The Paradox of Russel

symplectic_manifold

The Paradox of Russell

I hope most of you are fairly well acquainted with Russell's paradox, which states that the concept of the set of all sets is contradictory within the framework of naive set theory.
Could anyone explain in the layman's language how this paradox is resolved in modern set theory?

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matt grime

Homework Helper
Ok, let's have a bash.

In Naive Set Theory, a set is something like: {x | P(x)} ie we can say that the collection of all sets that do not contain themselves is a set, whcih is what leads to the contradiction.

modern axiomatic set theories side step this by disallowing such unrestricted statements.

The first thing you must realize is that modern axiomatic set theories do not say "the collection of real numbers is a set", what they do is give a set of rules, and you must then find a collection of objects satisfying those rules. such a collection is a model of the axioms.

Are you aware of other axiomatic systems? For instance groups? If I were to ask you is 3 a group element then you can't answer yes or no without me saying in what group I'm considering elements.

So, when we talk about sets we are implicitly saying in some model of the axioms, and when we say there is no set of all sets, we mean that the collection of all the sets in a particular model is not a set in that model.

it's not very lay I admit but what makes sense and what doesn't?

if you could mention any axiomatic systems of which you are aware I might be able to say something in analogous terms

Dr.ThinkDeep

For physicists, let me give a pragmatic answer:
A set is defined by a list of rules that allow to determine whether anysome "element" belongs to the set or not.
I realize this may mean to sweep the underlying mathematical problem under the carpet of what are admissable objects - but - let us face it, that is something you should decide first of all anyhow.
You need that definition as a carpet to stand on.

Now ask whether the list of rules
1. it is a set
2. it does not contain itself.
allows to decide whether our objects belong in or out.
1. we do not know yet
2. a) consider a set that does contain itself, like the set of all sets.
this does not comply with rule 2 and is exluded.
b) consider a set that does not contain itself like {1,2,3} -> included.
Now see whether this set of rules defines a set?
NO, says Russel, because if it were, we should be able to decide whether it contains itself. We are not, ergo it is not a viable definition of a set.
Anything else?

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matt grime

Homework Helper
"A set is defined by a list of rules that allow to determine whether anysome "element" belongs to the set or not."

that is the exact way of naively defining a set and it is inconsistent.

also you do not justify why "sets that do not conatin themselves" is inadmissable, and the phrase "it contains itself" is odd. what does the it refer to? There is another it in the sentence and their referents don't seem to tally.

.

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Dr.ThinkDeep

matt grime said:
...it is inconsistent...
Could you prove this in some other way than by attempted intimidation?

On the rest I was still working when you posted your reply. So let us start at the beginning, then proceed to the end and then stop, ok?

matt grime said:
...the phrase "it contains itself" is odd. what does the it refer to? There is another it in the sentence and their referents don't seem to tally...
Is there anything "odd" with the set of all sets, in your opinion?
It obviously does contain itself does it not?
Do you have anything that could not decided about whether it belongs in or out?

If you just object to the wording, you are welcome to cast it more precisely.

.

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matt grime

Homework Helper
I see you've drsatically altered your orginal post, and it makes even less sense to me now (as a piece of English), and I'm still at a loss to decide what you're getting at. We agree that in modern axiomatic set theory we use axioms that preclude a set of all sets?

What you wrote: a set is a collection of objects with a rule for belonging is what "unrestricted comprehension" is, or at least that is how one can interpret it. This doesn't get round russell's paradox without further elucidation.

If there is a set of all sets, then there is a set of sets that do not contain themselves, and set of sets that do contain themselves, this is russells paradox, this is how it arises.

If there is a set of sets that do not contain themselves, then this both is and isn't an element of itself, which is a contradiction.

So, whilst there is a class of sets (whereby something is a set in some model) then, that class is not a set in that model.

There is a model of ZFC in which all sets are countable, so R is not a set in that model.

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Dr.ThinkDeep

matt grime said:
What you wrote: a set is a collection of objects with a rule for belonging is what "unrestricted comprehension" is, or at least that is how one can interpret it. I did not us the term "unrestricted comprehension" and the rest of this sentence I do not recognise as anything I said. I have a feeling, we better stop before sliding into flaming, as we cannot make sense of each others posts. Peace!

