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Get Full Access to Discrete Mathematics And Its Applications - 7 Edition - Chapter 2.1 - Problem 46e
Get Full Access to Discrete Mathematics And Its Applications - 7 Edition - Chapter 2.1 - Problem 46e

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# This exercise presents Russell’s paradox. Let S be the set ISBN: 9780073383095 37

## Solution for problem 46E Chapter 2.1

Discrete Mathematics and Its Applications | 7th Edition

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Problem 46E

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.

Step-by-Step Solution:

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.

Step 2 of 2

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