The other side of Russell's paradox

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Re: The other side of Russell's paradox

Postby Certainly real » Tue May 11, 2021 8:41 pm

obsrvr524 wrote:I wouldn't say that Existence is a member of itself (that confuses the language). Existence IS itself.


Yes, Existence is itself, but so is the collection of books I own. The collection of books I own is a member of Existence. Existence is a member of Existence. You say saying Existence is a member of itself confuses the language, whereas I think it is a necessary part of language without which the language is not as logically precise as it could be.

What you say is only true IF your list is a list of references (or titles) to the lists including a reference to itself.


That is what I was saying.

Your 5th list cannot contain itself except as only a title or reference. That has been this whole argument from the beginning.


I don't think I suggested otherwise. Maybe you were under impression that I was based on our previous discussion of a folder of all folders (wherein which I did say the folder contains itself, as opposed to just referencing itself).

In any case, as far as I can see, Russell's paradox is solved. This absurdity of rejecting an absolute universal set needs to end.
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Re: The other side of Russell's paradox

Postby obsrvr524 » Tue May 11, 2021 9:05 pm

Certainly real wrote:
What you say is only true IF your list is a list of references (or titles) to the lists including a reference to itself.


That is what I was saying.

So do you agree that the set \(A = \{1, 2, 3, A \}\) cannot exist?

Certainly real wrote:This absurdity of rejecting an absolute universal set needs to end.

I think that is just a part of the cancel culture politics. They don't care that it doesn't make sense.
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Re: The other side of Russell's paradox

Postby Magnus Anderson » Tue May 11, 2021 10:07 pm

Certainly real wrote:Yes, Existence is itself, but so is the collection of books I own. The collection of books I own is a member of Existence. Existence is a member of Existence.


I understand and agree with the first two sentences.

As for the third, I am not sure I understand it. There are two ways to interpret it. The first is "That which physically exists is a member of the set of everything that physically existed, that physically exists and that will physically exist". If that's what you mean then I agree (but that would mean the first occurence of the word "existence" in your statement has a different meaning than the second.) But if what you actually mean is "The set of everything that existed, exists and will exist is a member of itself" then I disagree.

I also think it's necessary to note that the word "existence" normally means "physical existence" but that it can also mean "conceptual existence". Unicorns are part of the latter but are not part of the former. Round squares, on the other hand, do not belong to any. What about your notion of existence (which is capitalized as "Existence")? Is it a reference to physical existence, conceptual existence, physical-and-conceptual existence or something else?

It's hard to follow.

Russell's paradox arises from the belief that you cannot have a set of all sets that are not members of themselves.


I think it's more accurate to say that Russell's Paradox shows that the set of all normal sets cannot exist (i.e. that it's an expression that contains a contradiction.)

Russell's Paradox:

1) A normal set is defined as any set that does not contain itself.

2) An abnormal set is defined as any set that contains itself.

3) Any given set is either normal or abnormal i.e. it cannot be neither of those.

4) Let R be the set of all normal sets.

5) If R is a normal set, then it contains itself, which means it's an abnormal set.

6) If R is an abnormal set, then it does not contain itself, which means it's a normal set.

7) Therefore, the set of all normal sets is an oxymoron (i.e. an expression containing a contradiction.)

As you can see, Russell's Paradox concludes that the set of all normal sets is an oxymoron. In other words, it does not use it as a premise.
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Re: The other side of Russell's paradox

Postby obsrvr524 » Wed May 12, 2021 8:03 am

Magnus Anderson wrote:I also think it's necessary to note that the word "existence" normally means "physical existence" but that it can also mean "conceptual existence". Unicorns are part of the latter but are not part of the former. Round squares, on the other hand, do not belong to any. What about your notion of existence (which is capitalized as "Existence")? Is it a reference to physical existence, conceptual existence, physical-and-conceptual existence or something else?

It's hard to follow.

Russell's paradox arises from the belief that you cannot have a set of all sets that are not members of themselves.


I think it's more accurate to say that Russell's Paradox shows that the set of all normal sets cannot exist (i.e. that it's an expression that contains a contradiction.)

Russell's Paradox:

1) A normal set is defined as any set that does not contain itself.

