# Arbitrary Set A a Function??

Assume you have an arbitrary set A, let RA be the relation defined on A × Power Set(A) by, for all a ∈ A and B ⊆ A,

"a RA B iff a ∈ B"

1.Let A = {0, 1}. Is RA a function? Justify?

2.Find a set A such that RA is a function.

does anybody have an idea of what to do? Im confused on trying to show its a function

The powerset(A) = {{0},{1},{EmptySet},{0,1}}

is RA a function then? because the X value is connected to a Y value?

## 2 Answers

A relation in $A\times B$ is a function if each element in $A$ is related to only one element of $B$.

1. $RA$ is not a function because $(1,\{1\})\in RA$ due to the fact that $1\in\{1\}$ and $(1,\{0,1\})\in RA$ because $1\in \{0,1\}$, and a function cannot have two values at a point.

2. Let $A=\emptyset$. Then every relation is a function.

• How did you know that (1,{1})∈RA and (1,{0,1})∈RA ? I didnt understand what you did to get the answer – RandomMath Nov 11 '14 at 14:55
• could you explain what you mean by "and a function cannot have two values at a point" – RandomMath Nov 11 '14 at 15:01
• You need to look up the definition. I can't make it any clearer, sorry. – Matt Samuel Nov 11 '14 at 15:03
• I don't think that a function defined over $\emptyset$ makes much sense. But the set $A=\{1\}$ also works. – ajotatxe Nov 11 '14 at 15:22
• Functions on the empty set are perfectly valid. For every set there is a unique function from the empty set into that set. – Matt Samuel Nov 11 '14 at 15:59