Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If $X$ is a set and $n \in \mathbb N$, then $[X]^n$ will denote the set of all subsets of $X$ with exactly $n$ elements.

For a set $X$ and natural numbers $n$ and $m$ define a relation $R$ on $[X]^n$ and $[X]^m$ by $R(A,B)$ holds if and only $A \cap B = \emptyset$.

Is $R$ transitive if $n = m$? Need an example or proof

share|improve this question
add comment

1 Answer 1

up vote 2 down vote accepted

No. Take $X = \{1,2\}$ and $n = m = 1$, $A = C = \{1\}$ and $B = \{2\}$. Then $R(A,B)$ and $R(B,C)$ but not $R(A,C)$.

share|improve this answer
1  
And more generally, $R$ is symmetric and totally irreflexive, so transitivity fails in this way whenever $\vert X\vert\ge 2n$. –  Brian M. Scott Sep 15 '11 at 23:56
    
@Brian: Yes, if $|X| < 2n$ then there exist no $A,B$ with $R(A,B)$ so by definition the relation is transitive. If $|X| \geq 2n$, then you can always take $A = C \subseteq X \setminus B$ to get a contradiction. –  TMM Sep 16 '11 at 0:00
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.