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.
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.