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Let $X$ be a set. Define a relation $\rm R$ on the power set $\mathcal P(X)$ by

$$A \operatorname{\rm R} B ~\iff~ A \cap B = \emptyset$$

for $A,B \in \mathcal P(X)$, do determine whether this relation is reflexive, symmetric, and/or transitive.

Ok I understand by an awesome person recently to think of relations literally as friends.

Is the above saying $A$ is friends with $B$ is equivalent to that a friend must be friends with both $A$ and $B$, and that friend cannot be equal to zero?

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J Crawford
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  • $A$ and $B$ are both sets in this context, and the relation is defined as "$A$ and $B$ are friends if they have no common elements". For example, ${1, 2}$ is friends with ${3}$, but it is not friends with ${1, 3}$. – Austin Mohr Jun 07 '17 at 03:52
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    Oh I see so, that is very odd. Hmm OK I will take a crack at it thank you. – J Crawford Jun 07 '17 at 03:54
  • This is what I have ....If A and B are friends but have no common friends (elements) in this case then ….. Reflexivity Is A friends with A, yes Symmetry If A is friends with B then B is friends with A, yes Transitivity If A is friends with B, B is friends with C, then A and C are friends but that would mean C would be a common element among A and B which would break the original equation. – J Crawford Jun 07 '17 at 04:01
  • I use the word "elements" to refer to the items within the sets (1, 2, 3 in my example). So when I say "$A$ and $B$ have no common elements", this is just another way to say that the intersection of $A$ and $B$ is empty. Think about reflexivity with that in mind. What is the intersection of $A$ with $A$? – Austin Mohr Jun 07 '17 at 04:04

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Is the above saying A is friends with B is equivalent to that a friend must be friends with Both A and B, and that friend cant be equal to zero?

No, you can interpret it as saying $A$ is $R$-related to $B$ if they don't share contents.   Indeed, rather than "is friends with", the relation $\rm R$ is "is disjoint from".

So, will every $A$ be disjoint from itself? Reflexivity requires it to be so, is it?

$$\forall A\in\mathcal P(X)~:~ A\cap A=\emptyset\qquad \mathrm{(T/F)?}$$

If any $A$ is disjoint from any $B$, then will that $B$ be disjoint from that $A$? Symmetry requires this to be so, is it?

$$\forall A\in\mathcal P(X)~~\forall B\in\mathcal P(X)~:~ [A\cap B=\emptyset \to B\cap A=\emptyset]\qquad\mathrm{(T/F)?}$$

Then there's transitivity.

$$\forall A\in\mathcal P(X)~~\forall B\in\mathcal P(X)~~\forall C\in\mathcal P(X)~:~ [(A\cap B=\emptyset \wedge B\cap C=\emptyset) \to A\cap C=\emptyset]\qquad\mathrm{(T/F)?}$$

Graham Kemp
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