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Suppose $R$ and $S$ are partial orders on $A$ and $B$ respectively. Define a relation $T$ on $A \times B$ as follows: $T=\{((a,b),(a',b')\in (A\times B)\times (A\times B)|aRa'$ and $bSb'\}$. Show that $T$ is a partial order on $A\times B$.

I am not having trouble mainly with the question - instead I don't understand how is $T$ a relation on $A\times B$, instead of $(A\times B) \times (A \times B)$.

Because here, $(a,b)\in A\times B$ and $(a',b')\in (A\times B)$, so surely $T$ is on the Cartesian product of the two sets $(a,b)$ and $(a',b')$ are from, i.e. $(A\times B) \times (A \times B)$....right?

Sorry guys I know this is a very basic and stupid question, thank you so much for your kind assistance!

2 Answers 2

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Generally a "relation on $M$" means a subset of $M\times M$.

So saying that $T$ is a relation on $A\times B$ is the same as saying that $T$ is a subset of $(A\times B)\times(A\times B)$.

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    Damn...I don't know how I missed the idea that...a relation on $A$ is the same as $A\times A$... Thank you for your answer; is there anyway I can delete this question? I feel that it is literally too stupid to be posted... – 2017-01-13
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    @DanielMak: Once there are upvoted answers, the asker loses the power to delete the question. – 2017-01-13
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A binary relation on a set $X$ is a subset of $X \times X$. That is, if $R \subseteq X \times X$, then we write $xRx'$ to abbreviate $(x,x') \in R$, and we say that $R$ is a binary relation on $X$.

In your case, $X = A \times B$. So the relation is a subset of $(A \times B) \times (A \times B)$, making it a binary relation on $A \times B$.