Set theory: Supremum and infimum sets proof

Question

Let $A$ be a partially ordered set and let $B \subset A$. Prove that $\upsilon(B)=\upsilon(\lambda(\upsilon(B)))$.

$\upsilon(B)$ is the set of all upperbounds of $B$ when $B$ is a subset of $A$.

$\lambda(B)$ is the set of all lower bounds of $B$.

My attempt:

By the theorem: if $B$ is a subset of $A$, then $B \subset \lambda(\upsilon(B))$,

we have, $B \subset \lambda(\upsilon(B))$

and by the theorem: If $B \subset C$, then $\upsilon(C) \subset \upsilon(B)$,

we have $\upsilon(\lambda(\upsilon(B))) \subset \upsilon(B)$

I have proved the reverse inclusion as above, but I have no idea of proving the forward inclusion. Any help would be appreciated.

Answer 1

Assume that $a\in \upsilon(B)$.

Pick any $y\in\lambda(\upsilon(B))$. Then $y\leq a$ by the definition of $\lambda$. Thus by the definition of $\upsilon$ we have $a\in\upsilon(\{y\})$. Since $y$ was chosen arbitrarly from $\lambda(\upsilon(B))$ then $a\in\upsilon(\lambda(\upsilon(B)))$. Thus

$$\upsilon(B)\subseteq \upsilon(\lambda(\upsilon(B)))$$

Answer 2

Let $a \in \upsilon(B)$.
This means that $a \geq b$, for all $b \in B$.
Now we want to prove that $a \in \upsilon( \lambda( \upsilon(B) ) )$, that is, $a \geq c$, whenever $c \in \lambda( \upsilon(B) )$.
So pick $c \in \lambda(\upsilon(B))$; by definition of $\lambda$, $c \leq a$, which was what we wanted to prove.