1. ## Zorn's lemma problem

I have this kind of problem:

Let $\displaystyle A$ be a non-empty set, where is valid

$\displaystyle \forall B (B \in A \Leftrightarrow b \in A$ with every finite $\displaystyle b \subseteq B)$.
Prove that there is the maximal element in the set $\displaystyle A$.

My solution (or the idea at least):
First, let's choose arbitrary $\displaystyle C \subseteq A$.
If $\displaystyle C$ is the chain, then $\displaystyle \forall B_i, B_j \in C$ holds $\displaystyle B_i \subseteq B_j$ or $\displaystyle B_j \subseteq B_i$.

Now I'm facing the problem. I know that in set $\displaystyle C$, there exists an element $\displaystyle B_m$, which has every finite $\displaystyle b \subseteq B_i$, where $\displaystyle B_i \in C$. (It occur me, that maybe $\displaystyle B_m = C$. Is that possible?)
How can I prove, that there exists this element $\displaystyle B_m$?

After I have solved the problem above, I can proceed with my solution:
$\displaystyle \forall b \subseteq B_m : \forall x \in b (x \in B_m) \Leftrightarrow \forall b \subseteq B_m: \forall x \in b (x \in \bigcup C) \Leftrightarrow \forall b \subseteq B_m: b \subseteq \bigcup C$.
Because $\displaystyle C \subseteq A$, then every $\displaystyle b \subseteq B_i$,where $\displaystyle B_i \in C$, also holds that every $\displaystyle b \subseteq A$. That means, that $\displaystyle b \subseteq A$ with every finite $\displaystyle b \subseteq \bigcup C$. So $\displaystyle \bigcup C \in A$.
And according to Zorn's lemma, there exists the maximal element in the set $\displaystyle A$.

2. Originally Posted by Ester
I have this kind of problem:

Let $\displaystyle A$ be a non-empty set, where is valid

$\displaystyle \forall B (B \in A \Leftrightarrow b \in A$ with every finite $\displaystyle b \subseteq B)$.

So A is a set whose elements are sets...?

Prove that there is the maximal element in the set $\displaystyle A$.

My solution (or the idea at least):
First, let's choose arbitrary $\displaystyle C \subseteq A$.
If $\displaystyle C$ is the chain,

You mean "a chain"

then $\displaystyle \forall B_i, B_j \in C$ holds $\displaystyle B_i \subseteq B_j$ or $\displaystyle B_j \subseteq B_i$.

Now I'm facing the problem. I know that in set $\displaystyle C$, there exists an element $\displaystyle B_m$, which has every finite $\displaystyle b \subseteq B_i$, where $\displaystyle B_i \in C$. (It occur me, that maybe $\displaystyle B_m = C$. Is that possible?)
How can I prove, that there exists this element $\displaystyle B_m$?

After I have solved the problem above, I can proceed with my solution:
$\displaystyle \forall b \subseteq B_m : \forall x \in b (x \in B_m) \Leftrightarrow \forall b \subseteq B_m: \forall x \in b (x \in \bigcup C) \Leftrightarrow \forall b \subseteq B_m: b \subseteq \bigcup C$.
Because $\displaystyle C \subseteq A$, then every $\displaystyle b \subseteq B_i$,where $\displaystyle B_i \in C$, also holds that every $\displaystyle b \subseteq A$. That means, that $\displaystyle b \subseteq A$ with every finite $\displaystyle b \subseteq \bigcup C$. So $\displaystyle \bigcup C \in A$.
And according to Zorn's lemma, there exists the maximal element in the set $\displaystyle A$.

The above is a mess. Let's try the following: first we must define/choose the partial order we want to work with in A, so I propose, since the elements of A are sets, to partial order them wrt to set inclusion.

Let C be a chain in A (i.e., for any two $\displaystyle X,Y\in C$ , either $\displaystyle X\subset Y\,\,\,or\,\,\,Y\subset X$ , and let us now take $\displaystyle M:=\bigcup_{X\in C}X$ . Clearly $\displaystyle M$ is an upper bound of $\displaystyle C$ and what you now have to prove is that $\displaystyle M\in A$ in order to be able to use Zorn's lemma...so you have to prove that for any finite $\displaystyle t\subset M\,,\,also\,\,\,t\in A$...and this isn't hard.

Tonio