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.

Maybe join and meet : "In a partially ordered set $P$, the join and meet of a subset $S$ are respectively the supremum of $S$, denoted $\lor S$, and infimum of $S$, denoted $\land S$" ? — Mauro ALLEGRANZA, Apr 19 '17 at 14:42
You say you have proven the forward inclusion, $\upsilon(B) \subseteq \upsilon(\lambda(\upsilon(B)))$. So what follows the "My attempt" line is to prove the reverse, but it seems like you proved that as well... — amrsa, Apr 19 '17 at 14:46
@amrsa oops. I mean I have proved the reverse inclusion. Sorry that was a typo. Thank you — Cruso James, Apr 19 '17 at 14:55

score 3 · Answer 1 · answered Apr 19 '17 at 15:02

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

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

score 1 · Answer 2 · answered Apr 19 '17 at 15:03

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.

Set theory: Supremum and infimum sets proof

2 Answers2