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Let $A=\{a,b,c,d\}$ and $B = \{w,x,y\}$, then a non-empty relations on $A$ is: $\{ (b,c), (b,d）\}$

Can someone explain why this is true? I thought that the requirements for any relations of a set has to be such that for $(x,y)$, $x$ has to be a subset of $y$, and $x,y$ have to be a subset of $A$. How is $b$ a subset of $c$?

Shawn O'Hare
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jack
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    Does ${ b, c }$ denote the subset consisting of $b$ and $c$ or the ordered pair $(b, c)$? In the context of the question, the latter makes a lot more sense. Moreover, $({b, c}, {b, d})$ typically would denote an element of $2^A \times 2^A$, that is, the Cartesian product of the power set of $A$ with itself. – Shawn O'Hare Jul 14 '14 at 19:19

2 Answers2

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A relation on $A$ is simply a subset of the Cartesian product $A \times A$. For $A=\{a,b,c,d\}$, both $(b,c)$ and $(b,d)$ are contained in $A \times A$; therefore, $\{(b,c),(b,d)\}$ is a relation on $A$.

hexaflexagonal
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  • How come when A={{1},{2},{1,2}}, then （{1},{2}) is not a relation? – jack Jul 14 '14 at 19:14
  • If $A={{1},{2},{1,2}}$, then ${{{1},{2}}}$ is a relation on $A$, not ${{1},{2}}$. – hexaflexagonal Jul 14 '14 at 19:18
  • In my textbook it doesn't show that as a relation when it said it listed every possible subset relation – jack Jul 14 '14 at 19:22
  • Apologies, my notation was way off. I used ${}$ where I should have used $()$. But anyway, the set ${({1},{2})}$ is in fact a relation on $A$. Remember, relations are sets, so an ordered pair $({1},{2})$ by itself cannot be a relation, though it can be (and, here, is) an element of a relation. – hexaflexagonal Jul 14 '14 at 19:26
  • If your textbook does not list ${({1},{2})}$ as a relation, then either you have copied $A$ from the textbook incorrectly or it is simply an error in the book. – hexaflexagonal Jul 14 '14 at 19:35
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Any set $$R \subseteq A \times A = \{(x,y) \mid x \in A ,\,y \in A\} $$ is a relation on $A $. Since $$\{(b,c), (b,d)\} \subseteq A \times A$$ holds, it is indeed a relation on $ A$.

hexaflexagonal
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Adam
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