I've been having trouble finding the domain and the image of this set: {<$a,b$> : $a,b$ $\in \Bbb N \land a>b$} (For those who don't get it, it is an infinite relation, made of ordered pairs, for which $a$ and $b$ are natural numbers, and for each of them, $a$ is bigger than $b$)
Logically I think that the domain should be the set of Natural numbers excluding $0$ a.k.a the lowest number possible because $a>b$ (assuming $0$ is in them).
And the image should be all the natural numbers? We know that the natural numbers set is infinite, thus there are infinite numbers that are bigger than $b$.
I'm not really sure if I'm correct and I would be happy if someone could check this.
Thanks.
As customary, let us call that relation $R$.
The domain of $R$ is $\{ m \in \Bbb N \mid \exists n \in \Bbb N \ m > n \}$. Notice that if $m = 0$, there exist no $n \in \Bbb N$ such that $0 > n$, so $0$ does not belong to the domain. On the other hand, every other natural number has another number strictly smaller than it (its predecessor, for instance). Therefore, the domain of $R$ is $\Bbb N \setminus \{ 0 \}$.
The range of $R$ is $\{ n \in \Bbb N \mid \exists m \in \Bbb N \ m > n \}$. It is clear that for every natural number there exist another one strictly greater (its successor, for instance), therefore the range of $R$ is $\Bbb N$.
This means that your argument is correct.