquely satisfying that ∀x x∉S. This does not mean that some spooky metaphysical concept of nothingness/nonexistence/emptiness/whatever is itself a member of the set, rather, literally that the set has no members.

No equivocation at all between "is a member of some set" and "exists", it's not a matter of conflating the concepts rather simply a matter of logical entailment. It's incoherent (and inconsistent) for anything to be a member of a set but also simultaneously not exist.

No equivocation at all between "is a member of some set" and "exists", it's not a matter of conflating the concepts rather simply a matter of logical entailment. It's incoherent (and inconsistent) for anything to be a member of a set but also simultaneously not exist. — Kuro

The "existence" of mathematical objects in mathematical anti-realism is different to the "existence" of mathematical objects in mathematical realism. I took the "set of all that exists" as referring to existence in the realist sense. If this is correct, and if mathematical anti-realism is true, then no member of the power set exists in the realist sense (every member of the power set is a set and sets don't exist in the realist sense), and so that the power set has a greater cardinality is not a proof that there isn't a set of all that exists.

So could you clarify what you mean by "exists" when you consider the set of all that exists, and whether or not you're arguing for mathematical realism.

The "existence" of mathematical objects in mathematical anti-realism is different to the "existence" of mathematical objects in mathematical realism. — Michael

Correct, hence why platonism and nominalism about mathematics here is far-reaching and beyond the closer phenomenon at hand, being just that universal set itself. Surely any ordinary set, even a coherent one like {1, 2, 3} does not exist independently for a nominalist about abstract objects (though, they might still insist it exists as a concept or, for Field, as a fiction abstracted from concrete reality) though it will for the platonist. This is not to say that there's some different theory of existence necessarily being employed by the platonist or the nominalist (of course, there can), but the nominalist and platonist can perfectly disagree in using the same sense of existence (say, as a second-order predicate of concepts a la Frege, or as an instantiation of properties a la Russell)

There's one thing that the platonist and nominalist would still agree on, in that contradictory sets, like the Russell set, or this universal set, do not exist because they're incoherent (and so would their existence). Certainly the nominalist needs not raise the issue of whether any sets exist at all to just say that this one set does not exist, which is the first point I made in this post: the fact that this universal set, the set of all that exists, is contradictory.

The "existence" of mathematical objects in mathematical anti-realism is different to the "existence" of mathematical objects in mathematical realism. — Michael

Correct, hence why platonism and nominalism about mathematics here is far-reaching and beyond the closer phenomenon at hand, being just that universal set itself. — Kuro

This is why I think you need to clarify your argument. Is the set of all that exists the set of all that realist-exists or the set of all that antirealist-exists? Because if it's the former, and if mathematical anti-realism is true, then your set of all that realist-exists isn't a universal set, as the universal set contains members that don't realist-exist.

To make my point clearer, assume physicalism and mathematical anti-realism. Everything that exists is a physical object. Therefore, the set of all that exists is the set of every physical object. That its power set has a greater cardinality doesn't entail that there are physical objects that are not in the set of every physical object, because no member of the power set is a physical object.

. Everything that exists is a physical object. The set of all that exists is the set of every physical object. — Michael

If you assume physicalism, the set of all that exists, let alone the set of anything, since sets are not physical objects neither identical to their physical members nor the collection of their physical members (the proof of this is simple: suppose it is the case, then submerge that same set under a further set, which is mathematically non-identical!)

So physicalism just entails mathematical anti-realism, and it goes back to what I said earlier here:

There's one thing that the platonist and nominalist would still agree on, in that contradictory sets, like the Russell set, or this universal set, do not exist because they're incoherent (and so would their existence). Certainly the nominalist needs not raise the issue of whether any sets exist at all to just say that this one set does not exist — Kuro

The fixing to what sense of existence is unnecessary, namely because of the above, and further because the platonist/nominalist can perfectly disagree while holding to the same theory of existence (though they can differ if they want to, obviously), as I clarified earlier as well:

