Define a well-order $\le$ in $P_\text{fin}(\mathbb{N})$ - (it's the set of all finite subsets of $\mathbb{N}$, the natural numbers) that: $ A, B \in P_\text{fin}(N) (A \subseteq B \rightarrow A \le B $)
Could someone solve this for me? I don't have an idea how to do it.
On
For every $k$, the set of $k$-element subsets of $\Bbb N$ is countable, so fix arbitrary well-order $\leq_k$ on it. Now order elements of $P_{fin}(\Bbb N)$ by $A\leq B$ if $|A|<|B|$ or $|A|=|B|=k$ and $A\leq_kB$. I leave it for you to check that this satisfies all the requirements (including the fact that it's a well-order).
On
One way to do this: enumerate the set of all primes $p_i: i = 0,1,2,3,\ldots$. Then define $P: F \rightarrow \mathbb{N}$ by $P(F) = \prod_{n \in F} p_n$ (the product is well-defined as we have a finite set), and show this is an injective (by unicity of prime factorisations) and increasing function (a superset has a larger $P$ value). Then $F \le G$ is defined by $P(F) \le P(G)$. This defines a linear order on your set isomorphic to an infinite subset of $\mathbb{N}$, so this will be a well-order for that reason.
Define $f\colon P_{\text{fin}}(\Bbb N)\to \Bbb N$ as $$ f(A):=\sum_{n\in A}2^n$$ and declare $$ A\le B\;:\Leftrightarrow\; f(A)\le f(B).$$ Note that $f$ is injective (in fact, bijective) and so relation defined is a well-order because $\Bbb N$ is well-ordered. Also, if $A$ is the disjoint union of $B$ and $C$ then $f(A)=f(B)+f(C)$, hence $B\subseteq A$ indeed implies $B\le A$, as desired.