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I have a question that I've been struggling with: Define a relation $M(A,B)$: $A \cap B = \varnothing$. Domains for $A$ and $B$ are all subsets of $\Bbb Z$. What properties does the relation $M$ satisfy? I think I'm confused because I keep thinking of $A$ and $B$ as individual sets of integers. For example, I thought that $M$ was not symmetric because there can never be the same elements in sets $A$ and $B$ and still satisfy $A \cap B =\varnothing$. However I was told that $M$ is symmetric because $A \cap B =\varnothing$ and $B \cap A = \varnothing$. I was thinking about it more like defining $M$ with example sets $A$ and $B$, where no elements could be in both sets and still satisfy an empty set. Any help to get me through this block?

Brian M. Scott
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Misha
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Consider the following subsets of $\Bbb Z$: $A=\{-1,0,1\}$, $B=\{0,2,3\}$, and $C=\{4,5,6\}$. You can check that $$A\cap B=B\cap A=\{0\}\ne\varnothing\;,$$ $$A\cap C=C\cap A=\varnothing\;,$$

and $$B\cap C=C\cap B=\varnothing\;.$$

By the definition of $M$ this means that $M(A,C)$, $M(C,A)$, $M(B,C)$, and $M(C,B)$ are all true, but $M(A,B)$ and $M(B,A)$ are not.

Intersection is commutative: for any sets $A$ and $B$, $A\cap B=B\cap A$. Thus, if $M(A,B)$ holds, meaning that $A\cap B=\varnothing$, then $B\cap A=A\cap B=\varnothing$, and therefore $M(B,A)$ holds as well. This shows that $M$ is symmetric.

For reflexivity: if $B$ is as above, is it true that $M(B,B)$, i.e., that $B\cap B=\varnothing$?

For transitivity, note that with the sets above we have $M(B,C)$ and $M(C,B)$; do we also have $M(B,B)$?

Brian M. Scott
  • 616,228
  • Thanks. I'm beginning to start understanding this better. See if I'm correct with my thinking to your questions: Reflexivity: M is not reflexive since B intersect B = {0,2,3} does not equal empty set. M is not transitive since M(B,B) = {0,2,3} does not equal empty set. – Misha Nov 09 '13 at 17:15
  • @Misha: Yes, that's exactly right. – Brian M. Scott Nov 09 '13 at 20:46