2) An abnormal set is defined as any set that contains itself.

3) Any given set is either normal or abnormal i.e. it cannot be neither of those.

4) Let R be the set of all normal sets.

5) If R is a normal set, then it contains itself, which means it's an abnormal set.

6) If R is an abnormal set, then it does not contain itself, which means it's a normal set.

7) Therefore, the set of all normal sets is an oxymoron (i.e. an expression containing a contradiction.)

As you can see, Russell's Paradox concludes that the set of all normal sets is an oxymoron. In other words, it does not use it as a premise.

I agree with all of that but another way to look at it -

Consider -

\(NamesOfSets = \{ A \space set, \space B \space set, \space C \space set, \space NamesOfSets \space set \}\) - Abnormal Set

I think that is what an "abnormal set" is supposed to be - a set of references or names that includes a reference to itself. That is why I had asked for an example.

    \(AllNormalSets = \{ A \space set, \space B \space set, \space C \space set \}\) - XX Normal but Incomplete

    \(AllNormalSets = \{ A \space set, \space B \space set, \space C \space set, \space AllNormalSets \space set \}\) - XX Complete but Abnormal

    \(AllNormalSets = \{ Irrational \space Concept \}\) - Neither Normal nor Abnormal

    \(Universe = \{AllNormalSets, \space AllAbnormalSets\}\)

    \(Universe = \{ Irrational \space Concept, \space AllAbnormalSets \}\) - ??

That should seem like an irrational conclusion and paradox - Russel's Paradox.
But it is just a wording game.

And there are two sentences in this one post that explains the entire resolve. Both are required. Can you spot them?
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Re: The other side of Russell's paradox

Postby Certainly real » Wed May 12, 2021 8:31 am

obsrvr524 wrote:So do you agree that the set \(A = \{1, 2, 3, A \}\) cannot exist?


With reference to what Magnus defined as volume, I agree that it is clearly absurd for A to have volume x and volume 2x at the same time. 2 is not equal to 2 times 2.

Having said that, there are two ways in which set \(A = \{1, 2, 3, A \}\) exists:

p) The set A encompasses items 1, 2, 3, and A. Much like how list A, lists the lists 1, 2, 3, and itself (A). I don't think you disagree with this. But I think you disagree with the following:

q) The A in set \(A = \{A...1, 2, 3...A \}\) is such that A contains A, and A is contained by A. A... implies that no number begins before A, and ...A implies that no number comes after A.

Take x as your starting point on a computer. If your starting point on a computer consists of folders ...1, 2, 3..., and A, and you click A, you get ...1, 2, 3..., and A. You click A again, the same thing happens again. This happens ad infinitum. Going back to x, if you try to come out of the folder that you are in (which is A), by going up a folder to the folder that encompasses the folder that you are in, you find yourself with ...1, 2, 3..., and A. Go up another layer in an attempt to get the to the root folder, and you get the same thing again. This happens ad infinitum. A is that which is between any and every identified folder, number, or thing. A = infinity.

Given our previous discussions, you are strongly in opposition to q. Note that I am not saying A has volume x plus volume x + x + x...ad inifnitum. I am saying A's volume is infinite. So it's not a case of one thing having two different volumes at the same time. A is that which is between any and every identified item. You can either interpret set \(A = \{1, 2, 3, A \}\) as absurd by viewing A as being non-infinite, or you can interpret as non-absurd by viewing A as infinite.
set \(A = \{1, 2, 3, A \}\) is representative of how A is fully between items 1, 2, and 3.
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Re: The other side of Russell's paradox

Postby Certainly real » Wed May 12, 2021 8:53 am

Magnus Anderson wrote:I also think it's necessary to note that the word "existence" normally means "physical existence" but that it can also mean "conceptual existence". Unicorns are part of the latter but are not part of the former. Round squares, on the other hand, do not belong to any. What about your notion of existence (which is capitalized as "Existence")? Is it a reference to physical existence, conceptual existence, physical-and-conceptual existence or something else?


