Resolution of Russell's and Cantor's paradoxes

[micromass]

Dear Micromass,

If we could have a proof that "there is a greatest prime" (by "proof" I mean the widely accepted one - otherwise it is not a "proof" for math) we would have it! But we do not and you know it quite well!

Why don't we have such a proof in mathematics, Micromass?

Yours,

Dan

Well, "IF there exists a greatest prime THEN there exists a greatest prime" is true. This does not imply the existence of a greatest prime.

 

[DanTeplitskiy]

I was responding to



and I think that was micromass' statement.

Dear Disregardthat,

Sorry pal. I did not get it was not for me.

Yours,

Dan

P. S. By the way, do you agree with the following: If we get something using contradiction as a starting point it is not a proof for math as we can get literally anything this way (see "principle of explosion" )?

 

[DanTeplitskiy]

Well, "IF there exists a greatest prime THEN there exists a greatest prime" is true. This does not imply the existence of a greatest prime.

Dear Micromass,

Can I take it for the fact that you agree that if we get something using contradiction as a starting point it is not a proof for math as we can get literally anything this way (see "principle of explosion" )?

Yours,

Dan

 

[disregardthat]

P. S. By the way, do you agree with the following: If we get something using contradiction as a starting point it is not a proof for math as we can get literally anything this way (see "principle of explosion" )?

A contradiction is meaningless, just as asserting a false statement. We can however assume any statement we want, and by falling into contradiction (which we immidiately do by assuming a contradiction), we appeal to consistence and conclude that the assumption is wrong (it cannot be asserted).

 

[DanTeplitskiy]

A contradiction is meaningless, just as asserting a false statement. We can however assume any statement we want, and by falling into contradiction (which we immidiately do by assuming a contradiction), we appeal to consistence and conclude that the assumption is wrong (it cannot be asserted).

Dear Disregardthat,

I totally agree with you.

I would like to have your opinion on the following matter:

If we take the following statement:

“Dan is completely legless. If his right ankle is bleeding he should be taken to the nearest hospital
for completely legless people.”

Can we consider "...he should be taken to the nearest hospital for completely legless people." to be logically ungrounded for the following reasons:

1) To make a logically grounded conclusion “...he should be taken to the nearest hospital for completely legless people” two conditions should be met: 1. “completely legless Dan” has some health problem - this condition is met under the assumption “his right ankle is bleeding” and 2. “completely legless Dan” is completely legless - this condition is not met as under the assumption “his right ankle is bleeding” “completely legless Dan” is not actually completely legless.

2) Technically, when we get some contradiction (e. g. “Dan is completely legless” And “His right
ankle is bleeding ⇒ Dan is not completely legless”) we can only say that we got a contradiction –
we should not continue the same line of reasoning further any other way (otherwise we could get literally anything next – see “principle of explosion”).


Yours,

Dan

 

[disregardthat]

Yes, I agree with what you are saying. The condition cannot be met. Given the information that he is legless and his right ankle is bleeding leaves you with non-sense. But the statement itself is not a contradiction (if I understand you right you don't actually say that).

However the condition "Dan has a health problem" seem unecessary to me with regard to the logical statement itself. It can also be imagined that people without health problems must go to the hospital (maybe a suspected health problem).

 

[DanTeplitskiy]

Yes, I agree with what you are saying. The condition cannot be met. Given the information that he is legless and his right ankle is bleeding leaves you with non-sense. But the statement itself is not a contradiction (if I understand you right you don't actually say that).

However the condition "Dan has a health problem" seem unecessary to me with regard to the logical statement itself. It can also be imagined that people without health problems must go to the hospital (maybe a suspected health problem).

Dear Disregardthat,

Thanks a lot for reply!

Could you please answer two questions of mine:

1. Do I get it right that by saying "But the statement itself is not a contradiction..." you mean generally the principle of explosion: if we have a contradiction - Dan legless And Dan not legless - we can get anything then - even "Santa Clause exists"? Or you meant something else by that?

2. Regrding the necessity of the condition "Dan has a health problem": does “Dan is completely legless. If his right ankle is severely bleeding he should be taken to the nearest emergency department for completely legless people.” "repairs" the example for you ?

Anyway, the above example is only a simple analogy to the standard formulation of Russell's paradox .

Thanks a lot in advance!

