Vrftsjtryk

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
$$U(A):= { bin L | forall ain A: a leq b} \
L(A):= { bin L | forall ain A: b leq a}$$

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) $Asubseteq C(A)$

2) $C(C(A)) = C(A)$

3) $Asubseteq B implies C(A) subseteq C(B)$

My try:

1) $bin A implies forall ain U(A): bleq a implies b in L(U(A)) implies A subseteq L(U(A)) = C(A)$

2) ?

3) We have property: $A subseteq B implies U(B) subseteq U(A)$ and $L(B) subseteq L(A)$.

Thus
$ Asubseteq 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.