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Thread: Zorn's lemma problem

  1. #1
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    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 $\displaystyle A$.
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  2. #2
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    Quote Originally Posted by Ester View Post
    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
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