Yours,

Dan

 

[DanTeplitskiy]

Yes, I agree with what you are saying. The condition cannot be met. Given the information that he is legless and his right ankle is bleeding leaves you with non-sense. But the statement itself is not a contradiction (if I understand you right you don't actually say that).

However the condition "Dan has a health problem" seem unecessary to me with regard to the logical statement itself. It can also be imagined that people without health problems must go to the hospital (maybe a suspected health problem).

Dear Disregardthat and other guys,

My point in regard to Russell's paradox is the following one:

"If R is included in itself then R is not included in itself. If R is not included in itself then R is included in itself"

to make a logically grounded conclusion “...then R is not included in itself” two conditions should be met: 1. R does not have the required property - this condition is met under the assumption “if R includes itself” and 2. R is a set that includes all those sets that are not included in themselves and only them - this condition is not met as under the assumption “if R includes itself” R includes a set that is included in itself (R itself is such a set).


to make a logically grounded conclusion “...then R is included in itself” two conditions should be met: 1. R has the required property - this condition is met under the assumption “if R does not include itself” and 2. R is a set that includes all those sets that are not included in themselves and only them - this condition is not met as under the assumption “if R does not include itself” R does not include a set that is not included in itself (R itself is such a set).

Technically, when we get a contradiction (e. g. “R is a set that includes all those sets that are not included in themselves and only them” And “R includes itself ⇒ R is not a set that includes all those sets that are not included in themselves and only them”) we can only say that we got a contradiction – we should not continue the same line of reasoning further any other way (otherwise we could get literally anything next – see “principle of explosion”). Though, it is not the main point here. The main point here is the following one: it is only when we do not notice that one of the conditions required to draw some conclusion is not met, we may still draw it in our line of reasoning.

Comments are welcome especially from Disregardthat.

Thanks a lot in advance!

Yours,

Dan

 

[DanTeplitskiy]

Now you seem to be missing my point. My point is that there is no logical problem with treating R as {x: x∉x} throughout, even though R≠{x: x∉x}.

Dear Citan,

First of all I am sorry for the late response.

When we reason like “let R = {x: x∉x} then R ∈ R ↔ R ∉ R” we talk about different things using one and the same identifier R (without realizing the fact though if we consider it as a correct reasoning). When we say “let R = {x: x∉x}” we talk about R = {x: x∉x}. When we say “if R ∈ R/R ∉ R...” (talking about R as about an element with some property), we actually talk about “another” R - R≠ {x: x∉x}. When we say “...then R (as an element) ∉ R(as a set)” or “...then R (as an element) ∈ R (as a set)” we talk about both Rs: R ≠ {x: x∉x} (as an element) and R = {x: x∉x} (as a set).

That is, if one treats R as {x: x∉x} throughout, even though R≠{x: x∉x} one breaks the law of identity (R=R).

The point is we can not consider something as "legitimate" if we get it with breaking the laws of logic.

Yours,

Dan

P.S. I improved the paper. Hope it will be better understood now.

https://docs.google.com/viewer?a=v&...YTZlMy00NDJhLWJjN2MtMDAzNDUzOWQ2Y2Ew&hl=en_US

 

[DanTeplitskiy]

Dear gyus,

Though risking to be regarded as a boring one:

It is not an everyday thing that a long-lasting problem that had a serious impact on math is solved (in regard to paradoxical problems maybe it is better to use the word "resolved" to avoid confusion with axiomtic "solutions" - "ways around" paradoxes).

I improved the paper thoroughly. As to me, the current version of the proof of Proposition2 (the proof is pp. 12-13, the proposition itself is p. 13) - the root part of the paper, is quite understandable and almost "easy reading" thing. If you are not sure what the term "reslove" really means it would make sense to glance through the end of p. 14 as well to get the idea better. Maybe I am wrong about "easy reading" thing. Anyway my English is far from perfect - I am quite aware of that. :shy:

https://docs.google.com/viewer?a=v&...YTZlMy00NDJhLWJjN2MtMDAzNDUzOWQ2Y2Ew&hl=en_US

I realize quite well that not receiving any positive comment on the paper itself does not mean that none of you understood it correctly and considered it the one actually containing the result I declare.

In case there are those who really got the idea, I am asking you to share your opinion because such feedback is really important to me - the paper was written mostly for amateurs' reading.

Yours,

Dan