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Suppose naïve set theory, let's do a tought experiment:

Informally, let's define a set € such that € contains all the sets that don't contain themselves.(yes, all but not necessarily only those), let's also specify that € must have as minimal members as possible.

Now the question is: must € contain itself? If it does contain itself then we can agree that it is the only set that does contain itself that is in €; if there were more then it would violate the definition of € having as minimal members as possible. If it doesn't contain itself and given the specifications given above then it certainly must contain itself: a contradiction.

Do we create Russel's "paradoxical" (or not so paradoxical 😉) set just by excluding € from itself ? Does Russel's set have less then minimal members as it possibly could have?

  • If € contains all the sets that don't contain themselves, and as few as possible, then wouldn't it just contain only those? "If there were more then it would violate the definition of € having as minimal members as possible". No, it wouldn't. If we are allowing € to both contain and not contain itself then there could be others, and they would have to be there. Or, since this is inconsistent, neither they nor € exist. Either way, it is unclear what you are asking. – Conifold Aug 26 at 6:27
  • No, it wouldn't only contain those sets that don't contain themselves , since € must contain itself if it contains all the sets that don't contain themselves if not then € is inconsistent since it would qualify for being included in this set of all the sets that don't contain themselves, there is a nuance there. therefore it includes all of them it CANNOT include ONLY those, since € is minimal in addition it must be the only set that includes itself in € – Noname Aug 26 at 10:56
  • Your nuance is lost on me, and € is inconsistent anyway. The problem is already in "all the sets" (unrestricted comprehension), the further specifications are moot. The "all" here is inherently ambiguous since it is left open whether sets formed in the process will be included or not. The set of all sets is just as inconsistent as the Russell's set. – Conifold Aug 26 at 23:01
  • @Conifold "The set of all sets is just as inconsistent as the Russell's set." That's not true - plenty of consistent set theories, like NFU, permit the existence of a universal set. And I think the OP's idea is actually quite coherent (see my answer, and in particular I think the theory GPK^+_\infty does a good job of capturing it). – Noah Schweber Aug 28 at 3:42
  • @NoahSchweber So does NBG in its way. Both restrict comprehension structurally, either directly or by excluding what is being formed as a non-set, not by making a special exception because what comes out happens to be inconsistent. – Conifold Aug 28 at 3:52
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Let me try to capture what you're doing with a general comprehension (= set-formation) principle:

(C) For any property p, there is a set [p] such that (i) everything with property p is in [p] and (ii) whenever X is any set containing everything with property p, we have X is a subset of [p].

Basically, [p] is the smallest set containing everything with property p.

Here by "property" I mean "first-order formula in the language of set theory," and - while it's not material in this case - the right way to formulate (C) is with parameters allowed as well. That is, properly expressed, (C) is an axiom scheme consisting, for each formula p(x, y1,..., yn), of the axiom "For all y1,...,yn, there is a set s such that whenever p(x,y1,...,yn) holds we have x in s, and s is minimal with that property." This isn't really relevant, but I want to emphasize that the obvious vagueness in the above isn't the issue here; it's easily dealt with, and still leaves the key problem.

We can formulate a consistent set theory based on this approach. However, it has some drawbacks. Most obviously, basic set operations become problematic. In particular, consider the following additional two axioms:

  • (S) For each set X, the singleton {X} exists.

  • (D) For all sets X, Y, the difference set X-Y={a: a in X, a not in Y} exists.

Both (S) and (D) are pretty unobjectionable, but they wind up contradicting (C):

  • Let p be the property "Does not contain itself." Let [p] be the corresponding set guaranteed by (C).

  • As usual, we must have [p] in [p].

  • But now consider the new set U=[p]-{[p]}. Clearly U contains every set which doesn't contain itself, and U is a proper subset of [p], which is a contradiction.


OK, so that's a bit annoying. But is it fatal?

Well, the immediate takeaway is this: since singletons and differences are things we want to use from time to time (to put it mildly), in order to whip up a usable theory based on (C) we'd need to somehow pin down an "algebraically well-behaved" class of sets - those sets which one can do basic operations on without worry. The idea would then be that "usual" sets that we actually tend to work with in mathematics would live in this core, while weird Russell-type sets might be relegated to the less-algebraically-well-behaved parts of our universe.

There is indeed a theory which takes this approach: namely, Esser's GPK^+_\infty. Roughly speaking, in this theory the universe of sets is best thought of as a topological space, with (C)-style comprehension being an appropriate closure operation on this space. Any model of Esser's system comes with a "discrete core" of sets where the topological aspects basically become trivial, and this core forms a model of ZF. Indeed, the "core" of a model of this theory is actually much more than a model of ZF, and the result is a theory with significantly greater consistency strength (if I recall correctly, around the level of a weakly compact cardinal). The Holmes/Forster/Libert article in Sets and Extensions in the 20th Century may be of interest here.

So there is indeed an interesting set theory which follows the lines you're looking at. However, the issue above means that it, and any other similar theory, must have a "working part" which solves Russell's paradox in a different way - namely, the fragment of the universe which allows basic set-theoretic algebra must not contain anything like the Russell set, for whatever reason - and so in some sense the problem hasn't vanished entirely.

