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3 comments:
A pretty good informal explanation of Russel's paradox from wikipedia: Let us call a set "normal" if it does not contain itself as a member. For example, take the set of all squares. That set is not itself a square, and therefore is not a member of the set of all squares. So it is "normal". On the other hand, if we take the complementary set of all non-squares, that set is itself not a square and so should be one of its own members. It is "abnormal".
Now we consider the set of all normal sets – let us give it a name: R – and ask the question: is R a "normal" set? If it is "normal", then it is a member of R, since R contains all "normal" sets. But if that is the case, then R contains itself as a member, and therefore is "abnormal". On the other hand, if R is "abnormal", then it is not a member of R, since R contains only "normal" sets. But if that is the case, then R does not contain itself as a member, and therefore is "normal". Clearly, this is a paradox: if we suppose R is "normal" we can prove it is "abnormal", and if we suppose R is "abnormal" we can prove it is "normal". Hence, R is neither "normal" nor "abnormal," which is a contradiction.
A pretty good informal explanation of Russel's paradox from wikipedia: Let us call a set "normal" if it does not contain itself as a member. For example, take the set of all squares. That set is not itself a square, and therefore is not a member of the set of all squares. So it is "normal". On the other hand, if we take the complementary set of all non-squares, that set is itself not a square and so should be one of its own members. It is "abnormal".
Now we consider the set of all normal sets – let us give it a name: R – and ask the question: is R a "normal" set? If it is "normal", then it is a member of R, since R contains all "normal" sets. But if that is the case, then R contains itself as a member, and therefore is "abnormal". On the other hand, if R is "abnormal", then it is not a member of R, since R contains only "normal" sets. But if that is the case, then R does not contain itself as a member, and therefore is "normal". Clearly, this is a paradox: if we suppose R is "normal" we can prove it is "abnormal", and if we suppose R is "abnormal" we can prove it is "normal". Hence, R is neither "normal" nor "abnormal," which is a contradiction.
Opps, looks like I should learn some patience when clicking buttons.
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