# Prove L.U.B. of $\mathcal{F} = \cup\mathcal{F}$

The title refers to an exercise in Velleman's "How to Prove It" that I'm trying to work through for self-study. My proof doesn't feel right, so am hoping that the wisdom of the group can help me correct my mistakes. Specifically, I'm suspect of the proof because:

1) I haven't used all the givens;

2) I'm assuming a least upper bound exists in the left direction, and;

3) I only use the fact that the least upper bound is an upper-bound.

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

($\rightarrow$) Suppose $F'$ is the least upper bound of $\mathcal{F}$. Then by definition $F' \in \mathcal{F}$. Next, let $f \in F'$. Then clearly $f \in \cup\mathcal{F}$. So $F' \subseteq \cup F$.

($\leftarrow$) Suppose $\mathcal{F}$ has a least upper bound $F'$, and let $x \in \cup \mathcal{F}$. It follows that $\exists A \in \mathcal{F}(x \in A)$. But since $F'$ is an upper bound, then $A \subseteq F'$. So $x \in F'$ and $\cup \mathcal{F} \subseteq F'$.

Thanks for any and all comments!

• We don't necessarily have $F' \in \mathcal F$ in your $(\to)$ paragraph. For instance, take $A = \{a, b\}$ with $\mathcal F = \{\{a\}, \{b\}\}$. In that case, $A$ is the only upper bound for $\mathcal F$, and we don't have $A \in \mathcal F$. – Arthur Nov 10 '15 at 19:48

For all $F \in \mathcal{F}$, we have that $F \subseteq \cup \mathcal{F}$, or $F \le \cup \mathcal{F}$ (to use the order notation). This means that $\cup \mathcal{F}$ is an upper bound of $\mathcal{F}$. This is one part of the definition.
Suppose that $G$ is another upper bound for $\mathcal{F}$. Then for each $F \in \mathcal{F}$ we have that $F \le G$, or otherwise put $F \subseteq G$. So the union of all such $F$, $\cup \mathcal{F} \subseteq G$. So $\cup \mathcal{F}$ is less than or equal to any upper bound for $\mathcal{F}$, so the latter is the least upper bound, by definition.