This is not to say that there's some different theory of existence necessarily being employed by the platonist or the nominalist (of course, there can), but the nominalist and platonist can perfectly disagree in using the same sense of existence (say, as a second-order predicate of concepts a la Frege, or as an instantiation of properties a la Russell) — Kuro

Alas, I think your request for clarity was preemptively satisfied unless you hold to the presumption that realists and anti-realists must be using different senses or theories of existence. I don't take you to be claiming this, since you didn't object to it when I asserted its negation, but just on the possibility this is a claim of dispute it can be falsified using a trivial example (i.e. Quine vs Field), a pair of platonist / nominalist(fictionalist) who disagree on the existence of mathematical objects while using 'existence' in the same sense. Obviously though, the reverse claim of them having to use the same sense is not true either, as there are indeed platonists/nominalists who disagree while using different senses, examples being along the lines of Meinong

If you assume physicalism, the set of all that exists, let alone the set of anything, since sets are not physical objects neither identical to their physical members nor the collection of their physical members (the proof of this is simple: suppose it is the case, then submerge that same set under a further set, which is mathematically non-identical!) — Kuro

I don't understand how this addresses my argument. Can you specify which step you disagree with?

1. Physicalism is true (assumption)
2. Everything that exists is a physical object (from 1)
3. The set of all that exists is the set of all physical objects (from 2)
4. The power set has more members than the set of all that exists
5. No member of the power set exists (from 2)
6. Therefore, that the power set has more members than the set of all that exists does not prove that some things exist which are not in the set of all that exists (from 5)

I don't understand how this addresses my argument. Can you specify which step you disagree with?

1. Physicalism is true (assumption)
2. Everything that exists is a physical object (from 1)
3. The set of all that exists is the set of all physical objects (from 2) — Michael

(1) entails that no sets exist, including that set in (3) regardless of its incoherent status. It could be any ordinary set, like a set of an apple, someone's toenail & an ant. A set whose members are physical objects is not itself, as a set, physical (for obvious reasons: it'd entail infinite interpenetration)

The other issue is that, obviously, this still instantiates the same contradiction (4-5) in my initial argument: the powerset is either not larger than its set because there aren't more members in it than in the set, falsifying either the powerset axiom or this set's status as a powerset, or there is no set of all that exists.

FWIW, this is technically not a valid argument since 6 seems to only follow from (what I assume are crudely skipped steps for brevity's sake) the assumption in 5, that the powerset is literally empty, which not only is an issue in the terms I explained earlier (falsifying the powerset axiom or just giving up the nonexistence of that set), this still never means that the powerset is empty. That falsifies the axioms we use for set building, namely in that we can join any urelement or set into a further set containing just that set or urelement (being a singleton), but if the members of that set can't be joined into singleton supersets, let alone the powerset itself, then we've falsified several of our basic set theoretic apparatus just to suppose the existence of this set (which still manages to be incoherent, regardless: this doesn't actually make a powerset of a non-empty set empty!)

(1) entails that no sets exist, including that set in (3) — Kuro

I think there's a difference between saying "there is a set of all that exists" and saying "the set of all that exists, exists". The mathematical anti-realist will assert the former but reject the latter.

the assumption in 5, that the powerset is literally empty — Kuro

I didn't say that it's empty. Similar to the above, there's a difference between saying "the set has members" and saying "the members of this set exist". The mathematical anti-realist will assert the former but reject the latter.

So, the power set isn't empty, but as all of its members are sets, and as sets don't exist, none of its members exist. As such, it doesn't follow from the fact that the power set has more members that there exist things which aren't in the set of all that exists.

As I said before, I think you're equivocating on the word "exists". Being a member of a set isn't the same thing as existing (if physicalism and mathematical anti-realism are true).

I think there's a difference between saying "there is a set of all that exists" and saying "the set of all that exists, exists". The mathematical anti-realist will assert the former but reject the latter. — Michael

Incorrect. This is a Meinongian there-is/exists distinction which has been proven inconsistent by Russell and largely abandoned ever since (I can elaborate on that further if you'd like), and like I said earlier, the presumption that an anti-realist is forced to use this outdated theory of existence is nonsense: anti-realists and platonists can perfectly disagree using the same theory of existence, as I said about 3-4 times earlier (and even gave examples, in case you don't believe me). Let's not be silly.

