Equivalence classes

Let p, p' be equivalence relations on a set A. Let n, n' be the number of equivalence classes pf p, p' respectively.... Please look at the attached doc for rest of question.

Prove the Transitive theory

Prove the following theory: 1) R1 is a subset of R2 => All of R3, R1R3 is a subset of R2R3 and 2) R1 is a subset of R2 => All of n, (R1)^n is a subset (R2)^n 3) Suppose R is transitive, then for all of n, R^n is a subset of R.

Partitions on a set

We denote the number of partitions of a set of n elements by P(n). Suppose the number of partitions of a set on n elements into k parts is denoted by P(n,k). Then obviously P(n) = P(n,1) + P(n,2) + ….. + P(n,n) Show that P(n,2) = 2^(n-1) - 1

There is no bijection between any set A and its power set P(A) of A.

There is no bijection between any set A and its power set P(A) of A. For finite sets, proof is trivial since |A| = n and |P(A)| = 2^n. For finite sets, this is done by contradiction. Suppose there is a bijection $ between a set A and its power set P(A). Consider the set B={x|x is a member A where x is not a member $(x)}For each element x A, since $ is a function from A to the power set of A, &(x) ia a subset of A. By our earlier assumption on set theory that every element is either in a set or ot in a set, for each x A, w can certainly ask if x is a member of $(x). Therefore, the set B is well definedB. We know B is a subset of A and $ is a bijection from A to its power set P(A). For... click for more

Discrete. Send answer as attachment

Please see the attached file for full problem description. The double bracket notation is pronounced " n multichoose k". The doubled parentheses remind us that we may include elements more than once.