This is a concept that seems very intuitive to me, but I feel my proof is messy. Could someone more experienced than I perhaps offer some criticism/tell me if I'm even correct?
Let $A\subset\omega$. I want to show that $A$ is infinite if and only if $ \forall_m(m\in\omega\implies\exists_n(n\in A\land (m\lt n))). $
Here $a for $a,b\in\omega$. Also, $A\sim B$ means two sets are in bijection.
I think my proof is too roundabout. I prove $\implies$ by the contrapositive. Suppose the above property is not true. So there exists an $m$ such that there is no $n\in A$ such that $m
For $\impliedby$, I feel things get even worse. I try to prove the contrapositive by contradiction. I suppose $A$ is finite, but the property holds. Now if $A$ is empty, the property is false, a contradiction, so the contrapositive holds as needed. So assume $A$ is nonempty. Since $A$ is finite, $A$ must have a greatest element. (I try to justify this by applying the well-ordering principle on the reverse ordering. This is probably my biggest concern, is it acceptable?) Denote this element $k$. Consider $k^+$. Then $k^+\in\omega$, and by the property, we have that there is a $p\in A$ such that $k^+ , but $k , a contradiction. Apologies for this messy proof. How can I clean it up/correct it? Many thanks.