1. solution: subsets of set Sn can be divided into two categories: ① subsets containing 1; ② Does not contain a subset of 1. These two types each have two subsets (n- 1). For any subset A in ②, there must be a unique subset A ∨{ 1} in ①. If A is an odd subset, A ∨{ 1} is an even subset. If a is an even subset, then A ∨{ 1} is an odd subset. Therefore, if there are x odd subsets and y even subsets in ②, there must be x even subsets and y odd subsets in ①.

Therefore, the number of odd subsets and even subsets of Sn is the same.

2. Solution: Let A be any odd subset of Sn, and construct the mapping F as follows:

A→A

-

{1} if 1∈A

A→A ∨{ 1 }，

If 1? A

(1)

-

{1} represents the set obtained by removing 1 from the set a).

Therefore, the mapping f is a mapping that maps an odd subset to an even subset.

It is easy to know that if A 1, A2 is two different odd subsets of Sn. Then f(A 1)≠f(A2), that is, f is injective.

I hope you know what a single shot is.

For every even subset b of Sn, if 1∈B, there exists A=B\{ 1} (meaning b = {x ∈ x ∈ 1 and x∈B), so that f (a) = b. B, then A = B ∨{ 1} exists, so f(A)=B, so f is surjective.

(Do you know Man Ge ...)

Therefore, f is a one-to-one correspondence between the set composed of odd subsets of Sn and the set composed of even subsets of Sn, so that the number of odd subsets and even subsets of Sn is equal, and both are 1/2× 2 n = 2 (n- 1).

A * * * involves the contents of sets and functions, and typing is very painful. I hope to adopt.