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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?

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

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  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.

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

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