Question

Prove that power set is a field?

Answer 1

The power set of a set is a commutative ring under the natural operations of union and intersection, but not a field under those operations, since it lacks inverse elements.

Explanation:

Given any set `S`, consider the power set `2^S` of `S`.

This has natural operations of union `uu` which behaves like addition, with an identity `O/` and intersection `nn` which behaves like multiplication with an identity `S`.

In more detail:

• `2^S` is closed under `uu`
  If `A, B in 2^S` then `A uu B in 2^S`

• There is an identity `O/ in 2^S` for `uu`
  If `A in 2^S` then `A uu O/ = O/ uu A = A`

• `uu` is associative
  If `A, B, C in 2^S` then `A uu (B uu C) = (A uu B) uu C`

• `uu` is commutative
  If `A, B in 2^S` then `A uu B = B uu A`

• `2^S` is closed under `nn`
  If `A, B in 2^S` then `A nn B in 2^S`

• There is an identity `S in 2^S` for `nn`
  If `A in 2^S` then `A nn S = S nn A = A`

• `nn` is associative
  If `A, B, C in 2^S` then `A nn (B nn C) = (A nn B) nn C`

• `nn` is commutative
  If `A, B in 2^S` then `A nn B = B nn A`

• `nn` is left and right distributive over `uu`
  If `A, B in 2^S` then `A nn (B uu C) = (A nn B) uu (A nn C)`
  and `(A uu B) nn C = (A nn C) uu (B nn C)`

So `2^S` satisfies all of the axioms required in order to be a commutative ring with addition `uu` and multiplication `nn`.

If `S = O/` then `2^S` has one element, namely `O/`, so it fails to have distinct additive and multiplicative identities and is therefore not a field.

Otherwise note that `S` has no inverse under `uu` and `O/` has no inverse under `nn`. So `2^S` does not form a field due to lack of inverse elements.