Russell's paradox and logical errors in the proof

[DanTeplitskiy]

I need your opinion/help, guys.

Well-know American logician H. Curry once expressed the opinion that in spite of the fact that it seemed to be absolutely impossible to explain Russell's paradox in terms of conventional 19th century logic, it may happen in modern days that some error would be identified.

I consider my paper (link below) to give the ultimate answer to this challenge, identifing the one.

https://docs.google.com/file/d/0B_tihhgZ1L4wOUN3aTVFN05sXzQ/edit?usp=sharing

I hope you will become interested and give your valuable opinion on the paper. Such feedback will help me a lot!

The paper is rather small and written in a clear organized way – I do not think it will take more than 30 minutes from a person knowing the very basics of math to get it all.

Thanks a lot in advance!

Dan

P.S. The paper is in English, though, of course, it is not English of an English speaking person.

 

[ImaLooser]

I need your opinion/help, guys.

Well-know American logician H. Curry once expressed the opinion that in spite of the fact that it seemed to be absolutely impossible to explain Russell's paradox in terms of conventional 19th century logic, it may happen in modern days that some error would be identified.

I consider my paper (link below) to give the ultimate answer to this challenge, identifing the one.

https://docs.google.com/file/d/0B_tihhgZ1L4wOUN3aTVFN05sXzQ/edit?usp=sharing

I hope you will become interested and give your valuable opinion on the paper. Such feedback will help me a lot!

The paper is rather small and written in a clear organized way – I do not think it will take more than 30 minutes from a person knowing the very basics of math to get it all.

Thanks a lot in advance!

Dan

P.S. The paper is in English, though, of course, it is not English of an English speaking person.

Russell's Paradox seems quite straightforward to me. I don't understand what the alleged obscurity is.

 

[DanTeplitskiy]

Dear ImaLooser,

Have you read pages 2 through 4 of the paper?

Yours,

Dan

 

[Fredrik]

I had a quick look at pages 2-4. What is wrong with the following proof: Suppose that ##\{x\,|\,x\notin x\}## is a set, and denote it by R. Since R is a set, the statement ##R\in R## must be either true or false.

If it's true, then the definition of R tells us that ##R\notin R##, and we have a contradiction.

If it's false, then the definition of R tells us that ##R\in R##, and we have a contradiction.

So regardless of the truth value of ##R\notin R##, we have a contradiction. This forces us to conclude that the statement ##R\notin R## is neither true nor false. This forces us to reject the assumption that ##\{x\,|\,x\notin x\}## is a set. ##\square##

You seem to be rejecting both of colored sentences above, and I don't understand why. To say that ##R\notin R## has a truth value seems very different from saying that "Dan is a legless man who's bleeding severely from his ankle".

 

[DanTeplitskiy]

Dear Fredrik,

You missed the point. However I can not put it clearer then it is put in the paper. Sorry.

Yours,

Dan

 

[Fredrik]

We shouldn't have to read a paper to get a single point. If you want someone to help you, you need to make it easy to do that.

I should also tell you that the forum has a rule against "original research". One of the reasons for that is that we don't want to spend half our time reading through people's papers to see what they did wrong. If your paper is original research, you're already breaking the forum rules. If it's not, and you just want to discuss a detail in a proof, then you should just describe that detail. There's no need to post the whole paper.

 

[DanTeplitskiy]

Dear Fredrik,

On details:

1. Both assumptions (R∈R is true and R∈R is false) contradict the definition of R (why it is so is elaborated end of page 2 through beginning of page 3).
2. However, in Russell's paradox we use this definition along with these assumptions to make what you call "tells us that R∉R" and "tells us that R∈R" - that is, conclusions in two parts of reasoning.
3. That is, in Russell's paradox we use "contradictory premises". The known logical error, example of which is put in my paper (you mentioned this example).

Yours,

Dan

 

[Fredrik]

I still don't understand. Are you saying I made a logical error in the colored sentences above? Or are you just saying that there's a logical error in this:

If ##R=\left\{x\,|\,x\notin x\right\}##, then ##R\in R\leftrightarrow R\notin R##.

