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So I have a question about relations. In particular, here is the formal question:

Let $\beta$ be the relation "is a brother of" and let $\sigma$ be the relation "is a sister of". Describe $\beta\cup\sigma, \beta\cap\sigma, \beta-\sigma$.

So we can think of say $P$ as the set of people on Earth. Then both $\beta$ and $\sigma \in P\times{P}$. So intuition is telling me that

1) $\beta\cup\sigma$ is the set of all ordered pairs of people with the appropriate sibling whether sister or brother

2) $\beta\cap\sigma=\varnothing$ set since no one can be both the brother of someone AND the sister of someone.

$\mathbf{EDIT}$, this is actually the set $b$. See comments below.

3) $\beta-\sigma=\beta$, since these sets are disjoint.

Is this the correct intuition and answer to the problem?

Finally, as an addendum, (was not asked), these relations can not be equivalence relations since they are not reflexive (you really can't be the brother of yourself can you?). They are symmetric and transitive though i think, so is there a name for the type of ordering that is just symmetric and transitive?

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  • $\begingroup$ Your reasoning looks good to me. But symmetry and transitivity fail: If Bob is the brother of Alice, then Alice is the sister, not the brother, of Bob. But if Bob is the brother of Charlie then Charlie is the brother of Bob, so the brother relation is symmetric if restricted to males. Not transitive, though, since transitivity would imply that Bob is his own brother (still assuming he has a brother). $\endgroup$ Commented Sep 9, 2013 at 18:52
  • $\begingroup$ ha ha ha! I was totally not thinking about Alice!! In my head I said, if Tom is my brother, then i am tom's brother.... Yeah I was definitely restricting myself to the males when I had no license to do so. Thanks for the help! $\endgroup$ Commented Sep 9, 2013 at 18:54
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    $\begingroup$ I know people in $\beta \cap \sigma$; unfortunately, people treat them like they belong to $\emptyset$. $\endgroup$ Commented Sep 9, 2013 at 18:58
  • $\begingroup$ Actually, one more question @HaraldHanche-Olsen, the next problem says if $A$ is the set of students in my school, what is $\beta(A)$ and $(\beta\cup\sigma)(A)$. I think this would be the ordered pairs of students and their brothers, and students and their siblings (brother or sister), correct? $\endgroup$ Commented Sep 9, 2013 at 18:59
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    $\begingroup$ Not if you include bigendered people (with siblings) in the set of people on Earth (and likely other transgendered people). $\endgroup$ Commented Sep 9, 2013 at 19:02

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You may also proceed as follows:

\begin{align}\beta \cup \sigma &= \{(x,y)\in P \times P : (x,y)\in \beta \text{ or } (x,y)\in\sigma\}\\ &=\{(x,y)\in P \times P : (x \text{ is a brother of y}) \text{ or } (x \text{ is a sister of y})\} \\ &=\{(x,y)\in P \times P : (x \text{ is a sibling of y})\} \\ &=\{(x,y)\in P \times P : (x \text{ and } y \text{ are siblings})\} \\ \end{align}

Similarly,

\begin{align} \beta \cap \sigma &= \{(x,y)\in P \times P : (x,y)\in \beta \text{ and } (x,y)\in\sigma\}\\ &=\{(x,y)\in P \times P : (x \text{ is a brother of y}) \text{ and } (x \text{ is a sister of y})\} \\ &=\{(x,y)\in P \times P : (x \text{ is a brother and a sister of y}) \} \\ \end{align}

In the same manner, you may proceed with $\beta-\sigma$ where $A-B$ is defined as $\{x : x\in A \text{ and } x\not\in B\}$.

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  • $\begingroup$ That is real nice. Thank you! $\endgroup$ Commented Sep 9, 2013 at 19:25

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