  • Incidentally, the fact that GPK^+_\infty has such high consistency strength is arguably an indicator that making a (C)-style set theory which actually implements classical mathematics cleanly is a surprisingly loaded task, and that we should view (C) with more-than-usual suspicion. I'm not sure how much I buy that, but it is something to consider. – Noah Schweber Aug 25 at 18:36
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The heart of Russell's paradox is that a set must contain all of the sets that do no contain themselves. If € does not contain itself, then it must (ipso facto) contain itself, because it satisfies the criteria for inclusion. You're trying to avoid a paradox here by redefining the set to contain all sets that do not contain themselves, plus any set that might cause a paradox, but there are two problems with this kind of €+ set:

  • It's an arbitrary (ad hoc) rule, not an analytical one, which goes against the grain of analytic philosophy.
  • It's counter-intuitive, like saying that the set 'cats' contains all cats plus a few dogs that look like cats.

You're doing the opposite of what (early) Wittgenstein tried to do. He asserted an ad hoc rule that the 'set of all sets that do not contain themselves' (SOAS) was a different set on each side of the function: i.e. that the SOAS that contains is different from the SOAS that is contained. It's a good intuition either way, but it doesn't really resolve the paradox as much as sweep it under a rug.

-1

"You're trying to avoid a paradox here by redefining the set to contain all sets that do not contain themselves, plus any set that might cause a paradox, but there are two problems with this kind of €+ set:

It's an arbitrary (ad hoc) rule, not an analytical one, which goes against the grain of analytic philosophy.It's counter-intuitive, like saying that the set 'cats' contains all cats plus a few dogs that look like cats."

First thanks for replying, but I think you misunderstood the main idea of my proposition, it is not as much that I include a set to try to avoid a paradox , that would indeed be adhoc because it would lack explaining power it is more that I include the set € in itself out of analytic necessity , since otherwise it would give rise to a contradiction. Key difference: a contradiction is not a paradox. avoiding contradictions in defining terms is the essence of the analytic tradition and certainly not adhoc.

It is more like: saying that the set of all and only cats doesn't include itself, since it including itself does imply that it doesn't include itself (since it is not a cat) hence defining it as such that it includes itself would give rise to a contradiction, not a paradox. because the premises leading to the contradiction are wrong. This is a better analogy for the what I proposed about €. if we try to exclude € from € it gives rise to a contradiction I.e Russel's set including € in € doens't

So if we start by excluding € in the definition of € i.e. Russel's set

It seemingly gives rise to a paradox, which I argue is not a paradox but a contradiction. since we defined € as such that it both does and doesn't contain itself ( that it contains itself if and only if it doesn't contain itself). A good analysis would reveal that the contradiction resolves by including € into itself as such creating the set of all sets with as minimal as members as POSSIBLE. I'm not in the business of trying to the impossible.. creating a set that has less as possible minimal amount of members ( Russel's set).

Kind regards,

-1

"We can formulate a consistent set theory based on this approach. However, it has some drawbacks. Most obviously, basic set operations become problematic. In particular, consider the following additional two axioms:

(S) For each set X, the singleton {X} exists.

(D) For all sets X, Y, the difference set X-Y={a: a in X, a not in Y} exists.

Both (S) and (D) are pretty unobjectionable, but they wind up contradicting (C):

Let p be the property "Does not contain itself." Let [p] be the corresponding set guaranteed by (C).

As usual, we must have [p] in [p].

But now consider the new set U=[p]-{[p]}. Clearly U contains every set which doesn't contain itself, and U is a proper subset of [p], which is a contradiction."

First all thanks for taking your time articulate your answerin the way you did.

So if I understand it well, We could derive a contradiction as soon as we try to remove the set from itself via U= [p]-{[p]}. via axiom (S) and (D).

Well consider this;

Suppose we define a set [p]'

such that it contains almost all the sets that don't contain themselves expect for that set itself. (kinda like the set € but in the reverse direction) clearly that set doesn't contain itself also if [p]'would contain itself and it contains almost all the sets that don't contain themselves (from the definition we know that [p]' was the only set making that collection incomplete) Then we could derive a contradiction, namely [p]' does contain itself and [p]' doesn't contain itself

can the collection of almost all the sets that don't contain themselves in that set be completed?

Well , suppose that [p]' contains itself Then we have a set of all the sets that don't contain themselves since it was the only set that made it as such that it was that collection was incomplete

how did adding [p]' to itself complete it ?

NOT by satisfying the property of "being not contained by itself" since that would be a contradiction. well that is exactly why it is complete now exactly because [p]' doesn't satisfy that property no more but it DOES satisfy the property of "being contained by itself"

As we have already seen above with the set € that is not inconsistent. Since it is not a set of all and only the sets that don't contain themselves. but it is a set of all the sets that contain themselves + [p]' similary [p]' can be removed from itself again so [p]' now contains almost* all the sets that don't contain themselves. (except for [p]')

Therefore it can be consistently contain itself and be removed from itself , to have an analogy we can have a set off all cats + that set containing itself. but we can also remove that set from itself and therefore there would be only a set off all and only cats left over.

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