# 2 solutions to Russell's paradox

Consider meaningfulness. For the purposes of this proof, the meaning of ‘meaning’ will be compared with the meaning of ‘set’.

All meaningful things are meaningful (just as all triangular things are triangular). Being meaningful and being the meaning meaningful are two different truths. All meanings that do not mean meaningful, are still members of being meaningful despite not being the meaning meaningful. ONLY the meaning ‘meaningful’ means ‘meaningful’ and all meanings are members of the meaning ‘meaningful’ in the following way:

The meaning/set ‘triangle’, encompasses all triangular things: If t has three sides, then t is a member of at least all the following sets/meanings: triangle, shape, meaningful. t is not all those things. t is an instance/member of all those things. This distinction is crucial. One might be tempted to say “where t is three sided, t is not just a member of triangle, t is both a triangle and a member of triangle. t is a member of itself.” This is mistaken because t is not the meaning ‘triangle’ just as t is not the set of triangle. t does not encompass triangle. Triangle encompasses t. Crucially, t does not mean triangle. t means an instance of triangle/shape/meaning.

The meaning/set ‘meaningful’ encompasses all meaningful things. Just as the set of all meanings is just one set, the meaning ‘meaning’ is just one meaning. Meaning does not mean all meanings (for example, meaning does not mean ‘triangle’). Meaning means ‘meaning’ but it encompasses all meanings (including itself). ‘meaning’ is the only meaning that is exclusively a member of itself. Every other meaning is a member of at least one meaning other than itself and never itself.

From the above we can conclude that just as the meaning ‘meaning’ cannot be a member of any meaning other than itself (because it is the meaning that all meanings are members of including itself), the set that all sets are members of cannot be a member of any set other than itself. The meaning ‘meaning’ is one meaning and all meanings are members of it. The set that all sets are members of, is one set and all sets are members of it. Consider all sets that are not members of themselves. Also, consider all meanings that do not mean ‘meaning’. Are any of them members of themselves? No. Only meaning means ‘meaning’ and only the set of all sets is a member of itself. Clearly, the set of all sets that are not members of themselves is definitively a member of the set of all sets.

The phrase all meanings except the meaning ‘meaning’, is a member of the meaning ‘meaning’. It is also a member of the meaning ‘phrase’. Note that we cannot say ‘meaning’ is not a member of ‘meaning’. Nor can we say meaning does not = to meaning. But we can say phrase is not a member of phrase. We cannot say phrase does not = to phrase. Thus meaning = meaning and phrase = phrase but meaning is a member of itself whilst phrase is not. Thus the set of all sets that are not members of themselves, is definitively not a member of itself, because it is a member of the set of all sets (this is provided that this set is actually a set and is different to the set of all sets)

If one insists that there are other sets that are members of themselves such that phrase = phrase and phrase is a member of itself, then one must accept that all sets are members of themselves. Where this line of reasoning is pursued, even then, the following will hold true:

There is only one set that is not a member of any set other than itself (the set of all sets…or the meaning of ‘meaning’). All other sets/meanings are members of it whilst not being equal to it. In this scenario, there is no such thing as a set that is not a member of itself. And so we have no Russell’s paradox as we clearly have a set of all sets and we do not have to answer: is the set of all sets that are not members of themselves, a member of itself?.

There exists meanings. All meanings are meaningful. Consider the meaning ‘triangle’. A triangle is a meaningful shape. Thus, the meaning triangle, is a member of the meaning ‘shape’ and the meaning ‘meaningful’. The meaning triangle means triangle. As in it means what it means. So it is a member of itself.

There is no such thing as a meaning that does not mean what it means (there is no meaning that does not mean itself). All meanings mean themselves. So there is no meaning that is not a member of itself.

Consider the meaning ‘meaningful’. It is clear that all meanings that are meaningful, are members of the meaning ‘meaningful’. The meaning ‘meaningful’, is the ONLY meaning that is EXCLUSIVELY a member of itself. It is not a member of any meaning other than itself, whereas ALL meanings are members of it (including itself).

All meanings are the meanings that they are. All sets are the sets that they are. Thus, when Russell asked:

Is the set of all sets that are not members of themselves, a member of itself?

The answer is either yes (because the set of all sets is a member of itself, and no other set can contain ALL sets that are not members of themselves) or no (because there is no such thing as a set that is not a member of itself). Depends on how you want to look at it.

There is one set/meaning that is not a member of any set/meaning other than itself. That set is the set of ‘all sets’. That meaning is the meaning of 'meaningful '. All sets/meanings are members of it. Including itself.

I think Russell’s paradox is just another instance of “This statement is false” - nonsense in a sensible looking package.

I felt the package looked sensible too (as in I understood why he put the question to Frege). But I knew there must’ve been some semantical misunderstanding somewhere.