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matt grime

Homework Helper
I don't think that would happen, since we both agree that there is no set of all sets (don't we?), but the question asked how modern theories get roud Russell's paradox, and I'm attempting to understand how your reply fits in with that question.

Dr.ThinkDeep

ok, forget my answer. I am not stubborn.

Can you please explain in a first step:
How can I check whether something is a set or not?

matt grime

Homework Helper
It entirely depends on what model you're talking about.

This is exactly the same as asking is 3 an element of a group.

symplectic_manifold

Now...hold...hold on, Guys!!! :yuck: I think I have real understanding problems.
Now let me check it all through:
The Russell's Paradox states that, given there exists an object called "set" with the following properties

1. it may consist of any distinguishable objects (or it may not, then it's an empty set...hm, well if does not consist of any objects, are we allowed to perceive set as objects?...because it would mean that there are no objects that can determine the object "set"...well, in case "set" or its existence determines itself, but it follows that a set is an element of itself, so it indeed consists of some object, mainly itself...see further for details),

2. it is unambiguously determined by the collection of objects that comprise it,

3. any property (may) define(s) the set of objects having that property...right?

...so given this, suppose that for such an object there exists the following property - this same object is not an element of itself...so here I imagine that this very object is an element of another object called set with the above properties...right?...then I am personally inclined to try and imagine what would be if this object were an element of itself: I think it's very sticky...it would then follow (in my opinion) from the naive perception of such an object as "set" that the existence of this object or its creation in our minds by thought determines itself i.e. its existence determines its existence... ...

...on the other hand, is there no contradiction if I say that the distinguishable object "set" is not an element of itself?...because (in my opinion) it would then mean that the object set does not exist i.e. the not-belonging of the object to itself should actually prove the non-existence of the given object

...if I take a collection of such objects, which are not elements of themselves, and state that this collection is not an element of itself...do I not come to the result that this collection cannot be exactly determined by its elements because the property "the object is not an element of itself" fails to determine the objects contained in this collection?
...and in addition by saying "the collection of objects called sets which are not elements of themselves are contained in the object which in turn is not an element of itself too" I actually state that this collection is a single object...or that there are no initial "single" objects but only collections of objects... Well, Guys, no matter if my above thoughts on the subject are plausible or not, I think the real problem is to define what a set actually is...or what an object of our perception really is...
it actually reminds me of the question on the existence or non-existence of God..."the initial cause" and stuff...it has wider applications so to speak...I think

but first, please, read carefully my thoughts again and comment on it...
we'll return to groups and axiomatic set theory later on...I also think I should inform myself on this matter a bit...so I hope we'll continue with all that... Last edited:

matt grime

Homework Helper
A "set" is an element in a model of a set theory, that is all

Dr.ThinkDeep

matt grime said:
It entirely depends on what model you're talking about.
This is exactly the same as asking is 3 an element of a group.
I have my reservations about the "exactly" as I would answer "yes" to your question.
However, if you would ask me what a tensor is, I would answer "an element of a tensorspace".

what are the defining properties of a model?
The model determines what are the objects which are admissable as elements?

matt grime

Homework Helper
The axioms are what determines what may be a model.

A group is a set of elements that satisfy the axioms of a group, agreed?

Well, a set is an element of a class of objects that satisfy certain rules. One of which (in the axioms we now come to use usually) states that the class of all "sets" is not one of the sets in the class.

Interesting you should mention the example you do in the way you do. A tensor is an element of a tensor space, right? But a tensor space is not a tensor is it? Just as an element of a group is not a group, and a vector space is not a vector (all with the proviso, in that space, group or whatever is appropriate).

Perhaps "exactly" wasn't the best word. However, if you are to answer "yes" to the qestion "is 3 an element of a group", then what you are presumably doing is providing a group, that is a model of the axioms of a group, in which 3 is an element. Which is how we often interpret the question: is this a set - we find a model of the axioms of the theory we use (ZF, often) that that set is in.

phoenixthoth

Mind you this is for the layperson.

