Using zorn's lemma to prove aleph 0 is the least infinite cardinal I understand zorn's lemma, but it is confusing me on how to apply it to an infinite set. 
I am asked to prove that if A is an infinite set, then there exists an injection from $\mathbb{N}$->A using Zorn's lemma. 
The proof is easy using the axiom of choice, but I cannot figure out how to ensure that the set is always partially ordered and every totally ordered subset has an upper bound.
Thank you
 A: Hint: Let $\mathcal{F}$ be the set of all injective functions $f$ such that $f$ has domain a subset of $\mathbb{N}$, and $f$ has codomain $A$.
For any two such functions $f$ and $g$, write $f\lt g$ if the domain $D_f$ of $f$ is a subset of the domain $D_g$ of $g$, and $g$ restricted to $D_f$ is $f$.
Note that any chain in $\mathbb{F}$ has an upper bound. 
A: Consider the collection of finite subsets of $A$. If it has a maximal element, then $A$ must be finite, therefore there exists a chain without an upper bound. Therefore it must be infinite (as finite chains has a maximal element which is a bound).
Note that Zorn's lemma proves, quite easily, that every chain can be extended to a maximal chain. So we may assume that this chain is a maximal chain.
But this chain is a chain of finite subsets of $A$, each cardinality appearing in this chain is distinct. By maximality we have that every finite cardinality appears in that chain exactly once. 
Write $A_n$ for the unique set of size $n$ in this chain, can you come up with an injection from $\Bbb N$ into $\bigcup A_n$?