Furthermore, it's widely accepted by empirical linguistics that "is" has three senses, (1) predicative in the form of "x is an F", (2) identity in the sense of "x is y", and (3) existential in the sense of "there is x". This usage is also standard by logicians and mathematicians, in that the particular quantifier ∃ is understood as the existential quantifier and translated as 'there is' beyond the domain of heterodox systems like free & inclusive logics.

I didn't say that it's empty. Similar to the above, there's a difference between saying "the set has members" and saying "the members of this set exist". The mathematical anti-realist will assert the former but reject the latter. — Michael

Nonsense: I explained earlier why there's no such thing as "this set has members such that they do not exist". If you plan to assert negations of some of the claims I've formally elaborated on to explain them to you, you need to either substantiate them of a similar level or at the very least address what I said:

Sets do not have meontological members, because set-membership itself is a relation requiring that there are two relata of the set and the given member, yet the necessary condition can't be satisfied when one of the relata quite literally isn't there. Since Santa does not exist, {Santa} as a set doesn't exist in the real world (though there are possible, hypothetical universes out there where Santa does exist, and thus the singleton exists as well). — Kuro

So, the power set isn't empty, but as all of its members are sets, and as sets don't exist, none of its members exist. As such, it doesn't follow from the fact that the power set has more members that there exist things which aren't in the set of all that exists. — Michael

Nope! When we say "Pegasus does not exist", we're not referencing that there is a Pegasus such that it does not exist, as a Meinongian would, for that would be a contradiction, instead, we're quantifying over our most general domain to say that there is no x such that x is identical to Pegasus. It's a trivial inference in first-order logic, the language of set theory, to infer from being predicated to existing (to be predicated is just such that the constant is a member of some predicate F's extension, or that the extension of F satisfies the existential quantifier in that it has at least one member, and so on.) Deviant logics like free/inclusive logics obviously preclude this result, but these are not the languages which set theory is built on.

If you wish to protest these logical results in virtue of a distinct metaphysical theory (one that has been almost universally done away with and proven inconsistent several times), that alone requires substantial motivation on your part which you have not provided whatsoever.

As I said before, I think you're equivocating on the word "exists". Being a member of a set isn't the same thing as existing (if physicalism and mathematical anti-realism are true). — Michael

I've already explained that "being a member" and "existing" is not the same thing, in the same way that (1) "P -> Q, P" and (2) "Q" are not "the same thing", but (1) logically entails (2) in the same way the former logically entails the latter (being predicated of anything entails existing). Just in case you really don't believe me and don't want to take me at my word (and I hate to use this, especially for really basic inferences, but this is about the second time I had to do this on this site), you can run this inference in any truth tree generator of your choice:



Also, the issue with this being a strawman is because it's after I've already clarified to you that my position, (I hate to call this a 'position' since it's literally a well-understood and universally uncontroversial logical inference), is of logical entailment and not of identity between the antecedent (being a member) and the consequent (existing), I've spelled this out for you here:

No equivocation at all between "is a member of some set" and "exists", it's not a matter of conflating the concepts rather simply a matter of logical entailment — Kuro

So I find this quite bad-faith on your end, especially in the sense that I've been more than happy to steelman your arguments and fix various technical or mathematical issues in some of them where I eagerly addressed the corrected versions to make for a fruitful discussion.

