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Theorem Suppose $A$ is a set, $F ⊆ P (A)$, and $F \ne ∅$. Then the least upper bound of $F$ (in the subset partial order) is $∪F$ and the greatest lower bound of $F$ is $∩F$.

Proof: Since any element of $F$ is a subset of $\cup F$, $\cup F$ upper bound of $F$. To prove $\cup F$ is the least upper bound, let $U$ be the set of all upper bounds. Suppose $x\in U$. Suppose $y\in \cup F$. Since $y\in \cup F$ there is a $B\in F$ such that $y\in B$. Since $x\in U$, then for all $h\in F$, $h\subseteq x$. So $B\subseteq x$. Thus $y\in x$. Thus $\cup F$ is the least upper bound of $F$.

Suppose $x\in F$. Suppose $y\in \cap F$. Then $y\in x$. Thus $\cap F$ is the lower bound. To show $\cap F$ the greatest lower bound, suppose $L$ is the set of all lower bounds. Suppose $x\in L$. Suppose $y\in x$. Suppose $z\in F$. Since $x$ is a lower bound and $z\in F$ then $x\subseteq z$. Thus $y\in z$. So $\cap F$ is the greatest lower bound. $\square$

  1. Is this proof correct?

  2. The assumptions state that $F\ne \emptyset $. I have not used this assumption in my proof thus was it a necessary assumption for this theorem to stay true. Or is it a redundant assumption?

Martin Sleziak
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Nameless
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  • Suppose $x\in U$. Suppose $y\in x$. Thus $y\in \cup F$. - There must be a typo. This implies that for all $x \in U$, $ x \subset \cup F$. The rest seems to be good to me... 2) It is not redundant because we usually don't consider the intersection of empty set. Usual definition of intersection set will make it be the whole set $A$.
    • – Asydot Sep 25 '15 at 14:34
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    Have you thought about what the intersection over the empty set, $\bigcap\emptyset$ is? I think that's the reason they assume that $F$ does indeed have elements. – Arthur Sep 25 '15 at 14:34
  • @Asydot I corrected it now. – Nameless Sep 25 '15 at 14:46
  • @Arthur $\cap \emptyset $ is a null set. Do you mean that when proving that $\cap F$ is the greatest lower bound the conclusion will always be false thus the theorem incorrect ? – Nameless Sep 25 '15 at 14:51
  • @Arthur But isnt proving $y\in \cap \emptyset$ mean $\forall A \in \emptyset (y\in A)$ which is vacuously true ? – Nameless Sep 25 '15 at 14:58
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    @Nameless exactly. Which means that $\bigcap\emptyset$ contains every element there is. That invites all kinds of trouble. Therefore it is best to assume that $F$ is non-empty. – Arthur Sep 25 '15 at 16:36
  • Some discussion of empty intersection is also in this post: Empty intersection and empty union. – Martin Sleziak Sep 21 '20 at 04:45