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**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 $\displaystyle A$.