Math Help - All Sets

Start to list all sets starting with A,B,C,…
you can’t get past A:
A, {A},{{A}}, {{{A}}}, …..
Let alone A, B:
A, B, {A,B}, {A,{A,B}}, {B,{A,B}}, {A, {A,{A,B}}}, {A, {B,{A,B}}},
{B,{A,{A,B}}}, {B,{B,{A,B}}},…..

True, the fact that you can’t list them all doesn’t mean they don’t exist. However, it appears to me this construction renders the notion of “all sets” meaningless, especially if you start with A any object or concept in the universe. It's not mathematics anymore, its just words and combinations of words- philosophy (in my opinion).

And each one of these sets has a different definiton that have to be combined.

If you are saying that we cannot have "sets of sets", yes, that's true.

But surely, not for the reason you state. It is not, for example, possible to list all of the rational numbers between 0 and 1 so you can't even "get past 1" "let alone 2". Does that show that rational numbers do not exist?

{A}

This is not in your list of sets. It is a set which contains a set that you listed, so its completely different.

I´m supposing you are saying one cannot have the set of all sets.
In this case, yes, one can´t. For example, in ZFC the axiom of regularity forbids it by not allowing in the language to exist a set that is a member of itself (solving the Russel´s paradox).

However, the "set" of all sets can be very treated mathematically. For this, on needs a stronger language that encompasses the language of ZFC and build a stronger axiomatic. For example, in Neumann-Bernays-Gödel set theory the "set" of all sets is not a set but it is a proper class and you still can formally work with classes and sets in the same formal theory (in the case, NBG).

Unfortunately or not, the boundaries of philosophy and mathematics are not well defined.

Originally Posted by FelipeAbraham

I´m supposing you are saying one cannot have the set of all sets.
In this case, yes, one can´t. For example, in ZFC the axiom of regularity forbids it by not allowing in the language to exist a set that is a member of itself (solving the Russel´s paradox).

However, the "set" of all sets can be very treated mathematically. For this, on needs a stronger language that encompasses the language of ZFC and build a stronger axiomatic. For example, in Neumann-Bernays-Gödel set theory the "set" of all sets is not a set but it is a proper class and you still can formally work with classes and sets in the same formal theory (in the case, NBG).

Unfortunately or not, the boundaries of philosophy and mathematics are not well defined.

I was thinking there might be a way in that direction. But I was thinking of categories. What is the difference between a class and a category?

-Dan

Hi Dan,

Sorry being late.

A category is a diferent kind of mathematical object and it can be more general than the usual notion of set. It is closer notion to groups than to sets, since it was formerly conceived as an algebraic structure. They are usually defined in NBG set theory, but we can weaken it to ZFC.

You can found a good explanation on Wikipedia.
A category is always a pair: the class of the objects and the class of the morphims. So only makes sense of saying the category of all sets if we add morphisms to it. For example, the satisfying functions of model theory.
There you can found there is no category of all large categories. Analogously, there cannot be the class of all classes.

Originally Posted by Hartlw

Start to list all sets starting with A,B,C,…
you can’t get past A:
A, {A},{{A}}, {{{A}}}, …..
Let alone A, B:
A, B, {A,B}, {A,{A,B}}, {B,{A,B}}, {A, {A,{A,B}}}, {A, {B,{A,B}}},
{B,{A,{A,B}}}, {B,{B,{A,B}}},…..

True, the fact that you can’t list them all doesn’t mean they don’t exist. However, it appears to me this construction renders the notion of “all sets” meaningless, especially if you start with A any object or concept in the universe. It's not mathematics anymore, its just words and combinations of words- philosophy (in my opinion).

And each one of these sets has a different definiton that have to be combined.

All you are saying is that sets cannot be listed linearly- they are not countable. Nothing new in that.

Originally Posted by FelipeAbraham

in ZFC the axiom of regularity forbids [the set of all sets] by not allowing in the language to exist a set that is a member of itself (solving the Russel´s paradox).