I’m still undecided as to answer no to his question (because there are no sets that are not members of themselves) or to answer yes to his question (because if there are sets that are not members of themselves, then the set of all these sets must be a member of itself for it to semantically encompass ALL sets that are not members of themselves).

If he focuses the point that the set of all ducks, is not a duck, then the answer to his question is yes. If he focuses on what it is for a set to be a set, then the answer to his question is no.

In trying to account for:

The meaning of (all meanings that are not the meaning ‘meaningful’), which is like saying the set of (all sets that are not the set of ‘all sets’), there is an answer:

The meaning of (all meanings that are not the meaning ‘meaningful’) is both meaningful, and means what it means. Which is like saying it is both a set and a member of itself. But it is NOT exclusively a member of itself (because it is a member of the meaning ‘meaningful’…because it is a meaningful thing). However, when we view ‘a set that is not members of itself’ as being a meaningful thing, then the following happens:

The set of (all sets that are not members of themselves) is clearly a set. To deny this is to say there is no set of all sets that are not members of themselves (which makes Russell’s point wholly irrelevant). If we assume his point to be relevant, then the following follows:

There are sets that are not members of themselves. By definition, ALL such sets are not members of themselves,. So by definition, the set of ALL such sets, can’t be a set that is not a member of itself. Thus, the set of ALL such sets, is a member of itself. Only the set of ‘ALL sets’ is a member of itself. It encompasses all sets other than itself, as well as itself. No other set does this. No other set is a member of itself, by these standards.

So based on the above, either all sets (that are actually sets) are members of themselves, or only ONE set is a member of itself.

The paradox isn’t about a set being a member of itself. And even when he says that, he is referring to the sets name being included in the set - not the entire set itself within itself.

The paradox is about specifying that a set that includes every name of every set that does NOT include its own name must include its own name. And then because now it includes its own name, it must not include its own name. That is exactly like the statement - “this statement is false”. If it is true then it must be false and if it is false then it must be true.

The only issue is the assumption that every statement must be either true or false. That is a false assumption. Some statements are simply senseless - irrational.

And Russell’s specified set is merely an irrationally defined set. Basically he said “I have a set that contains only items that are not in the set.”

It is easy to make irrational claims (such as “there was no fraud” ).

And such claims shadow the prisoners’ dilemma in trying to sort through the paradoxical nature of simply ambigious propositions that could determine a course of action by rote.

In other words, a man may or may not act in accordance to his best interest, or that of the group all together. In what sense does data impinge on the interpretation, quite quixotic, and gestalt like.

Is the glass half full, or half empty? There certainly is a difference.

The difference in answering the question whether “a glass is half full or half empty” is only significant if the speaker of this statement is included. For the statement itself there is no difference, at least not in the mathematical or mathematical-logical sense. That is just the difference between a purely mathematical statement and a purely linguistic statement. And the logic mediates between the mathematical language and the linguistic language. You must use your words (your lexemes) correctly if your statement has to be logically correct; you must use your numbers correctly if your statement has to be logically correct. Only then, if you include yourself or another person (a subject) in your statement, you have to leave mathematics to this very part and to the other part you have to take care that you say what you want to say linguistically correct (well-formed) and logically correct.

The problem of gaming behavior in the sense as I am one of the prisoners, not particularly keen on the quantifiable logic, but am on the intuitively mathematical basis of logic, and it is a matted of life and death , wether to adopt one of the two ways to solve the ( my) puzzle, I can not construct or deduct from prior premises of constructed reality.

Wethed to put a primary focus on that self serving, evolutionary logical assessnent, or, direct that focus on the welfare of probable logic, to advance the mutual benifit regardless of my own maximized welfare, then this dies present either a sensible or non sensible proposition that may be checked by inducing the veracity one way or another.

To make this clear( er) , let’s say the probibility of a Covid19 infection corresponds to a more far reaching, need of making sure, loved ones are not passed to the infection.

The self quaranteen validity should overcome throw probable certainty of such a trade off.

In such a case, a pro sense-data construction would override a non sensible deconstruction od sense toward day a. The holes connected in a situation like that, elongate with the prisoner into unforeseeable mathematical uncertainty.

The prisoner is connected to the the paradoxical nature of both, yet the game can not equivocate both, from an existential point of view. The decesion is not bound by assumptions about either choice, since no assumptions can project the other prisoners infectious state.

There can not arise an ethical paradigmn, for lack of a mathematical/ logical assurance.

The glass will be more filled with doubt than the assurance.(or, conversely)

The situation gets a lot more imbalanced, when probabilities sink to minimum levels of occurabce, while the levels of insensibly induced uncertainty rise hyperbolically.

Translation - “Fear biases the calculation of reasoning”?

Yes, but fear exists on either side of the equasion, and where the fulcrum lies determines how much to fear that middle and how to go to balance the other side.

A super duper computer could perhaps figure it out, maybe, but not yet.