I already have proved in a previous first assuming that $A$ is a set of sets: $A$ is a finite set whose elementes are finite sets if and only if $\bigcup_{B \in A}$ is finite.
Here I will use $P_n$ to denote the set $\{x : x \in \mathbb{N} \land x <= n \}$, $P(A)$ to denote the power set of $A$ and $A \approx B$ to say that $A$ and $B$ are equinumerous.
Then first trying to show that if $A$ is finite then any subset of $P(A)$ have a minimal element, I first assumed that $A$ is finite, then since $\bigcup_{B \in P(A)} = A$, we know that $P(A)$ and all its members are finite sets (by the previous exercise)
Now take any nonempty subset $S$ of $P(A)$, then let $G = \{n : n \in \mathbb{N} \land (\exists X)(X \in S \land X \approx P_n)\}$, Since $S \neq \emptyset$, we know that there is some set $X$ in $S$ and $X \approx P_k$ for some $k \in \mathbb{N}$, thus $k \in G$.
Then we know that $G$ is a nonempty subset of $\mathbb{N}$, then by the least number principle $G$ have a least number $m$. Now we define a set $H = \{Y : Y \in S \land Y \approx P_m\}$.
Now if we take any element $Y$ in $H$, and assume that $Y$ is not minimal in $S$, then we have some set $Y'$ such that $Y' \subset Y$ and $Y' \in S$, but then $Y' \approx P_u$ such that $u<m$ and then $u \in G$, which contradicts $m$ as the least number.
Therefore any element in $H$ must be minimal in $S$.
I dont tried the start the converse part of the proof, because Im very confuse about the partial order relation established by inclusion relation. From other examples in internet I have seen that there are noncomparable elements, so in a set of sets with just noncomparable elements, what can be said about maximal and minimal?
I want to know if my approach is correct until this part, any tip on the converse will appreciated also.
Edit to insert the converse part
First I tried to assume that every nonempty subset $P(A)$ have a minimal element with respect to $\subseteq$ and then to think I way to show that $A$ must be finite, but I got stucked and at this point I dont have understood the given answers also, thus I choosed to try to put what Im trying to show in symbols.
I choosed to denote "S have a minimal element with respect to $\subseteq$" using the notation $M(S)$, then I rewrote what I need to prove as follows:
$$(\forall S)([S \neq \emptyset \land S \subseteq P(A)] \Rightarrow M(S)) \Rightarrow A \approx P_n$$
Then since I was stucked after just trying to use the assumption of the hypothesis, I wrote the contrapositive of the formula:
$$A \not\approx P_n \Rightarrow (\exists S)([S \neq \emptyset \land S \subseteq P(A)] \land \lnot M(S))$$
From reading the formula this way I understood the given answer, then I proceeded as follows, first assume $A \not\approx P_n$, then define the set S = $\{ X : X \in P(A) \land X \not\approx P_n \}$, by definition of $S$ we have $A \in S$, then we have $S \neq \emptyset \land S \subseteq P(A)$
We just need to show that $\lnot M(S)$ holds, we can do this by contradiction, first assuming that $M(S)$ and that $Y$ is a minimal element in $S$, but then if we take any element in $y$ in $Y$, and define the set $Y' = Y – \{y\}$, such that $Y' \not\approx P_n \land Y' \in P(A)$. But then $Y' \in S$ and $Y' \subset Y$, which contradict $Y$ as a minimal element in $S$, therefore $\lnot M(S)$ holds true.
Best Answer
Suppose $A$ is not finite. Then $A$ has a countably infinite subset $\{a_n | n \in \mathbb{N}\}$. Consider the family $\{\{a_n | n \in \mathbb{N}, n \geq m\} | m \in \mathbb{N}\}$. This family has no $\subseteq$-minimal element.