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Each case below gives a relation on the set of all nonempty subsets of the natural numbers.

Determine whether the relation is transitive,symmetric, or reflexive.

Case 1:

$R$ is defined by $ARB$ if and only if $A\cap B \neq \emptyset$

:

This seems to be reflexive because the intersection of a set with itself is not going to be disjoint (that is have nothing in common).

This seems to be symetric because B intersection A not being disjoint, is same thing as A intersection B not being disjoint.

Transitive no it probably is not transitive.

Case 2

R is defined by by $ARB$ if and only if $1\in A \cap B$

it is reflexive because if A and B have the element 1 in common then A and A will also have element on in common.

It it transitive because if A and B have 1 in common, B and C have same 1 in common, then A and C will have one in common

Reflexive not sure.

Fernando Martinez
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1 Answers1

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For case $1$ you are correct, it is reflexive since $A\cap A= A$ and $A\neq \emptyset$ since we are working on non-empty sets.

It is also symmetric since $A\cap B=B\cap A$.

it is not transitive since we can take $A=\{1,2\}, B=\{2,3\}, C=\{3,4\}$. We have $ARB$ and $BRC$ since $A\cap B=2$ and $B\cap C=3$. However we do not have $ARC$ since $A\cap C$ is empty. A wordy explanation is that just because $A$ and $C$ both have elements in $B$ it doesn't mean these are the same elements.

For problem $2$ you are correct, it is symmetric, this is because if $1\in A\cap B$ then $1\in B\cap A$ since $A\cap B=B\cap A$

You are also correct, it is transitive. if $1\in A\cap B$ and $1\in B\cap C$ then $1$ is in $A$ and $1$ is in $C$, therefore $1$ is in $A\cap C$.

It is not symmetric, if it was we would have that $1\in A\cap A$ for all $A$. Since $A\cap A=A$ we would have $1\in A$ for all $A$. This is clearly false if we take $A=\{2\}$ for example.

Asinomás
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