1. ## Zorn's lemma problem

I have this kind of problem:

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

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

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

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

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

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

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

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

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

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

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

You mean "a chain"

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

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

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

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

Tonio