I don't see any logical errors. This if-then statement can't be illogical, since the notation ##R=\left\{x\,|\,x\notin x\right\}## by definition means ##\forall x~\left(x\in x\leftrightarrow x\notin x\right)##. So the assumption implies that ##x\in x\leftrightarrow x\notin x## holds for all x. This implies that it holds when x=R. This is a perfectly valid way to find out that the statement ##R=\left\{x\,|\,x\notin x\right\}## can't be true for any set R.

Edit: I made a mistake when I wrote ##\forall x~\left(x\in x\leftrightarrow x\notin x\right)##. It should be ##\forall x~\left(x\in R\leftrightarrow x\notin x\right)##. So the assumption implies that ##x\in R\leftrightarrow x\notin x## holds for all x. In particular, it holds when x=R.

 

[DanTeplitskiy]

Dear Fredrik,

Yes, you made a logical error in the colored sentences above.

As I am talking about a logical error, that fact that predicate calculus axioms (the language you just used) give us R∈R↔R∉R does not invalidate my reasoning.

Do you understand why both assumptions (R∈R is true and R∈R is false) contradict the definition of R (why it is so is elaborated end of page 2 through beginning of page 3)?

Yours,

Dan

 

[Fredrik]

Do you understand why both assumptions (R∈R is true and R∈R is false) contradict the definition of R (why it is so is elaborated end of page 2 through beginning of page 3)?

Yes, I do. I haven't studied your argument for it, but it follows from what I meant to say in my previous post. What I actually said was wrong. The notation ##R=\left\{x\,|\,x\notin x\right\}## by definition means ##\forall x~\left(x\in R\leftrightarrow x\notin x\right)##. (In my previous post, I put an x where the R is supposed to be). So the definition of R implies that ##R\in R\leftrightarrow R\notin R##, which is a false statement regardless of the truth value of ##R\in R##.

 

[DanTeplitskiy]

Dear Fredrik,

If you say you understand my argument without reading it, it makes no sense in further discussion.

Anyway, thanks for your attention!

Yours,

Dan

P.S. I noticed your small error in the previous message. Though, as I understood the whole point it made no sense for me to mention it.

 

[AlephZero]

I'm not sure what point the OP is trying to make, but the point that Russell and Whitehead made was that the notion of "defining" includes the notion that the thing being defined exists.

If I say "let S be a square circle", I haven't defined anything. And the basic point of R&W's book is that the string of symbols ##\{x\, |\, x \notin x\}## also doesn't define anything.

I can't be bothered to look up a section reference in R&W's Principia Mathematica, but one of their examples is the meaning (if any) of statements like "the present king of France is bald". I don't see much logical difference between buying the king a toupee or sending the OP's legless man to hospital...

 

[Fredrik]

I haven't said that I understand your argument without reading it. I don't even understand what point you're trying to make about Russell's paradox, and this is something I would need to know before I consider studying your paper in detail.

Let me combine what I said in posts #4 and #10, and then you can tell me what you think is wrong with it.

Let ##R## be an arbitrary set. We will prove that ##R\neq\{x\,|\,x\notin x\}## by deriving a contradiction from the assumption that this statement is false. So suppose that ##R=\{x\,|\,x\notin x\}##. This notation by definition means that ##\forall x~\left(x\in R\leftrightarrow x\notin x\right)##. This implies that ##R\in R\leftrightarrow R\notin R##. This contradicts the fact that ##R\in R\leftrightarrow R\notin R## is a false statement regardless of the truth value of ##R\in R##.

 

[DanTeplitskiy]

Dear AlephZero,

Though I could not see your point the point of the example in the paper was to make the example of "contradictory premises" logical error.

Yours,

Dan

 

[DanTeplitskiy]

Dear Fredrik,

If you do not read the argument you can not see the point. I am not a wizard :).

Anyway, thanks for your attention!

Yours,

Dan

 

[Fredrik]

If you do not read the argument you can not see the point. I am not a wizard :).

If you're not willing to explain what the point of the article is, then why should anyone read it?

