How to proove that smallest upper bound exists und it is cleary determined?

Question

Let X be a set. Then a relation '$\le$' on $\mathcal P(X)$ is defined by: $A \le B :\Leftrightarrow A \subset B$ .

Let $\mathcal A \subset \mathcal P(X)$. One set $B \in \mathcal P(X)$ for which A$\le$B for all $A \in \mathcal A$ is valid, is one upper bound from $\mathcal A$. B is here smallest upper bound from $\mathcal A$, if B is upper bound from $\mathcal A$ and B$\le$C for any upper bound C from $\mathcal A$ is valid. Show that one smallest upper bound for $\mathcal A$ exists and it is clearly determined. And that set would be marked with sup$\mathcal A$.

Answer 1

Given any $\mathcal A \subseteq \mathcal P(X)$, we claim that: $$ \sup\mathcal A = \bigcup_{A \in \mathcal A}A $$ To see this, first note that $\bigcup_{A \in \mathcal A}A \in \mathcal P(X)$ is an upper bound for $\mathcal A$. Indeed, given any $S \in \mathcal A$, we have from the properties of union that: $$ S \subseteq \bigcup_{A \in \mathcal A}A $$ It remains to show that $\bigcup_{A \in \mathcal A}A$ is the smallest upper bound. To this end, choose any upper bound $T \in \mathcal P(X)$ so that $Y \subseteq T$ for all $Y \in \mathcal A$. We want to show that $\bigcup_{A \in \mathcal A}A \subseteq T$.

• With this in mind, choose any $a \in \bigcup_{A \in \mathcal A}A$ so that $a \in A$ for some $A \in \mathcal A$. Then by taking $Y = A$, we know (since $T$ is an upper bound) that $A \subseteq T$ so that $a \in T$, as required.

Thus, $\sup\mathcal A = \bigcup_{A \in \mathcal A}A$, as desired. $~~\blacksquare$