Perhaps a clearer example:

1. Physicalism is true (assumption)
2. The set of all that physically exists is {apple, pear, ...}
3. The power set of this is {{}, {apple}, {pear}, {apple, pear}, ...}
4. No member of the power set physically exists
5. Therefore, the power set is not proof that there are things which physically exist and are not members of the set of all that physically exists

1. Physicalism is true (assumption)
2. The set of all that physically exists is {apple, pear, ...}
3. The power set of this is {{}, {apple}, {pear}, {apple, pear}, ...}
4. No member of the power set physically exists
5. Therefore, the power set is not proof that there are things which physically exist and are not members of the set of all that physically exists — Michael

(2) is no different than (3) or a contradictory Russell set or some ordinary {a, b, c} set. The members of the set in (2) physically exist, but the set itself doesn't per physicalism. Recall what I said earlier:

(1) entails that no sets exist, including that set in (3) regardless of its incoherent status. It could be any ordinary set, like a set of an apple, someone's toenail & an ant. A set whose members are physical objects is not itself, as a set, physical (for obvious reasons: it'd entail infinite interpenetration) — Kuro

In even asserting that the set is anything, like having the property of "containing apple as a member", you get back to existential commitments. There's not that much middle ground between accepting the full force of the claim that sets, even consistent ones, do not exist, because sets, even impure ones, are not themselves physical objects and physicalism entails that all that exists is physical.

Your alternative is a type of Quinean physicalism, though I'm not sure if you'd actually want to take a position like this. Quine's view was that mathematics was indispensable to the natural sciences, and his naturalism in that we should commit to whatever our best theories committed to (thus commit to mathematics). In that sense, Quine is a physicalist platonist.

The members of the set in (2) physically exist, but the set itself doesn't per physicalism. — Kuro

I'm aware. What is the relevance of that? I'm not saying that the set physically exists. I'm only saying that the power set doesn't prove that some things physically exist which are not in the set of all that physically exists.

In even asserting that the set is anything, like having the property of "containing apple as a member", you get back to existential commitments. — Kuro

Mathematical anti-realists and physicalists are quite capable of doing maths with sets.

Mathematical anti-realists and physicalists are quite capable of doing maths with sets. — Michael

Of course, so is Hartry Field, and so is my capability to pray while not assenting to "God exists". Reasoning about mathematics and believing in a philosophical claim are distinct.

I'm aware. What is the relevance of that? I'm not saying that the set physically exists. I'm only saying that the power set doesn't prove that some things physically exist which are not in the set of all that physically exists. — Michael

The set itself asserted by that premise doesn't exist. Not its powerset: just the set of all that physically exists.

The set itself asserted by that premise doesn't exist. — Kuro

So you're saying that if mathematical anti-realism is true then there is no set of all that exists, because there are no sets? And so your very argument, which uses sets, depends on mathematical realism being true?

So you're saying that if mathematical anti-realism is true then there is no set of all that exists, because there are no sets? And so your very argument, which uses sets, depends on mathematical realism being true? — Michael

No? I'm saying that the non-existence of the set of all that exists is an issue far prior to the philosophy of mathematics (namely because it's an issue provable in mathematics): the existence of the set is contradictory, so both platonists, who are realists about other sets, and nominalists, realists about no sets whatsoever, would agree it doesn't exist.

Arguing that it doesn't exist because no sets exist, while not invalid (in the sense that it follows), is an odd choice of argument because it relies on controversial philosophical premises in comparison to the well-accepted mathematical reasons for why the set of all that exists doesn't exist (which I explored in the OP).

By the way, surprise surprise, this was also all stated earlier:

There's one thing that the platonist and nominalist would still agree on, in that contradictory sets, like the Russell set, or this universal set, do not exist because they're incoherent (and so would their existence). Certainly the nominalist needs not raise the issue of whether any sets exist at all to just say that this one set does not exist, which is the first point I made in this post: the fact that this universal set, the set of all that exists, is contradictory. — Kuro

If you look back to my reply with AgentSmith, I explored the mathematical and philosophical considerations of prioritizing competing intuitions (the axioms of set theory, and the result that a universal set doesn't exist) based on this result.

While the issue of mathematical realism and nominalism is interesting in its own right, certainly even so in relation to this very topic, it's not at all necessary for just this theorem being the universal set not existing (hence why no mathematician cites it)

No? I'm saying that the non-existence of the set of all that exists is an issue far prior to the philosophy of mathematics (namely because it's an issue provable in mathematics): the existence of the set is contradictory, so both platonists, who are realists about other sets, and nominalists, realists about no sets whatsoever, would agree it doesn't exist. — Kuro

Where’s the contradiction here?