 

[DanTeplitskiy]

Dear Fredrik,

The point of the article is in the abstract :) As usual.

I put here main details of my proof and asked to read a very small portion of the paper for little elaboration.
You did not manage to do it.

Anyway, thanks for your attention!

Yours,

Dan

 

[Stephen Tashi]

contradict the definition

What do mean by "contradict the definition"? I think you mean that no set exists that satisfies the definition of your paper's [itex] R [/itex]. That isn't controversial and, in fact, it is part of the content of Russell's Paradox.

Russell's Paradox is only a paradox if we assert that sets whose definitions are given in a very unrestricted way are always mathematical objects that exist. What you need to examine is whether "19 century" set theory did assert that. Or whether it expressly prohibited the definition of a set from being self-referential in some ways.


-------------

As to variants of "contradicting" a definiton:

For an assertion to be a definition, it must be possible to express it in the form:
[statement involving the things to be defined] if and only if [statement involving things that have been previously defined].

One may contradict a definition, in the sense that one may disagree with cultural conventions and have the private opinion that [statement involving the things to be defined] should be set equivalent to a different statement.

One may prove that no things exist that make [statement involving things to be defined] true by showing that no things exist that make [statement involving things that have been previously defined] true. Such a proof is not a contradiction of the definition. It merely shows that no things exists that satisfy the definition.

One may assert that a statement involving undefined things cannot be put in the form: [statement involving the undefined things] if an only if [ statement involving only things that have been previously defined]. This asserts that the statement is not a definition. It doesn't contradict a definition that is granted to exist.

 

[DanTeplitskiy]

Dear Stephen Tashi,

What do mean by "contradict the definition"?

Have you read the small portion of the paper I pointed at before during my discussion with Fredrik (end of page 2 through the beginning of page 3)?

Yours,

Dan

 

[DanTeplitskiy]

Dear Fredrik, Stephen Tashi and others,

I will try to elaborate on my point a little more.

1. Both assumptions (R∈R is true and R∈R is false) contradict the definition of R (Let R be the set of all sets that are not members of themselves):

1) Assumption R ∈ R contradicts the definition of R because if R ∈ R, R includes a member that is included in itself (R itself is such a member).
Or, symbolically: R ∈ R → ∃ y: y ∈ R ∧ y ∈ y → R ≠ {x: x∉x}: indeed, such y exists as we can take y = R

2) Assumption R ∉ R contradicts the definition of R because if R ∉ R, R does not include a member that is not included in itself (R itself is such a member).
Or, symbolically:R ∉ R → ∃ y: y ∉ R ∧ y ∉ y → R ≠ {x: x∉x}: indeed, such y exists as we can take y = R

2. However, in Russell's paradox we use this definition along with these assumptions to make conclusions in two parts of reasoning:
Suppose R ∈ R. Then, according to its definition, R ∉ R.
Suppose R ∉ R. Then, according to its definition, R ∈ R.

Argument 1
Premise 1: Let R be the set of all sets that are not members of themselves R = {x: x∉x}
Premise 2 (assumption): Suppose R ∈ R.
Conclusion: Then, according to its definition, R ∉ R.

Argument 2
Premise 1: Let R be the set of all sets that are not members of themselves R = {x: x∉x}
Premise 2 (assumption): Suppose R ∉ R.
Conclusion: Then, according to its definition, R ∈ R.

3. That is, in Russell's paradox we use "contradictory premises". The known logical error, example of which is put in my paper:
Example Argument.
Premise 1: Let Dan be a completely legless man
Premise 2: Suppose, Dan’s right ankle is severely bleeding
Conclusion: Then, according to his definition, Dan should be taken to an emergency for legless people (for his ankle bleeding).

The reasoning on `legless Dan' contains the same logical error as the Russell's paradox does: the second premise contradicts the denition which is the first premise though to make the conclusion both premises are used.

Yours,

Dan

P.S. I did not mean to be rude or arrogant or something...just wanted someone to read the paper. :)

 

[yossell]

Do you think that Russell's argument rests on the assumptions 'R is a member of R' and 'R is not a member of R'?

