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I refer to Exercise 23 of Section 4.4 of Velleman's 2nd edition book. It's also Theorem 4.4.11 in the Section. This post is not a duplicate of this (I include the posts that are mentioned there). I noticed that people start by assuming some arbitrary $F\in \mathscr{F}$, but this is clearly false since, in my perspective, we need to prove that $\cup\mathscr{F}$ is indeed an upper bound of $\mathscr{F}$ and that it is the lowest upper bound.

Suppose $A$ is a set, $\mathscr{F}\subseteq \mathscr{P}(A)$, and $\mathscr{F}\neq\emptyset$. Then the least upper bound of $\mathscr{F}$ (in the subset partial order) is $\cup\mathscr{F}$ and the greatest lower bound of $\mathscr{F}$ is $\cap\mathscr{F}$.

I was first trying to prove the least upper bound. This is what I have:

  • We need to prove that $\cup\mathscr{F}$ is the least upper bound (lub) of $\mathscr{F}$. This means that $\cup\mathscr{F}$ is the smallest element of $U$, the set of upper bounds for $\mathscr{F}$. Also, we can say that $U$ is a family of sets
  • The definition of the smallest element of $U$ would give us the followind: $\forall M\in U (\cup\mathscr{F}\subseteq M)$ and $\cup\mathscr{F}\in U$
  • We then proceed to assume some arbitrary $M\in U$ and we need to prove that $\cup\mathscr{F}\subseteq M$ and $\cup\mathscr{F}\in U$
  • We know that $\cup\mathscr{F}\subseteq M$ means that $\forall x\in\mathscr{F}(x\in M)$
  • We proceed to assume some arbitrary $x\in\cup\mathscr{F}$ and we need to prove that $x\in M$ and $\cup\mathscr{F}\in U$

This is as far as I can go. I know that $x\in \cup\mathscr{F}$ means that there is some set $N\in\mathscr{F}$ so that $x\in N$, but I don't see how this helps me to prove any of my goals.

I would appreciate any hints or help you could provide.

Alex
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  • It’s not clear what your objection to them is, but the arguments at your link that start with an arbitrary element of $\mathcal{F}$ are correct and accomplish exactly what you say is needed. – Brian M. Scott Apr 13 '21 at 20:37

2 Answers2

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To prove that $\bigcup \mathscr F$ is an upper bound of $\mathscr F$ it is sufficient to show that $F \subseteq \bigcup \mathscr F$, for all $F \in \mathscr F$.
I suppose that that is clear enough.

Now, if $G$ is another upper bound of $\mathscr F$, then $F \subseteq G$, for all $F \in \mathscr F$, whence $$\bigcup \mathscr F = \bigcup \{ F : F \in \mathscr F \} \subseteq G.$$ So $\bigcup \mathscr F$ is the least upper bound of $\mathscr F$.

The proof that $\bigcap \mathscr F$ is the grates lower bound of $\mathscr F$ is analogous.

amrsa
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  • Just one question, to prove that $\cup\mathscr{F}$ is a lower bound, are we not supposed to prove also that it is an element of $\mathscr{P}(A)$? – Alex Apr 13 '21 at 18:45
  • @AlejandroRuiz You mean, to prove that $\bigcap\mathscr F$ is a lower bound. No we're not. That's obvious: by definition $\mathscr F$ is a collection of subsets of $A$; if $(F_i){I\in I}$ are subsets of $A$, then so is $\bigcap{I\in I}F_i$. Why? Just pick $x$ in the intersection; can you see why it must belong to $A$? – amrsa Apr 13 '21 at 19:44
  • @AlejandroRuiz or maybe you meant to prove that $\bigcup\mathscr F$ is an upper bound. Anyway, it's a similar reasoning. – amrsa Apr 13 '21 at 19:50
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You have made a good start on this proof. Here's a hint for the next step: You know that $M \in U$, but you don't seem to have thought about what that tells you. You defined $U$ to be the set of all upper bounds for ℱ, so $M \in U$ tells you that $M$ is an upper bound for ℱ. That should help you.

Also: You might find this website useful for practicing proofs like this one: https://djvelleman.people.amherst.edu/pd.html

Dan Velleman
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  • Absolutely right! I had completely ignored that detail. Just one question, then when trying to prove that $\cup\mathscr{F}$ is the l.u.b., do I have to prove that $\cup\mathscr{F}\in U$ (i.e. $\cup\mathscr{F}$ is an element of $\mathscr{F}$ as part of the definition for an u.b.)? – Alex Apr 13 '21 at 19:41
  • Yes, you have to prove that $\bigcup\mathscr{F} \in U$, and since $U$ is the set of all upper bounds for $\mathscr{F}$, that means you have to prove that $\bigcup\mathscr{F}$ is an upper bound for $\mathscr{F}$. Now write out the definition of "upper bound" to see how to prove that. – Dan Velleman Apr 13 '21 at 19:49