– Explain Russell’s paradox using the set ( R = x \mid x \notin x ). Why is this not a set in ZFC?
5.1: ( A \times B = (a,1),(a,2),(a,3),(b,1),(b,2),(b,3) ); ( B \times A ) has 6 pairs reversed. 5.2: ( |A \times B| = m \cdot n ), so ( |\mathcalP(A \times B)| = 2^mn ). Chapter 6: Functions and Relations Focus: Function as a set of ordered pairs, domain, codomain, image, preimage. set theory exercises and solutions pdf
8.1: If ( R \in R ) → ( R \notin R ) by definition; if ( R \notin R ) → ( R \in R ). Contradiction → ( R ) cannot be a set; it’s a proper class. Epilogue: The Archive Opens Having solved the exercises, the apprentices returned to Professor Caelus. He smiled and handed them a single golden key—not to a building, but to the understanding that set theory is the foundation upon which all of modern mathematics rests. – Explain Russell’s paradox using the set (
– Draw a Venn diagram for three sets ( A, B, C ) and shade ( (A \cap B) \cup (C \setminus A) ). Contradiction → ( R ) cannot be a
He handed each student a scroll. On it were exercises that grew from simple membership tests to the paradoxes that lurked at the foundations of mathematics. “Solve these,” he said, “and the keys shall be yours.”
– Prove ( (A \cup B)^c = A^c \cap B^c ) using element arguments.
– Which of these relations from ( 1,2,3 ) to ( a,b ) are functions? (a) ( (1,a),(2,b),(3,a) ) (b) ( (1,a),(1,b),(2,a) ) (c) ( (1,b),(2,b) )