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Math Help - Zorn's lemma problem

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by Ester View Post
    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
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