Set theory task Let F be a set of sets.
We say a set is n-pretty  (it's writen in my text book, I don't know the exact name)(n is fixed nat. number) if
$$\forall X (X \in F \iff \forall Y (Y \subseteq X \land |Y| <= n \implies Y \in F)$$.
And we say that a set is finite-pretty if:
$$\forall X (X \in F \iff \forall Y (Fin(Y) \land Y \subseteq X \implies Y \in F)$$
So the task is to find if every n-pretty set is finite-pretty
and if every finite-pretty set is n-pretty.
I think that every finite-pretty set is n-pretty. Since it's true for every finite set, it will be true for every set with cardinality n. But I can't seem to  prove it in terms of set theory. And what about the other direction ? Any tips?
 A: (i) It is not true that every finite-pretty set is $n$-pretty.
Consider for example the set $F$ of all $X\subseteq \mathbb{N}$ such that $|X|\le n$. $F$ is closed under taking finite subsets, which shows the $\Rightarrow$ direction of finite-prettiness. On the other hand, let $X$ be any set such that all its finite subsets $Y$ are in $F$. Then $X$ cannot have subsets of size $>n$, and so $X$ itself has size $\le n$, which in turn implies $X\in F$. This shows the $\Leftarrow$-direction of finite-prettiness.
However $F$ is not $n$-pretty, as $Y\in F$ for all sets $Y\subseteq \mathbb{N}$ such that $|Y|\le n$, but $\mathbb{N}\notin F$.
(ii) All $n$-pretty sets are finite-pretty.
Let $F$ be $n$-pretty. Assume first that $X\in F$ and $Y\subseteq X$ is finite; we have to show $Y\in F$. Let $Z\subseteq Y$ have size $\le n$. Since $Z\subseteq X\in F$ and $F$ is $n$-pretty, it follows that $Z\in F$. Since $Z$ was arbitrary, all subsets of $Y$ of size $\le n$ are contained in $F$. Again by $n$-prettiness, this implies that $Y$ itself is in $F$. This is the $\Rightarrow$ direction of finite-prettiness. For the other direction, assume that $X$ is a set such that all finite $Y\subseteq X$ are in $F$. Then in particular $Y\in F$ for all $Y\subseteq X$ of size $\le n$, and so $X\in F$ by $n$-prettiness.
A: It's indeed true that all finite-pretty $F$ is $n$-pretty for any $n \in \Bbb N$ because (trivial set theory) $|Y| \le n \implies Y \in \textrm{Fin}(Y)$. This is almost true by definition: a set is finite iff there is some $n \in \Bbb N$ such that its cardinality is $=n$ or $\le n$, definitions vary by text...
The reverse fails in general: the powerset of $\Bbb N$ without the singletons is a boring example of a $2$- and $3$-pretty set that is not finite pretty.