That is incorrect. It is true that if ZFC is consistent (I'll not mention that condition again), then ZFC does not prove that there exists a set of all sets, but the axiom of regularity has nothing to do with it. Even though the axiom of regularity bars the existence of a set that is a member of itself, the axiom of regularity does not bar the derivation of Russell's paradox from the axiom of unrestricted comprehension. The reason ZFC does not allow Russell's paradox nor the existence of a set of all sets is that ZFC eschews the axiom of unrestricted comprehension; the axiom of regularity has nothing to do with it. Indeed, one cannot ADD an axiom to make a theory consistent. If, without the axiom of regularity, ZFC proved that there exists a set of all sets then it would still prove that even with the axiom of regularity added back it.

It is a misconception that the axiom of regularity plays any role in blocking Russell's paradox or in blocking the existence of a set of all sets; and, to my knowledge, such an assertion is not found in any textbook or informed paper on set theory.

MoeBlee,

Thank you very much helping clarifying the subject. I probably used the word "forbids" without further explanations. Likely my bad. I dind´t mean axiom of regularity makes ZFC coonsistent. It is why I put the condition "if ZFC is consistent, then...". I thought it wouldn´t generate missunderstandings. Sorry.

The axiom of regularity directly proves the negation of the existence of the universal set builded in Russell´s Paradox.

I guess I didn´t say Russell´s paradox cannot follow from the unrestricted comprehension together with axiom of regularity. Unrestricted comprehension is an axiom of the naive set theory (which is inconsistent). ZFC currently uses the restricted form: There is a set B such that (x belongs to B <=> Phi(x)) only if B is already a subset of a set A (the formal statement is a little diferent).

So I believe you meant by unrestricted comprehension the other ZFC axioms (the restricted comprehension and the other 7).
In that matter, forgeting the unrestricted form, I guess I didn´t say russell´s paradox´s univeral set cannot arise from the other ZFC axioms (the restricted comprehension and the other 7) without axiom of regularity. Of course it can, but if one proves it does then one proves ZFC is inconsistent.

I just tried to say the axiom of regularity proves the negation of the existence of the universal set builded in Russell´s Paradox, not that the axiom of regularity proves one can´t prove this existence.

Further explanations:
logic - Confusion regarding Russell's paradox - Mathematics Stack Exchange


Tks!

Originally Posted by FelipeAbraham

I just tried to say the axiom of regularity proves the negation of the existence of the universal set builded in Russell´s Paradox, not that the axiom of regularity proves one can´t prove this existence.

Okay. But the axiom schema of separation itself proves the negation of the existence of a universal set. We don't need to go to the axiom of regularity for this.

Originally Posted by FelipeAbraham

So I believe you meant by unrestricted comprehension the other ZFC axioms

No, I meant unrestricted comprehension. My point was that it is by eschewing unrestricted comprehension that ZFC avoids proving that there is a set of all sets.

Originally Posted by FelipeAbraham

Further explanations:
logic - Confusion regarding Russell's paradox - Mathematics Stack Exchange

The first post there says virtually what I said in my own post above.

Originally Posted by MoeBlee

Okay. But the axiom schema of separation itself proves the negation of the existence of a universal set. We don't need to go to the axiom of regularity for this.

No, I meant unrestricted comprehension. My point was that it is by eschewing unrestricted comprehension that ZFC avoids proving that there is a set of all sets.

The first post there says virtually what I said in my own post above.

Sure.

Originally Posted by FelipeAbraham

In this case, yes, one can´t. For example, in ZFC the axiom of regularity forbids it by not allowing in the language to exist a set that is a member of itself (solving the Russel´s paradox).
.

Originally Posted by FelipeAbraham

For example, in ZFC the axiom of regularity forbids it by not allowing in the language to exist a set that is a member of itself (solving the Russel´s paradox).