All. Numbers exist as numbers. Imaginary humans exist as imaginary humans. Real humans exist as real humans (by real here I am talking about our standard of reality. I am talking about the things that we describe as "physically seeing in our waking reality"). All such things exist in Existence (as opposed to non-Existence). They exist as a result of Existence. Existence exists as a result of Existence. This is why existentially speaking, I describe the latter as a member of itself, whilst I describe the former as not a member of itself.

But if what you actually mean is "The set of everything that existed, exists and will exist is a member of itself" then I disagree.


The way you've worded implies three different things (the past, the present, and the future, all of which are members of Existence because they are not members of non-Existence). So I can see why you don't see the above as a member of itself.

I think it's more accurate to say that Russell's Paradox shows that the set of all normal sets cannot exist (i.e. that it's an expression that contains a contradiction.)


Take V to be any set (the V of all Vs = the set of all sets). Take V' to be any set other than the set of all sets. Take -V to be any set that is not a member of itself. Take -V' to be any V' set that is not a member of itself.

No V, or V’, or -V, can encompass all -Vs and no other set. But a V can encompass all -Vs and another set. Note that no V’ can encompass all -Vs because whilst the V’ of all V’s encompasses all -V’s, it will not encompass all -Vs. Only the V of all Vs encompasses all -Vs.

Two Vs encompass all V’s (as well as all -V’s):

One (which is a V') encompasses all V's and nothing more. The other (which is not a V') encompasses all V's and something more. The latter is the set of all sets (the V of all Vs), the former is the V' of all V's (the not-the-set-of-all-sets set of all not-the-set-of-all-sets sets). Thus, only one V can encompass all Vs and nothing more (the V of all Vs). Only one V' can encompass all V's and nothing more (the V' of all V’s).

The above shows that whilst there can be no -V that encompasses all -Vs, there is a V that encompasses all -Vs. Whilst there can be two Vs that encompass all -V’s, there can only be one V’ that encompasses all -V’s.

Russell asks, is the set of all sets that are not members of themselves, a member of itself? There are two ways to interpret what he is asking:

1) Is the V of all -Vs a member of itself?
2) Is the -V of all -Vs a member of itself?

2 is an absurd question. 1 is not an absurd question. The answer to 1 is yes because the V of all Vs is the V that encompasses all -Vs, and it is a member of itself (because it encompasses itself). If Russell would disagree with this and say he wants a set that only encompasses all -Vs and nothing more, then Russell wants a round square (I don't think that's what he wanted).

If I asked is the -V' of all -V's a member of itself, the answer is: Such a question is absurd. If I asked is the V' of all -V's a member of itself, the answer is yes. Alternatively, there's no such thing as the V' of all -V's. There is the V' of all V's, and only it encompasses the all -V's and nothing more. Depends on how you want to word it.
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Re: The other side of Russell's paradox

Postby obsrvr524 » Wed May 12, 2021 10:17 am

Certainly real wrote:Take x as your starting point on a computer. If your starting point on a computer consists of folders ...1, 2, 3..., and A, and you click A, you get ...1, 2, 3..., and A. You click A again, the same thing happens again. This happens ad infinitum. Going back to x, if you try to come out of the folder that you are in (which is A), by going up a folder to the folder that encompasses the folder that you are in, you find yourself with ...1, 2, 3..., and A. Go up another layer in an attempt to get the to the root folder, and you get the same thing again. This happens ad infinitum. A is that which is between any and every identified folder, number, or thing. A = infinity.

The volume or size of that A - is undefinable. You cannot say that it is infinite because if it was infinite then it would have to be twice infinite. And if it was twice infinite then it would have to be thrice infinite. And so on.

And something you might be missing is that all of the appropriate folders have to already exist while every A is exactly the same as every other A.

That particular A cannot exist conceptually or physically. It has no definable volume.

Certainly real wrote: Note that I am not saying A has volume x plus volume x + x + x...ad inifnitum.

I think that you are saying that.
Certainly real wrote:I am saying A's volume is infinite. So it's not a case of one thing having two different volumes at the same time.

