Proof Help-- Basis of Vector in F2.

amm345 · #1 $Old$ November 3rd, 2009, 11:05 PM

Let X be a finite non empty set and V = P(X) = {S | S ⊆ X} the set of all subsets of X. For S, T ∈ V , we define

S + T = (S ∪ T) − (S ∩ T).

Suppose X={x1, x2,...,xn} with distinct xi. Prove that B=({x1}, {x2},..., {xn}) is a basis of V.

HallsofIvy · #2 $Old$ November 4th, 2009, 04:01 AM

Quote:

Originally Posted by amm345 $View Post$

Let X be a finite non empty set and V = P(X) = {S | S ⊆ X} the set of all subsets of X. For S, T ∈ V , we define

S + T = (S ∪ T) − (S ∩ T).

Suppose X={x1, x2,...,xn} with distinct xi. Prove that B=({x1}, {x2},..., {xn}) is a basis of V.

A basis for a vector space has two properties: the vectors in the basis span the space and they are independent. It is easy to see how to write any "vector" as a "linear combination" of the x_n and it should be just as easy to see that they are independent.

(At first I missed the "in F2" which is only in your title, not in the body of the question, and wondered about scalar multiplication. Of course, the only members of F2 are 0 and 1. 0 times any vector is 0 and 1 times any vector is the vector itself.)