The other side of Russell's paradox

Only Existence is exclusively a member of itself as an existent because it is not a member of anything other than itself (as an existent). All other existents are members of other than themselves (they are members of Existence, whilst they are not Existence Itself).

That would mean that I am Existence too. Whereas to my understanding, I belong to Existence. I am not Existence. If I cease to exist, Existence does not cease to exist.

I think it’s contradictory to say all existing sets are Existence. Only one existing set is Existence in my opinion. That is the set of all existents (which I have called ‘Existence’).

So you have said that a list of all lists is absurd. That would mean that if I have four lists in my room, and I wanted to make a fifth list titled “a list of all lists in my room”, this list cannot contain itself. Whilst I agree that this list cannot physically contain itself in addition to itself because it is physically itself once as opposed to twice, I do think that this list can list itself. I also think that this list is such that it lists items that are not members of themselves (because they are members of it), and it lists one item that is a member of itself (the list lists itself).

Where is the contradiction in the above? Rejecting the above is contradictory (and you have rejected a list that lists all lists).

I’ll try to present an argument in favor of the idea that a set that contains itself is a contradiction in terms.

However, in order to that, I’ll have to introduce a new term. I have to do this because I have trouble finding an existing term in mathematics that captures the concept that I want to express.

The notion that I am going to introduce will be represented by the expression “the volume of a set”.

obsrvr524 is using the word “size” to capture the same concept but that leads to a minor problem since the word “size” is already defined with respect to sets and it means something else – it refers to the number of elements. (A = {1, 2, 3, A}) has exactly (4) elements. No more and no less. The fact that (A) contains itself does not mean it has more than that.

So here we go. I’m going to kick it with a recursive definition.

Let the volume of a set be the sum of each member’s volume.

Let the volume of anything that is not a set be equal to (1).

For example, the volume of (B = {1, 2, 3}) is (3) because the set contains three elements, each one of which is not a set, which means, the volume of each element is exactly (1). Thus, the volume of (B) is (3 \times 1) which is (3).

The volume of (C = {1, 2, 3, {1, 2, 3}}), on the other hand, is equal to (6) even though its size (= cardinality) is equal to (4).

Let (n) be a number that represents the volume of (A = {1, 2, 3, A}).

What’s (n) equal to?

Let’s see. (A) contains (3) non-set elements and (1) set-element. This means the volume of (A) is (3 \times 1) plus the volume of the set-element it contains. We can simplify this to (3) plus the volume of the set-element. The set element it contains is (A) which means its volume is (n). Thus, the volume of (A) is (3 + n). This contradicts our earlier claim that the volume of (A) is (n). The volume of a set cannot be both (n) and (n + 3).

I think I get your point and I agree with it, but I don’t think it applies to what I’m saying.

Because we cannot ignore that when a list lists other lists, those items are not members of themselves (because they are members of this list whilst they are in reference to other lists), and that when a list lists itself, that item is a member of itself (because it is a member of this list whilst it is in reference to this list itself), I think it necessary for us to have a distinction between elements in a list that are not members of themselves, and an element in a list that is a member of itself (provided that a list has elements of such a nature). No list can contain more than one element as a member of itself (the same is true of sets).

So given the above, I don’t think your post highlights a contradiction in what I said.

The list of all lists is a member of itself because it is a list. Similarly, the set of all sets is a member of itself because it is a set. If you ascribe the volume x to the set of all sets, and then say the set of all sets is the volume x, plus it contains the volume x in addition to it being the volume x, then yes, I think that is contradictory. But I am not saying that the set of all sets = volume x plus volume x when I say that it is a member of itself.

No because you are NOT the entire collection. Existence is the entire collection as a whole.

The list of all lists (in your room) would - even in concept - be longer than itself - for the reason Magnus just pointed out.

• If A had a “volume” of 4 then A would have to have a volume of 7.
• And if A had a volume of 7 then A would have to have a volume 10.
• And if A had a volume of 10 then A would have to have a volume of 13.

.
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If it is a set then it has a volume/cardinality/size.
But any statement of the volume (or “size”) of A would be false.

The only kind of set that could fully contain itself would be a null set = zero volume.

I don’t think there is such a thing as a set that contains only itself. That too is an oxymoron. It’s just that my argument does not handle that case. Change “Let the volume of a set be the sum of each member’s volume” to “Let the volume of a set be the sum of each member’s volume plus one” and the problem should be resolved.

If an empty set contains itself then it is not an empty set.

It seems to me (though I am not exactly sure) that what you’re saying is that, when you say that a set contains another set that you actually mean that it contains a reference to another set and not the set itself. If that’s the case, then I agree: what I said is unrelated to what you said. But then, I’m not sure how what you said is related to Russell’s Paradox.

A set is whatever is in the set. Whatever is in the set is the set.
An empty set is the emptiness - not a container of it.

So every set has its entire contents within it which is having the set within it. It just can’t be a separate copy of the set.

That is the difference between a set containing itself inclusively vs exclusively.

I think he keeps going back and forth - confusing a set or list with a reference to it – sometimes.

There is a difference between a set and its contents (or members.) They are not one and the same. In mathematics, two sets are identical if and only if they have exactly the same elements. (\emptyset) (which stands for an empty set) and ({\emptyset}) are not identical sets. The former has no elements, the latter has one element.

So Existence is the collection that encompasses all collections then. Existence is the collection of all collections then. All other collections are members of Existence, whereas Existence is not a member of anything other than itself.

That’s because you’re focused on a different thing to me. I said if a list lists itself, it is a member of itself. When I have only have four lists in my room and I’m making a fifth titled “the list of all lists in my room”, that list will be finite. It will consist of five items. Four of which are not members of themselves, one of which is a member of itself.

Russell’s paradox arises from the belief that you cannot have a set of all sets that are not members of themselves. To which I thought:

The set of all sets encompasses all sets that are not members of themselves, as well as itself. To which a math professor told me:

By a set of all sets that are not members of themselves, we mean a set that encompasses all sets that are not members of themselves, and no other set. To which I thought:

In that case, you cannot have a set of all sets that are members of themselves because it would amount to one set being a member of itself twice (which is contradictory. See the op for proof of this).

In truth, there is one absolute universal set (which I call Existence or Infinity or God). Everything is a member of this set/collection/existent/thing. It is a member of itself (which is the same as saying it is itself. This is not the same as saying it is itself + another of itself combined). Think of it this way: Everything is contingent on God. God is not contingent on anything other than Himself. God is the set of all contingents. God is self-contingent, whilst all other things are not self-contingent.

I disagree but it is just a matter of semantic definition.

In my language every item is a set of 1 item.

I wouldn’t say that Existence is a member of itself (that confuses the language). Existence IS itself.

What you say is only true IF your list is a list of references (or titles) to the lists including a reference to itself. Your 5th list cannot contain itself except as only a title or reference. That has been this whole argument from the beginning.

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.

That is what I was saying.

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.

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

I think that is just a part of the cancel culture politics. They don’t care that it doesn’t make sense.

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.

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?

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.

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.

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.

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.

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.

I think that you are saying that.

But that is incorrect. If its volume was infinite then its volume would have to be twice infinite.

I don’t believe that is true. The future and the past do not exist. If they existed then they would be the present.