But that is incorrect. If its volume was infinite then its volume would have to be twice infinite.
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              You have been observed.
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Re: The other side of Russell's paradox

Postby obsrvr524 » Wed May 12, 2021 10:31 am

Certainly real wrote:the past, the present, and the future, all of which are members of Existence because they are not members of non-Existence

I don't believe that is true. The future and the past do not exist. If they existed then they would be the present.
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              You have been observed.
    Though often tempted to encourage a dog to distinguish color I refuse to argue with him about it
    It's just same Satanism as always -
    • separate the bottom from the top,
    • the left from the right,
    • the light from the dark, and
    • blame each for the sins of the other
    • - until they beg you to take charge.
    • -- but "you" have been observed --
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Re: The other side of Russell's paradox

Postby Magnus Anderson » Wed May 12, 2021 10:54 am

obsrvr524 wrote:\Consider -

\(NamesOfSets = \{ A \space set, \space B \space set, \space C \space set, \space NamesOfSets \space set \}\) - Abnormal Set

I think that is what an "abnormal set" is supposed to be - a set of references or names that includes a reference to itself. That is why I had asked for an example.


There are three different kinds of sets that must not be confused with each other.

1) There is a type of set that contains a reference to itself. This is the kind of set you seem to be (but I am not exactly sure you are) talking about. This is certainly what I meant when I spoke of "a reference to a set". A reference to a set is an object, not necessarily a set, that represents that set. It can be any object e.g. it can be a word. A set of all references (including a reference to itself) is thus not necessarily an oxymoron. \(A = \{1, 2, 3, refA\}\) is not an oxymoron because \(refA\) is just a word representing \(A\), not a set identical to \(A\).

2) There is a type of set that is a subset of itself. Every set belongs to this type -- every set is a subset of itself. The definition of subset relation is "\(A\) is a subset of \(B\) if and only if every element in \(A\) is present in \(B\)". There is no need for \(B\) to be bigger than \(A\). \(A = \{1, 2, 3\}\) is thus a subset of itself and there is nothing wrong with this.

3) There is a type of set that contains an element that is a set identical to the set it belongs to. \(A = \{1, 2, 3, A\}\) is an example. The containing \(A\) and the contained \(A\) are two identical sets. This is what I think is meant by "abnormal set". The containing \(A\) and the contained \(A\), however, cannot be identical because 1) if two sets are identical their volume is also identical, and 2) the volume of the contaning \(A\) is greater than the volume of the contained \(A\).
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Re: The other side of Russell's paradox

Postby obsrvr524 » Wed May 12, 2021 11:15 am

Magnus Anderson wrote:3) There is a type of set that contains an element that is a set identical to the set it belongs to. \(A = \{1, 2, 3, A\}\) is an example. The containing \(A\) and the contained \(A\) are two identical sets. This is what I think is meant by "abnormal set". The containing \(A\) and the contained \(A\), however, cannot be identical because 1) if two sets are identical their volume is also identical, and 2) the volume of the contaning \(A\) is greater than the volume of the contained \(A\).

That is the one I thought you agreed was an oxymoron because the internal A cannot be identical to the external A yet in logic they must be exactly identical. It is claiming that A is larger than A.

So that isn't a set at all. It isn't "abnormal". It is inconceivable.
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              You have been observed.
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    It's just same Satanism as always -
    • separate the bottom from the top,
    • the left from the right,
    • the light from the dark, and
    • blame each for the sins of the other
    • - until they beg you to take charge.
    • -- but "you" have been observed --
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Re: The other side of Russell's paradox

Postby Magnus Anderson » Wed May 12, 2021 6:01 pm

Yes, it is an oxymoron. But also, it is an abnormal set. The definition of an abnormal set is a set that contains itself as an element. See here.
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Re: The other side of Russell's paradox

Postby obsrvr524 » Wed May 12, 2021 7:32 pm

Magnus Anderson wrote:Yes, it is an oxymoron. But also, it is an abnormal set. The definition of an abnormal set is a set that contains itself as an element. See here.

Doesn't it have to qualify as a set before it can qualify as an abnormal set? :D
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              You have been observed.
    Though often tempted to encourage a dog to distinguish color I refuse to argue with him about it
    It's just same Satanism as always -
    • separate the bottom from the top,
    • the left from the right,
    • the light from the dark, and
    • blame each for the sins of the other
    • - until they beg you to take charge.
    • -- but "you" have been observed --
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Re: The other side of Russell's paradox

Postby Magnus Anderson » Wed May 12, 2021 7:34 pm

Certainly real wrote:q) The A in set \(A = \{A...1, 2, 3...A \}\) is such that A contains A, and A is contained by A. A... implies that no number begins before A, and ...A implies that no number comes after A.