If true, that would be bad. And I think i can see, from your summary, why it looks that way. Russell's argument seems to have two parts, and the contradictory assumptions appear in the two parts. But when you understand the logical structure of the argument you should see that these are not premises of his argument.

The relevant "premise" is a disjunction: either R is a member of R or R is not a member of R. This is an instance of a general logical truth. Then the logical structure of the proof is an ' disjunction elimination': if you can show that both disjuncts lead to the same conclusion, then that conclusion follows from the disjunct itself. So first Russell examines what follows from 'R is a member of R' and shows we get a contradiction. Then he examines what follows from 'R is not a member of R' and gets a contradiction. Thus, in either case, there is a contradiction. So 'either R is in R or R is not in R' leads to a contradiction. Problem.

Disjunction elimination can look confusing when written out in linear style, as it can seem as though the disjuncts themselves both figure as premises in the argument.

 

[DanTeplitskiy]

Dear Yossell,

Russell's argument seems to have two parts

The reasoning that leads to Russell's paradox (mathematically speaking, the proof of Russell's paradox) :

Suppose R ∈ R. Then, according to its definition, R ∉ R.
Suppose R ∉ R. Then, according to its definition, R ∈ R.

Or like Fredrik put it in this thread earlier:

the statement R∈R must be either true or false.
if it's true, then the definition of R tells us that R∉R, and we have a contradiction.
If it's false, then the definition of R tells us that R∈R, and we have a contradiction.

has two arguments.

I do not mean that there is nothing else besides these two arguments when we talk about Russell's paradox. We can say at first that the statement R∈R must be either true or false (like Fredrik did). However the two logical arguments in it do exist.

Yours,

Dan

 

[Stephen Tashi]

1. Both assumptions (R∈R is true and R∈R is false) contradict the definition of R (Let R be the set of all sets that are not members of themselves):

You still are using the idea of "contradicting" a definition. What you actually mean is that no set exists that satisfies the definition of R. There is no contradiction involved in writing definitions of things that cannot exist. (For example, define N to be the largest real number. This defines a number that does not exist, but it isn't a logical paradox.) A definition does not assert the existence of the thing defined. For example, you can define the "multiplicative inverse" of a matrix, but this doesn't assert that all matrices have multiplicative inverses. As assertion that something exists mathematically must be either an assumption or a theorem. Such an assertion is not in the content of what we call a definition.

The fact that no set R exists is only a "paradox" if the axioms you are using includes a statement that says that a set defined as R is defined must exist. Your arguments show that R cannot exist because assuming it exists, contradicts the axiom that "If S is a set and x is an element then either x is a member of S or x is not a member os S". That is a well known fact. However, you haven't dealt with the "paradox" aspect well. You need to look at the historical state of set theory prior to Russell and determine why people thought that an arbitrary definition of a set would always define a set that existed, even if it was only the null set.

 

[rubi]

The notation [itex]S = \{x : P(x)\}[/itex] is just a convenient version of writing [itex]\forall x (x\in S \leftrightarrow P(x))[/itex].

The point of Russels paradox is that if there were a set [itex]R = \{x : x \notin x\}[/itex], it would translate to [itex]\forall x (x\in R \leftrightarrow x\notin x)[/itex] and if this holds for all [itex]x[/itex], then it certainly holds for a particular choice of [itex]x[/itex]. Let this choice be [itex]x = R[/itex], then you immediately get [itex]R\in R \leftrightarrow R\notin R[/itex], which is a contradiction.

This proof uses only one single deductive step: From [itex]\forall x P(x)[/itex] deduce [itex]P(y)[/itex] for any [itex]y[/itex] you like (applied to the special case that [itex]P(x)[/itex] is [itex]x\in R \leftrightarrow x\notin x[/itex]). If you deny Russels paradox, you also deny that this single deductive step is valid reasoning.

