# Russell's paradox and logical errors in the proof

1. Apr 21, 2013

### DanTeplitskiy

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.

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.

Dan

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

Last edited: Apr 21, 2013
2. Apr 21, 2013

### ImaLooser

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

3. Apr 21, 2013

### DanTeplitskiy

Dear ImaLooser,

Have you read pages 2 through 4 of the paper?

Yours,

Dan

4. Apr 21, 2013

### Fredrik

Staff Emeritus
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".

5. Apr 21, 2013

### DanTeplitskiy

Dear Fredrik,

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

Yours,

Dan

6. Apr 21, 2013

### Fredrik

Staff Emeritus
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.

7. Apr 21, 2013

### 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

Last edited: Apr 21, 2013
8. Apr 21, 2013

### Fredrik

Staff Emeritus
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.

Last edited: Apr 21, 2013
9. Apr 21, 2013

### 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

10. Apr 21, 2013

### Fredrik

Staff Emeritus
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 $R\in R$.

11. Apr 21, 2013

### DanTeplitskiy

Dear Fredrik,

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

Yours,

Dan

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

12. Apr 21, 2013

### 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...

13. Apr 21, 2013

### Fredrik

Staff Emeritus
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 $R\in R$.

14. Apr 21, 2013

### 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

15. Apr 21, 2013

### DanTeplitskiy

Dear Fredrik,

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

Yours,

Dan

16. Apr 21, 2013

### Fredrik

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

17. Apr 21, 2013

### 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.

Yours,

Dan

Last edited: Apr 21, 2013
18. Apr 21, 2013

### Stephen Tashi

What do mean by "contradict the definition"? I think you mean that no set exists that satisfies the definition of your paper's $R$. 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.

19. Apr 21, 2013

### DanTeplitskiy

Dear Stephen Tashi,

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

20. Apr 28, 2013

### 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. :)

Last edited: Apr 28, 2013