0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ What is $\upsilon$ and $\lambda$? $\endgroup$ – amrsa Apr 19 '17 at 14:37
  • $\begingroup$ @amrsa I have added the information. $\endgroup$ – Cruso James Apr 19 '17 at 14:41
  • $\begingroup$ 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 $\land S$" ? $\endgroup$ – Mauro ALLEGRANZA Apr 19 '17 at 14:42
  • $\begingroup$ 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... $\endgroup$ – amrsa Apr 19 '17 at 14:46
  • $\begingroup$ @amrsa oops. I mean I have proved the reverse inclusion. Sorry that was a typo. Thank you $\endgroup$ – Cruso James Apr 19 '17 at 14:55
3
$\begingroup$

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

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.