1. The set of all that physically exists is {apple, pear, ...}
2. The power set of this is {{}, {apple}, {pear}, {apple, pear}, ...}
3. No member of the power set physically exists
4. Therefore, the power set is not proof that there are things which physically exist and are not members of the set of all that physically exists — Michael

Your response before was just to say that if physicalism is true then there are no sets.

Where’s the contradiction here? — Michael

The set of all that exists is contradictory.

Your response before was just to say that if physicalism is true then there is no set of all that exists. — Michael

Indeed. No sets exist if physicalism is true.

The set of all that exists is contradictory. — Kuro

The set of all that physically exists isn’t contradictory (at least with respect to the power set), which is what that argument shows.

The set of all that physically exists isn’t contradictory, which is what that argument shows. — Michael

Nope. This has already been addressed:

The other issue is that, obviously, this still instantiates the same contradiction (4-5) in my initial argument: the powerset is either not larger than its set because there aren't more members in it than in the set, falsifying either the powerset axiom or this set's status as a powerset, or there is no set of all that exists.

FWIW, this is technically not a valid argument since 6 seems to only follow from (what I assume are crudely skipped steps for brevity's sake) the assumption in 5, that the powerset is literally empty, which not only is an issue in the terms I explained earlier (falsifying the powerset axiom or just giving up the nonexistence of that set), this still never means that the powerset is empty. That falsifies the axioms we use for set building, namely in that we can join any urelement or set into a further set containing just that set or urelement (being a singleton), but if the members of that set can't be joined into singleton supersets, let alone the powerset itself, then we've falsified several of our basic set theoretic apparatus just to suppose the existence of this set (which still manages to be incoherent, regardless: this doesn't actually make a powerset of a non-empty set empty!) — Kuro

This "physicalism-constrained set theory" fails not only the powerset axiom but basic set-builders like joining sets into supersets, subsets, unions, and so on (which would not all exist per physicalism, and in your interpretation, in the literal sense of not existing in the same way the powerset's members don't). So in the same way my initial argument shows why the set of all that exists is inconsistent with set-theoretic axioms but particularly the powerset axiom, yours only scores more axioms!

This is not even considering the empty set, which, per whatever this "physicalism-constrained set theory" is, either (1) wouldn't exist, as it's a set, going back to what I said earlier (2) you might say it exists, if it does, you can generate N with set-builders and suppose E, then ask P(E) - same thing as what I said earlier

This "physicalism-constrained set theory" fails — Kuro

I’m not taking about that though. Use normal set theory. The set of all that physically exists is not contradictory. It might not be, within set theory, the set of all that exists, but if physicalism is true then everything that actually exists is a member of the set of all that physically exists.

Anti-realists can pretend that sets exist for the sake of the maths to then talk about everything that actually exists.

I’m not taking about that though. Use normal set theory. The set of all that physically exists is not contradictory. It might not be, within set theory, the set of all that exists, but if physicalism is true then everything that actually exists is a member of the set of all that physically exists. — Michael

Using "normal set theory" (suppose, it is ZF), then your argument is not even tenable

This set would violate the pairing axiom by being subsumed through a superset as well as the foundation axiom by there being sets disjoint from A. The axiom of empty set would also be false. This is all very trivial!

I've suggested using a very nonstandard "physical-constrained set theory" just to make your argument somewhat tenable to be charitable to you: if you agree on using ZF, ZFC, or other set theories that count as 'normal' then this is not even a topic for debate whatsoever. The non-existence of the universal set in ZF, ZFC and so on is a well established mathematical theorem in those set theories.

The non-existence of the universal set in ZF, ZFC and so on is a well established mathematical theorem in those set theories. — Kuro

I’m not talking about the universal set. I’m talking about the set of all that physically exists. These are not the same thing.

So, in ZFC, why is the set of all that physically exists contradictory?