2 solutions to Russell's paradox

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2 solutions to Russell's paradox

Postby Certainly real » Wed Jan 13, 2021 11:15 pm

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?.
Certainly real
 
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