Why is this binary relation symmetric but not reflexive or transitive

Question

Let $\def\Rthree{\,{\mathrm{R}_3}\,} \Rthree$ be the relation on sets $C$, $D$ of natural numbers such that $C \Rthree D$ iff $C \cap D$ is finite. Then $\Rthree$ is symmetric, but not reflexive or transitive.

I don't understand any of the 3. The example for why it was not reflexive was

$\Bbb{N} \Rthree \Bbb{N}$ is not true

If that's the case can't I say that because $\Bbb{C} \Rthree \Bbb{N}$ isn't true, it isn't symmetric either? Why is it symmetric?

I would appreciate insight onto why it's not transitive as well. Thank you.

Answer 1

Remember the definition of a relation on $\Bbb N$:

$R$ is a relation on the power set of $\Bbb N$ $\mathscr{P}(\Bbb N)$ (the set of all subsets of $\Bbb N$) if it is a subset of $\mathscr{P}(\Bbb N) \times \mathscr{P}(\Bbb N)$.

This means that for every pair $(A,B) \in R$ we have $A \subseteq \Bbb N$ and $B \subseteq \Bbb N$. We are really only talking about subsets of $\Bbb N$.

Alex S addressed the symmetry, so I will talk about the reflexivity and the transitivity.

$R$ is not reflexive because there exists $A \subseteq \Bbb N$ such that $A \not R_3 A$, or such that $A \cap A$ is infinite. For a counterexample you can take any infinite subset of $\Bbb N$ (even $\Bbb N$ itself, as you wrote).

$R$ is not transitive, because we can find subsets of $\Bbb N$ $A,B$ and $C$ such that $A \ R_3 \ B$ and $B \ R_3 \ C$, but $A \not R_3 C$. For a counterexample, try to find two infinite sets $A,C$ whose intersection is infinite, and let $B$ be any finite set.

Answer 2

First of all, the relation only applies to natural numbers, so $\mathbb C R_3\mathbb N$ doesn't mean any thing. Even so, the symmetric property states:

If $A\, R\, B$, then $B\, R\, A$.

If $A\,\not R\, B$, then the property doesn't apply (we might say it is vacuously satisfied). Symmetry is true for this realtion because if $A\cap B$ is finite, then $B\cap A$ is finite. Again, if $A\cap B$ is not finite, then symmetry doesn't apply, so we don't even worry about $B\cap A$.

Answer 3

As an intuition, the relation is similar to saying "not equal" or "disjoint", so it is almost the opposite of an equivalence relation. It is the relation of disjointness on sets of integers modulo finite sets (where we regard two sets as equivalent if they differ on only a finite number of elements).