Solution Found!
This exercise presents Russell’s paradox. Let S be the set
Chapter 1, Problem 46E(choose chapter or problem)
Problem 46E
This exercise presents Russell’s paradox. Let S be the set that contains a set x if the set x does not belong to itself, so that S = {x | x ∉ .x}.
a) Show the assumption that S is a member of S leads to a contradiction.
b) Show the assumption that S is not a member of S leads to a contradiction.
By parts (a) and (b) it follows that the set S cannot be defined as it was. This paradox can be avoided by restricting the types of elements that sets can have.
Questions & Answers
QUESTION:
Problem 46E
This exercise presents Russell’s paradox. Let S be the set that contains a set x if the set x does not belong to itself, so that S = {x | x ∉ .x}.
a) Show the assumption that S is a member of S leads to a contradiction.
b) Show the assumption that S is not a member of S leads to a contradiction.
By parts (a) and (b) it follows that the set S cannot be defined as it was. This paradox can be avoided by restricting the types of elements that sets can have.
ANSWER:
Solution:
Step 1:
a)In this problem we need to show the assumption that S is a member of S leads to a contradiction.
Russell’s paradox: Let S be the set that contains a set x if the set does not belongs to itself , so that .
If , then by using the Russell’s paradox condition for S we conclude that
, a contradiction.