Hopefully the layperson in mind will accept proof by contradiction. (For example, one can use it to prove there is no largest natural number. Suppose there is: n. But n+1 is a natural number and n<n+1, a contradiction implying there is no largest natural number.)

To proceed, denote the set of all sets by U. This is assuming U is a set.

In set theory, since U is a set, S={x in U : x is not in x} is a set.

In English, S is the set of all sets that are not elements of themselves.

The crucial step is to ask if S is in S. Either S is in S or S is not in S.

Note that S is in S, by definition of S, if and only if S is not in S! CONTRADICTION!

Note that S is not in S, by definition of S, if and only if S is in S! Again, a contradiction.

Therefore, our original assumption that U is a set is wrong. There is no set of all sets.

symplectic_manifold

matt grime said:
Ok, let's have a bash.

In Naive Set Theory, a set is something like: {x | P(x)} ie we can say that the collection of all sets that do not contain themselves is a set, whcih is what leads to the contradiction.

modern axiomatic set theories side step this by disallowing such unrestricted statements.

The first thing you must realize is that modern axiomatic set theories do not say "the collection of real numbers is a set", what they do is give a set of rules, and you must then find a collection of objects satisfying those rules. such a collection is a model of the axioms.

Are you aware of other axiomatic systems? For instance groups? If I were to ask you is 3 a group element then you can't answer yes or no without me saying in what group I'm considering elements.

So, when we talk about sets we are implicitly saying in some model of the axioms, and when we say there is no set of all sets, we mean that the collection of all the sets in a particular model is not a set in that model.

it's not very lay I admit but what makes sense and what doesn't?

if you could mention any axiomatic systems of which you are aware I might be able to say something in analogous terms
Yes, alright...what about the axiomatic system of Euclidean Geometry proposed by Hilbert? Please, tell me what are the analogies.

matt grime

Homework Helper
ok, that's not the best one, but let's see.

euclidean geometry is made of a space, the elements in the space are points. there are geodesics and given any two points there is a unique geodesic through them (straight line), given a line and a point not on the line there is a unique line through the point not intersecting the .... sorry, bored

so, a set theory is a collection of objects , these are the sets in the theory, satisfying certain rules. they are the points in the euclidean space.

the sets (points) satisfy certain rules. the rules depend on the set theory we want to use. just as we can change the rules of geometry and get hyperbolic geometry or spherical geometry.

that's about as far as the analogy goes in my head at the moment. I mean you don't ask "is this a point in euclidean space" very often do you?

but we should talk about a model.

just cos we've got a set of axioms doesn't mean that there is anything that actually is a representation of them.

in this context, for a long time people thought that the parallel postulate was a consequence of the other rules, but then someone came up with a model in which all the rules of euclidean goemetry are true except the parallel postulate.

the poincare disc is a model of hyperbolic geometry, as is the upper half plane. correspondignly there are different models of a set theory.

where from there?

Dr.ThinkDeep

phoenixthoth said:
In set theory, since U is a set, S={x in U : x is not in x} is a set.
I would rather sacrifice this step than sacrifice the set of all sets.
The self reference of this definition appears to me the problem and the set of sets does not appear self referential.
Just consider the Barber Paradox
The problem is not that the Barber shaves everybody who does not shave himself but the combination of the stipulations that
1. he does not shave anybody else
2. every male has to shave himself or get shaved by the Barber.
Every single stipulation is ok.

Similarly the combination of the two sentences:
1. The following sentence is true.
2. The negation of the previous sentence is true.

Both sentences alone are tolerable but the combination screws up.
Sure, you can exclude statements that state whether other sentences are true or false. But that seems to me to be too much.
Stipulating a hierarchy (without loops) in the statements that state truth or falsehood of others may be sufficient.

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matt grime

Homework Helper
well, you also introduce the problem that the following desirable statement is now false, in your logic:

If S is a set and T a collection of objects and each element in T is in S, then T is a set too.

The Barber problem is not really a mathematical one, but one about sentences and self reference (this statement is false kind of thing).

