- #1
Kniazi
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Prove Russel's paradox by contradiction and what does it tell us about sets?
I tried doing it like this and I am not sure it is right.
I supposed S was the collection of all sets and since S is a set S∈S.
Now we can split this universe S into two parts: U(for the unusual that are part of themselves like S) and N(for the normal sets).
So N={A∈S|A∉A}. In english this means that any set A will be part of N only if A is not an element of itself.
Now I plug in N in place of A, to apply the above statement for N.
N={N∈S|N∉N}.
So since N is not an element of itself, N belongs in N.
Is it correct ^?
Also what does this tell us about sets?
--
Khadija Niazi
I tried doing it like this and I am not sure it is right.
I supposed S was the collection of all sets and since S is a set S∈S.
Now we can split this universe S into two parts: U(for the unusual that are part of themselves like S) and N(for the normal sets).
So N={A∈S|A∉A}. In english this means that any set A will be part of N only if A is not an element of itself.
Now I plug in N in place of A, to apply the above statement for N.
N={N∈S|N∉N}.
So since N is not an element of itself, N belongs in N.
Is it correct ^?
Also what does this tell us about sets?
--
Khadija Niazi