And that is not well put, as written, even though I understand your subsequent disclaimers about it:

(1) Better to say that Z set theories (including extensions such as ZFC), from the central axiom (schema of separation, or in ZF set theories, the schema of replacement) that replaces the problematic axiom of unrestricted comprehension, yield the negation of the assertion that there exists a set of all sets, but there is no axiom whose presence can solve Russell's paradox as instead it is the LACK of presence of certain a axiom schema (unrestricted comprehension) that solves Russell's paradox. Meanwhile the axiom of regularity, added later, also negates the assertion that there exists a set of all sets; but that is merely an incidential result from the axiom of regularity since, as just mentioned, the axiom schema of specification does this already.

Moreover, as to Russell's paradox itself, there are no set theoretic axioms involved in deriving the negation of the assertion that there exists a set of all those sets that are not members of themselves. This derived purely by logic.

(2) "not allowing in the language" is off key at best. It should be, mutatis mutandis, "not allowing a theorem". The language itself is in to way restricted by axioms.

I get that you understand all this yourself. I'm just saying that your original statement of it was not well put.

No problem at all. Thanks.

Yes. I have already agreed it was not well put. And language was an informal way to say model (you are right, these are two very different things).

The reason I chose regularity here was that the sequence of sets from the original post leads to a direct opposite notion with the regularity. Russell´s antinomy is just directly negated and related with the notion of "all sets", but not the more important matter with the first post (that´s why I put it in parentheses).

Axiom of regularity does by negating the existence of set that contains itself (and maybe we can say it is a deeper axiom by not allowing several other undesirable properties) and, by "corolary", the universal set. The restricted comprehension does by negating the existence of the universal set, otherwise it would lead to Russel´s set { x | x not belongs to x } (which is not closely related to the main topic - in fact it is but, in my opinion, not as the axiom of regularity).

I think it's safe to say that the greatest importance of the axiom of regularity is that it entails that every set is in the cumulative hierarchy.

My take on all this (which, admittedly, is naive) is that the axiom of regularity "stratifies" set theory (by establishing a distinction between A and {A}). While this doesn't prevent Russell's paradox from occurring per se, it does prevent us from the paradoxes we would obtain by "mixing levels".

My first encounter with this "level mixing" was when i first took topology: a topology itself is a subset of a power set of another set (the "space"), and it becomes quite easy to get confused about which level the set you are concerned with (perhaps a certain open cover of an element (a neighborhood containing a certain point) of the power set of the space you're in) actually lies on. Many authors use different type-fonts to help with this. This becomes particularly bad when you are looking at, say, a Hilbert space comprised of certain functions (which adds even more "layers" to the mix) defined on sets which are themselves topological spaces.

All this layering seems necessary to keep within a logical framework that is (as far as we know) free from contradiction, but it's a bit unwieldy to take in at once.

What the original poster has shown, is that there does indeed exist a limit cardinal that is not finite (I think this is the modern term set-theorists like to use). One can take this idea and run with it, leading to unimaginably vast numbers. Even if one tries to get "ahead of the curve" by taking the "limits of all the limits", one can still push FARTHER, for if nothing existed beyond, we WOULD get a paradox. Using some strange technique called "forcing" apparently one can conceive of cardinals which are "strongly inaccessable": they must be there, but we can't quite get TO them. Even this has failed to give most logicians pause...

For the most part, working mathematicians have agreed to say: OK, I don't know what is "ultimately out there", but as long as we only go THIS FAR, we're "relatively safe". Opinions seem to differ a bit as to how far "this far" should be, but most are unwilling to settle for anything less than a framework that includes what are called "second countable" objects (the real number system, and the various objects spawned by it being chief amongst these). The motivation for this seems to be the great utility of such systems for modelling actual physical behavior of "real things". The accompanying philosophical quandries are by no means settled, and perhaps may not ever be.