# Prove that power set is a field?

Oct 31, 2017

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 $\cup$ which behaves like addition, with an identity $\emptyset$ and intersection $\cap$ which behaves like multiplication with an identity $S$.

In more detail:

• ${2}^{S}$ is closed under $\cup$
If $A , B \in {2}^{S}$ then $A \cup B \in {2}^{S}$

• There is an identity $\emptyset \in {2}^{S}$ for $\cup$
If $A \in {2}^{S}$ then $A \cup \emptyset = \emptyset \cup A = A$

• $\cup$ is associative
If $A , B , C \in {2}^{S}$ then $A \cup \left(B \cup C\right) = \left(A \cup B\right) \cup C$

• $\cup$ is commutative
If $A , B \in {2}^{S}$ then $A \cup B = B \cup A$

• ${2}^{S}$ is closed under $\cap$
If $A , B \in {2}^{S}$ then $A \cap B \in {2}^{S}$

• There is an identity $S \in {2}^{S}$ for $\cap$
If $A \in {2}^{S}$ then $A \cap S = S \cap A = A$

• $\cap$ is associative
If $A , B , C \in {2}^{S}$ then $A \cap \left(B \cap C\right) = \left(A \cap B\right) \cap C$

• $\cap$ is commutative
If $A , B \in {2}^{S}$ then $A \cap B = B \cap A$

• $\cap$ is left and right distributive over $\cup$
If $A , B \in {2}^{S}$ then $A \cap \left(B \cup C\right) = \left(A \cap B\right) \cup \left(A \cap C\right)$
and $\left(A \cup B\right) \cap C = \left(A \cap C\right) \cup \left(B \cap C\right)$

So ${2}^{S}$ satisfies all of the axioms required in order to be a commutative ring with addition $\cup$ and multiplication $\cap$.

If $S = \emptyset$ then ${2}^{S}$ has one element, namely $\emptyset$, so it fails to have distinct additive and multiplicative identities and is therefore not a field.

Otherwise note that $S$ has no inverse under $\cup$ and $\emptyset$ has no inverse under $\cap$. So ${2}^{S}$ does not form a field due to lack of inverse elements.