Lowest upper bound of a family of sets

Question

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.

Answer 1

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.

Answer 2

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