Handling empty sets in relations

Question

Let A consist of all 8 subsets of the set $\{1, 2, 3\}$, that is

A = $\{ \phi , \{1\}, \{2\}, \{3\}, ...\}$

The relation $R$ on $A$ is defined by:

$aRb$ iff some element of $a$ is larger than some element of $b$.

My question is, is $(\phi,\{1\})$ an element of $R$?

My answer is no, because there simply is no element of $\phi$. Similarly, $(\{1\}, \phi)$ is not in $R$. Is my thinking correct?

Answer 1

You are correct. This is a fundamental idea about the empty set that always deserves a stop and check.

$aRb$ iff some element of $a$ is larger than some element of $b$.

So is there an element of $\emptyset$ that is larger than an element of $\{1\}$? As you say, no.

By contrast, if the rule had been

$aRb$ iff all elements of $a$ are larger than all elements of $b$.

then $(\emptyset,\{1\})\in R$ vacuously true because there isn't an element of $\emptyset$ that is a counterexample to the definition.