 

[DanTeplitskiy]

Dear Rubi,

This proof uses only one single deductive step

I know well that in predicate logic it is one deductive step. I mention it on page 7 of my paper. :)

Rubi, when I talk about "contradictory premises" logical error I talk about logical arguments in Russell's paradox. See my message above (message #20). I do not deny the axioms of predicate logic :)On the fact that we get Russell's paradox in predicate calculus let me quote my own paper here:

It is well-known that we can get R ∈ R ↔ R ∉ R from ∀ x (x ∈ R ↔ x ∉ x) in predicate calculus.
Namely, through a quantifier axiom usually referred as Universal instantiation:
∀ x A → A (a/x), for some term a and where A (a / x) is the result of substituting a for all free
occurrences of x in A.

That is, it can be said that in predicate calculus Russell’s paradox is acquired as the conclusion R ∈ R ↔ R ∉ R from one premise - ∀ x (x ∈ R ↔ x ∉ x).

Though, we can easily see that ultimately R ∉ R is entailed by the set of two sentences: ∀ x (x ∈ R ↔ x ∉ R) and R ∈ R, while R ∈ R is entailed by the set of two sentences: ∀ x (x ∈ R ↔ x ∉ R) and R ∉ R. Indeed, let us use the importation:

∀ x (x ∈ R ↔ x ∉ x) → (R ∈ R → R ∉ R) is equivalent to (∀ x (x ∈ R ↔ x ∉ x) ∧ R ∈ R) → R ∉ R
∀ x (x ∈ R ↔ x ∉ x) → (R ∉ R → R ∈ R) is equivalent to (∀ x (x ∈ R ↔ x ∉ x) ∧ R ∉ R) → R ∈ R.

As we could clearly see from the proof of Contradictory premises theorem,
these pairs of sentences contain the "hidden" contradiction within each pair.

Yours,

Dan

 

[DanTeplitskiy]

Dear Stephen Tashi,

Your arguments show that R cannot exist because assuming it exists, contradicts the axiom that "If S is a set and x is an element then either x is a member of S or x is not a member os S".

My arguments are to demonstrate that there is a logical error in each of the two logical arguments in Russell's paradox. :) Please see my message#20 in this thread :)

Yours,

Dan

 

[rubi]

Dear Dan,

I don't understand what you are trying to say. I have shown (using conventional predicate logic) that the statement [itex]\exists R\forall x (x\in R \leftrightarrow x\notin x)[/itex] implies a contradiction ([itex](\exists R\forall x (x\in R \leftrightarrow x\notin x))\rightarrow\bot[/itex]).

Now i need you to answer the following question: Do you agree that this derivation is correct?

If the answer is no: Which step in the derivation is wrong? (You should be able to answer this question in one sentence, because you only need to point at one single step!) To make it even more clear, here are is the derivation again:
1. [itex]\exists R\forall x (x\in R \leftrightarrow x\notin x)[/itex]
2. [itex]\forall x (x\in S \leftrightarrow x\notin x)[/itex] (from 1; if there exists such an R, i can assign it to an unused variable)
3. [itex]S\in S \leftrightarrow S\notin S[/itex] (from 2; if it holds for all x, it also holds for R)
4. [itex]\bot[/itex] (from 3; it's a contradiction)
5. [itex](\exists R\forall x (x\in R \leftrightarrow x\notin x))\rightarrow\bot[/itex] (by the decuction theorem)

If the answer is yes, you have the following options: Either you accept that [itex]\exists R\forall x (x\in R \leftrightarrow x\notin x)[/itex] is false or you are willing to accept contradictory statements in your axiomatic system. Which one of these options do you choose?

 

[DanTeplitskiy]

Dear Rubi,

You correctly applied the axiom of predicate calculus.

Have you read my message #20 above? Have you understood everyithing there?

Yours,

Dan

 

[rubi]

Have you read my message #20 above?

Yes, i have.

Have you understood everyithing there?

No i haven't. That's why i want you to answer my question. I'm not willing to spend more time on this if you refuse to answer my questions.

 

[DanTeplitskiy]

Dear Rubi,

I answered your first question as well as I could.
"∃R∀x(x∈R↔x∉x) is false" - this is correct, of course :) There is no such R :). I am not making any alternative "my" axiomatic system )))).

