A solution to Russell’s Paradox

The paradox:
Consider the set of all sets that are not members of themselves.
Is that set a member of itself?

Here is a well made video explaining the paradox:

The answer:

itself‘ and a ‘set of itself‘ are not concrete finite types, they are infinity concepts that can be represented by infinity notations.

And just like any infinity concept, to reach to a concrete finite conclusion, you have to collapse its timeless state by specifying a moment / dimension.

A ‘set of itself‘ is always different than ‘itself‘ when you collapse its timeless infinite state.

Collapsing it is done by specifying a finite time / dimension unit.

Let’s represent a moment / dimension with the letter n, so that ‘itself in moment/dimension n‘ is represented by the function itself(n)

itself(n) = set of itself(n-1)

Removing time / dimension from the above is like asking for the answer to what’s the number for division by zero.

Division by zero number can only be witnessed if we can exist outside time. Hence it’s just a concept in our time bound dimension and not a number.

So a ‘set of itself at this moment‘ does not have a member of type ‘set of itself at this moment‘, it has a member of type ‘set of itself a moment ago‘.

Hence, a set of all sets that are not members of themselves would comfortably contain finite sets (e.g. ‘set of apples‘, ‘set of oranges‘) and infinite sets represented by infinity notations (e.g. a set of itself(n-1))

One Reply to “A solution to Russell’s Paradox”

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