Let $(L, \le)$ be a poset regarded as a lattice (s.t. $\forall a, b \in L$ both $sup(\{a,b\})$ and $inf(\{a,b\})$ exist in L).
Let $A \subseteq L$. We denote $$U(A):= \{ b\in L | \forall a\in A: a \leq b\} \\ L(A):= \{ b\in L | \forall a\in A: b \leq a\}$$
sets of all upper and lower bounds of $A$.
I need to prove that map $C(A):= L(U(A))$ defines a closure operator on $L$. That is I need to prove 3 properties:
1) $A\subseteq C(A)$
2) $C(C(A)) = C(A)$
3) $A\subseteq B \implies C(A) \subseteq C(B)$
My try:
1) $b\in A \implies \forall a\in U(A): b\leq a \implies b \in L(U(A)) \implies A \subseteq L(U(A)) = C(A)$
2) ?
3) We have property: $A \subseteq B \implies U(B) \subseteq U(A)$ and $L(B) \subseteq L(A)$.
Thus $ A\subseteq B \implies U(B) \subseteq U(A) \implies L(U(A)) \subseteq L(U(B))$
I'm strugling to prove idempotency property. I guess to prove it I need to show that $$U(A) = U(L(U(A)))$$ but I don't know how to do it formally.
Question: are these proofs valid and how do I prove idempotency?
Edit:
2) $U(L(U(A))) \subseteq U(A)$ from properties (1) and (3). $$a \in U(A) \implies \forall b \in L(U(A)): b \leq a \implies a \in U(L(U(A))) \implies U(A) \subseteq U(L(U(A)))$$ That is equality holds, thus $C$ is idempotent.
Thanks everyone!
As Kevin Long wrote in a comment, you already have that $U(A) \subseteq U(L(U(A)))$.
For the converse inclusion, using (1), and your observation that $A \subseteq B$ implies $U(B) \subseteq U(A)$, $$A \subseteq L(U(A)) \Longrightarrow U(L(U(A))) \subseteq U(A).$$ Notice also that this is true for any poset, which is quite clear since nowhere in the proof is any lattice property used.