finding a group that satisfy: $x,y\in A_n\Rightarrow x\in y \vee y\in x \vee y=x$ for all natural number $n$ need to show a set $A_n$ with $n$ terms which satisfy: 
$x,y\in A_n\Rightarrow x\in y \vee y\in x \vee y=x$
i tried to think recursively but i'm stuck.
thanks! 
 A: You are right to try to think recursively. With that in mind, let's start with $n=0$, the only set with $0$ elements is $\varnothing$, and this satisfies the property vacuously.
Now suppose we have $A_n$ as required, we want a set $A_{n+1}$, with $n+1$ elements satisfying the property you want. As $A_n$ is a set of $n$ elements we would hope that we could add just one more element to it, and be done. We have to be careful about what we pick though, as we still want $x,y\in A_{n+1} \Rightarrow ((x\in y)\lor (y\in x) \lor (y = x))$.
Lets call our new element $x$, so we want for every $y\in A_{n}$ that the disjunct above is satisfied. It seems hard to require that $x\in y$ as we don't know what the $y$ look like, however can you think of a set $x$ so that $y\in x$ for each $y\in A_n$? How about $A_n$ itself?
So we define $A_{n+1} = A_n \cup \{A_n\}$ and we have what we want.
Edit note: It probably bears mentioning that any superset of $A_n$ would work here. However if you pick the minimal $A_n$ then you get a type of set called an ordinal which is a very important type of object in set theory.
A: HINT: Use induction. Suppose that $A_n$ were defined, what element $x$ would you add that all the members of $A_n$ are elements of $x$ itself?