CrankFan

symplectic_manifold said:
I hope most of you are fairly well acquainted with Russell's paradox, which states that the concept of the set of all sets is contradictory within the framework of naive set theory.
Could anyone explain in the layman's language how this paradox is resolved in modern set theory?
Actually, that's not known as Russell's paradox. That's known as Cantor's paradox.

mathwonk

Homework Helper
Being fairly ignorant on the topic, I am therefore emboldened to leap right in.

To my knowledge, there are two problems posed by set theory.

1) which ones are defined legitimately by properties that are not self contradictory, as "the set of all sets not belonging to themselves" is.

2) which sets are small enough to be dealt with practically.

it seems to me to be usual practice to call all collections which have reasonable non contradictory definitions, as "classes". then in case a class happens to belong as an element of another class, the element class is called a set.

the idea is that sets are smaller than classes. in this way, one might speak of the collection of all sets, but that collection is not a set.

this may not be kosher doctrine.

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Owen Holden

symplectic_manifold said:
I hope most of you are fairly well acquainted with Russell's paradox, which states that the concept of the set of all sets is contradictory within the framework of naive set theory.
Could anyone explain in the layman's language how this paradox is resolved in modern set theory?
It is resolved by showing that the Russell class {x:~(x e x)} does not exist.

~Ey(Ax(x e y <-> ~(x e x))) is a theorem. That is to say,
~Ey(y={x:~(x e x)}) is a theorem.

More formally ...

Stanford Encyclopedia of Philosophy:

Russell’s paradox is the most famous of the logical or set-theoretical
paradoxes. The paradox arises within naive set theory by considering the set
of all sets that are not members of themselves. Such a set appears to be a
member of itself if and only if it is not a member of itself, hence the

Some sets, such as the set of all teacups, are not members of themselves.
Other sets, such as the set of all non-teacups, are members of themselves.
Call the set of all sets that are not members of themselves S. If S is a
member of itself, then by definition it must not be a member of itself.
Similarly, if S is not a member of itself, then by definition it must be a
member of itself. Discovered by Bertrand Russell in 1901, the paradox
prompted much work in logic, set theory and the philosophy and foundations
of mathematics during the early part of the twentieth century."

If S={x:~(x e x)}, then (S e S) <-> ~(S e S) is the paradox.

"If S is not a member of itself, then by definition it must be a member of
itself", is not true.
That is, Fy -> (y e {x:Fx}), is invalid.

~(S e S) -> (S e S), is false.

~({x:~(x e x)} e {x:~(x e x)}) -> ({x:~(x e x)} e {x:~(x e x)}), is false.

Proof:

If we grant first order predicate logic, and add Russell's contextual
definition of Classes,
determined by some predicate, then the antinomy does not occur.

D1. G{x:Fx} = EyAx((x e y) <-> Fx .& Gy) Df.

1. (z e {x:Fx}) <-> EyAx((x e y) <-> Fx .& (z e y))
<-> EyAx((x e y) <-> Fx .& Fz)
<-> Ey(Ax((x e y) <-> Fx) & Fz)
<->.EyAx((x e y) <-> Fx) & Fz valid

2. (z e {x:Fx}) -> Fz is valid

3. Fz -> (z e {x:Fx}) is invalid

4. EyAx((x e y) <-> Fx) is invalid

5. ~Ax((x e y) <-> ~(x e x))

Proof:

Ax((x e y) <-> ~(x e x)) ->. (y e y) <-> ~(y e y)

Ax((x e y) <-> ~(x e x)) -> contradiction

~Ax((x e y) <-> ~(x e x))

6. ({x:~(x e x)} e {x:~(x e x)}) <-> EyAx((x e y) <-> ~(x e x) .& (y e y))
by D1.
<->Ey(Ax((x e y) <-> ~(x e x) &
(y e y))
by 5.

Therefore 6. ~({x:~(x e x)} e {x:~(x e x)}), is a theorem.

The answer to Russell's question 'Is the class of those classes that are not
members of themselves,
a member of itself or not?' is NO it is not.

Frege's axiom V, {x:Fx}={x:Gx}<->Ax(Fx<->Gx), is not required for this
proof.

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