What exactly (which line) have you failed to understand in my message#20?

Yours,

Dan

 

[rubi]

"∃R∀x(x∈R↔x∉x) is false" - this is correct, of course :) There is no such R :).

Then we are done, because we came to the same conclusion as Russel did. There is no way to define the set [itex]R = \{x:x\notin x\}[/itex] in usual predicate logic without implying a contradiction. This is exactly Russels paradox.

What exactly (which line) have you failed to understand in my message#20?

"Both assumptions (R∈R is true and R∈R is false) contradict the definition of R" doesn't make sense, because you can't contradict a definition. Your use of language is wrong.

 

[DanTeplitskiy]

Dear Rubi,

Then we are done, because we came to the same conclusion as Russel did. There is no way to define the set [itex]R = \{x:x\notin x\}[/itex] in usual predicate logic without implying a contradiction. This is exactly Russels paradox.

Russell's paradox is like:
Let R be the set of all sets that are not members of themselves. Then it is a member of itself if and only if it is not a member of itself. - paradoxical incoherence.

The usual conclusion that we make from the paradox is that there is no such R.

"Both assumptions (R∈R is true and R∈R is false) contradict the definition of R" doesn't make sense, because you can't contradict a definition. Your use of language is wrong.

"Let ABC be a triangle. Suppose it has four angles." :) - well, this is a sort of joke ))
If seriously, enter "contradicts the definition" into google serach and find some math texts ;)

Yours,

Dan

 

[rubi]

Russell's paradox is like:
Let R be the set of all sets that are not members of themselves. Then it is a member of itself if and only if it is not a member of itself. - paradoxical incoherence.

We proved this and you already said that you accept the derivation.

The usual conclusion that we make from the paradox is that there is no such R.

That is true. If a statement leads to a contradiction, then it must be false. You also agreed on this.

If you really agree that one can derive a contradiction from the assumption of the existence of the Russel set, then I don't see what problems remain.

"Let ABC be a triangle. Suppose it has four angles." :) - well, this is a sort of joke ))
If seriously, enter "contradicts the definition" into google serach and find some math texts ;)

What I was trying to say was that it is irrelevant for the proof whether something contradicts the definition of the Russel set. [itex]R\in R[/itex] is a well-formed formula and thus you are allowed to use it according to the rules of predicate logic.

 

[DanTeplitskiy]

Dear Rubi,

I said you correctly used the axiom of predicate logic ))

Actually, I suspect you did not even try to read my message#20 after the sentence "Both assumptions (R∈R is true and R∈R is false) contradict the definition of R (Let R be the set of all sets that are not members of themselves)"
Which makes further discussion with you senseless ))

Yours,

Dan

 

[rubi]

I said you correctly used the axiom of predicate logic

So either you agree that the existence of R implies [itex]R\in R\leftrightarrow R\notin R[/itex] or you claim that using predicate logic is not a valid way of reasoning. I have still not found out, which of these two options you advocate.

Actually, I suspect you did not even try to read my message#20 after the sentence "Both assumptions (R∈R is true and R∈R is false) contradict the definition of R (Let R be the set of all sets that are not members of themselves)"
Which makes further discussion with you senseless

I have read it, but i don't see what the problem is. Let's look at this part for example:

Argument 1
Premise 1: Let R be the set of all sets that are not members of themselves R = {x: x∉x}
Premise 2 (assumption): Suppose R ∈ R.
Conclusion: Then, according to its definition, R ∉ R.

We can easily phrase this in predicate logic:
1. [itex]\forall x (x \in R \leftrightarrow x\notin x)[/itex] (this is your first premise)
2. [itex]R \in R[/itex] (this is your second premise)
3. [itex]R \in R \leftrightarrow R\notin R[/itex] (from 1 by setting [itex]x=R[/itex])
4. [itex]R \notin R[/itex] (from 2 and 3 by modus ponens)

This is completely valid reasoning. [itex](\forall x (x \in R \leftrightarrow x\notin x))\rightarrow R\notin R[/itex] is a theorem.