Ordering Relation

Question

Prove the following theorem: Suppose $A$ is a set, $F \subseteq P (A)$, and $F \neq \varnothing.\;$ Then the least upper bound of $F$ (in the subset partial order) is $\bigcup F.$

Answer 1

Hint: Show that for every $B\in F$ we have $B\subseteq\bigcup F$, so it is an upper bound; and if $C$ is such that for all $B\in F$ we have $B\subseteq C$ then $\bigcup F\subseteq C$.

Answer 2

Look at the definition of the least upper bound, first, it's an upper bound, and this is correct for $\bigcup F$ obviously. second, you should try to show that every set that includes all the members of $F$, should include $\bigcup F$.

Answer 3

Hint: You need to check

• $X\subseteq \bigcup F$ for all $X\in F$,
• if $Y\in P(A)$ and $X\subseteq Y$ for all $X\in F$ then $\bigcup F\subseteq Y$.