What does \(\{A...1, 2, 3...A\}\) mean? Note that mathematical sets have no order. \(\{1, 2, 3\}\) has no first element, no second element and no third element. Elements do not come before and/or after other elements. Perhaps you're speaking of a sequence that contains itself twice -- once at the start and once at the end. If that's what you're doing then I don't understand why. Me and obsrvr524 are claiming that self-containing sets are logically impossible. Of course, so are self-containing sequences, but . . . why bring them up?
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Re: The other side of Russell's paradox

Postby Magnus Anderson » Wed May 12, 2021 7:46 pm

obsrvr524 wrote:
Magnus Anderson wrote:Yes, it is an oxymoron. But also, it is an abnormal set. The definition of an abnormal set is a set that contains itself as an element. See here.

Doesn't it have to qualify as a set before it can qualify as an abnormal set? :D


I am merely trying to say that the term "abnormal set" means "a set that contains itself as an element" as opposed to "a set that contains a reference to (or a symbol of) itself as an element" and "a set that is a subset of itself".

As for \(A = \{1, 2, 3, A\}\), it is both an abnormal set and a normal set, which is a contradiction, which is why the possibility of such a set existing in physical reality is zero. Note that I am talking about what that symbol can be used to represent (= the meaning of the symbol) and NOT what those things that can be represented with it are.
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Re: The other side of Russell's paradox

Postby MagsJ » Wed May 12, 2021 10:17 pm

_
Isn’t this basic maths?

Do please correct me if I am wrong, but I do think I am right.

Where is the paradox?
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Re: The other side of Russell's paradox

Postby obsrvr524 » Thu May 13, 2021 9:10 am

MagsJ wrote:_
Isn’t this basic maths?

Do please correct me if I am wrong, but I do think I am right.

Where is the paradox?

Not maths. I think it is just very very subtle abuse of logic that leads to a conundrum where a certain type of set has to be declared as both one type and the opposite type.

It's actually just a subtle wording game - a riddle.

Magnus Anderson wrote:As for \(A = \{1, 2, 3, A\}\), it is both an abnormal set and a normal set

It is neither because it is not a set at all. It is a type of "square-circle" (which is not a shape at all).
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              You have been observed.
    Though often tempted to encourage a dog to distinguish color I refuse to argue with him about it
    It's just same Satanism as always -
    • separate the bottom from the top,
    • the left from the right,
    • the light from the dark, and
    • blame each for the sins of the other
    • - until they beg you to take charge.
    • -- but "you" have been observed --
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Re: The other side of Russell's paradox

Postby Magnus Anderson » Thu May 13, 2021 11:16 am

obsrvr524 wrote:It is neither because it is not a set at all. It is a type of "square-circle" (which is not a shape at all).


It is both. It is defined as a set that contains \(1\), \(2\), \(3\) and itself. It is explicitly stated that it is abnormal and implicitly that it is normal. By saying that it is neither as well as by saying that it is not a set, we are contradicting its definition.

As I said in my previous post, I am talking about what the expression means (i.e. what that expression can be used to represent regardless of whether or not such things exist) and not what the things that can be represented by that expression are. You are focusing on the latter. Obviously, nothing can be represented by the expression "A set that contains itself". Thus, those things that can be represented by that expression are certainly not sets -- or anything else -- because they do not exist so they can't be anything.
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Re: The other side of Russell's paradox

Postby obsrvr524 » Thu May 13, 2021 7:06 pm

I don't think that how obvious something is changes the truth of what it is. So I think we will just have to agree to maintain our own bubbles on that issue. O:)
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              You have been observed.
    Though often tempted to encourage a dog to distinguish color I refuse to argue with him about it
    It's just same Satanism as always -
    • separate the bottom from the top,
    • the left from the right,
    • the light from the dark, and
    • blame each for the sins of the other
    • - until they beg you to take charge.
    • -- but "you" have been observed --
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