I am really struggling with how to write down a proper proof (in general but particularly with this problem), the exact exercise states:
Let $X$ be a set and $\mathcal{P}(X)$ its powerset. Let $A \neq \emptyset$ be a subset of $\mathcal{P}(X)$. Show that $\sup A$ equals the union of $A$ and $\inf A$ the intersection of $A$.
Edit: While $\mathcal{P}(X)$ is ordered by $\subseteq$.
It makes intuitive sense to me. Since any set in $A$ is logically a subset of the union of all sets in $A$. And the intersection of $A$ would be a set that is a subset of all other sets in $A$: worst-case the empty set.
My only problem is how to write this down in the form of an acceptable proof.
Apologies for the maybe trivial question, but I feel like I never learned how to write a proper proof.
First of all, under inclusion, every non-empty $A\in\mathcal{P}(X)$ has both an upper bound and a lower bound. This is because $$u=\bigcup_{s\in A}s$$ is greater than or equal to every $s\in A$ (since $s\subseteq u$ for all $s\in A$ by the definition of $u$), and also $$l=\bigcap_{s\in A}s$$ is less than or equal to every $s\in A$ (since $s\supseteq l$ for all $s\in A$ by the definition of $l$). We will now show that $u$ and $l$ are the least upper bound and the greatest lower bound of $A$.
Suppose that $u'$ and $l'$ are an arbitrary upper bound and an arbitrary lower bound of $A$. Then $s\subseteq u'$ and $s\supseteq l'$ for every $s\in A$. Therefore, $$u=\bigcup_{s\in A}s\subseteq u'$$ and $$l=\bigcap_{s\in A}s\supseteq l'.$$ The claim follows.