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Suppose that $A$ is any set, $\mathcal{F}\subseteq\mathcal{P}(A)$ and $\emptyset\not\in\mathcal{F}$. Thus, the greatest lower bound of $\mathcal{F}$ in the partial order of subsets $\bigcap\mathcal{F}$ is lower and upper limit of $\mathcal{F}$ in the same partial order is $\bigcup\mathcal{F}$.

I have to prove it!

I appreciate if you can help me! :-D.

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What is your question? –  Trevor Wilson Mar 9 '13 at 6:10
    
I have to prove it! :-) –  Marcony Felipe Mar 9 '13 at 6:14
    
Example: proofwiki.org/wiki/… –  Marcony Felipe Mar 9 '13 at 6:15
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What you wrote is not quite grammatical, so I'm not sure how to prove it. But also you should tell us what you have tried so far and where you are stuck. –  Trevor Wilson Mar 9 '13 at 6:15
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Am I bad in english or the phrase "So, the greatest lower bound of F the partial order of subsets is ⋂F the partial order of subsets is F this same partial order is ⋃F" really makes no sense? –  Harold Mar 9 '13 at 6:16

1 Answer 1

HINT: You want to prove that if $\varnothing\notin\mathscr{F}\subseteq\wp(A)$, then the greatest lower bound of $\mathscr{F}$ in the partial order $\langle\wp(A),\subseteq\rangle$ is $\bigcap\mathscr{F}$, and the least upper bound of $\mathscr{F}$ is $\bigcup\mathscr{F}$. I’ll point you in the right direction for the first one; the second is very similar.

Let $L=\bigcap\mathscr{F}$. To show that $L$ is the greatest lower bound of $\mathscr{F}$, you must show two things:

  1. $L$ is a lower bound for $\mathscr{F}$, meaning that $L\subseteq F$ for every $F$ in $\mathscr{F}$.
  2. $L$ is the greatest lower bound of $\mathscr{F}$, meaning that if $B$ is any lower bound for $\mathscr{F}$, then $B\subseteq L$. In other words, you have to show that if $B$ is a subset of $A$ with the property that $B\subseteq F$ for each $F\in\mathscr{F}$, then $B\subseteq L$.

Both of these are easy to prove: they follow immediately from the definition of the intersection of a family of sets.

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