transitive relation question I need to show if the following relation is transitive :
$$
R\subseteq \mathcal{P}(\mathbb{N})\times\mathcal{P}(\mathbb{N}) \space \text{with} \space XRY \space :\Leftrightarrow \exists x \in\mathbb{N}:\space x\in X \space \land \space x\in Y
$$
The answer says that it is not transitive because there is:
$$ \{1,2\}R\{2,3\}\space \text{and} \space \{2,3\}R\{3,4\} \space \text{but}\space \space \text{not} \space
 \{1,2\} \cap \{3,4\}=\emptyset $$
what does intersection have to do with it?
 A: When you say you have a relation $R$ on a set "A" you're basically saying that $R \subseteq A \times A$ that follows some specific conditions:
$$R = \{(a_1,a_2) \in A \times A: \text{such that $a_1$ and $a_2$ satisfy a condition}\}$$
In your case you have $R \subseteq \mathcal{P}(\mathbb{N}) \times \mathcal{P}(\mathbb{N})$ and as you might be aware $\mathcal{P}(\mathbb{N})$ is the set of all subsets of natural numbers so every element $a \in \mathcal{P}(\mathbb{N})$ is necessarily a set of natural numbers.
Thus, in this case our relation is set subset of $\mathcal{P}(\mathbb{N}) \times \mathcal{P}(\mathbb{N})$ given by
$$R = \{(A,B) \in \mathcal{P}(\mathbb{N}) \times \mathcal{P}(\mathbb{N}):  \text{such that the sets $A$ and $B$ satisfy a condition}\}$$
In this case the condition for $(A,B) \in R$ (in other words $ARB$) is that theres is at least a natural number $x$ that is simultaneously in $A$ and in $B$ therefore in order to $(A,B) \in R$ we must have $$ A \cap B \neq \emptyset$$
You want to prove that $\{1,2\}R\{3,4\}$ but this only happens if and only if {1,2} and {3,4} have elements in common and they certainly don't: $$\{1,2\} \cap \{3,4\} = \emptyset$$
Therefore $R$ is not transitive because you found a counter-example in $\mathcal{P}(\mathbb{N})$.
