# upper set operation is a closure operator

###### Definition 1.

Let $P$ be a poset and $A$ a subset of $P$. The upper set of $A$ is defined to be the set

 $\uparrow\!\!A=\{b\in P\mid(\exists\,a\in A)\,a\leq b\}$

Now, we verify each of the properties which is required of a closure operator.

###### Theorem 1.

$\uparrow\!\!\emptyset=\emptyset$

###### Proof.

Any statement of the form “$(\exists\,a\in\emptyset)P(a)$” is identically false no matter what the predicate  $P$ (i.e. it is an antitautology) and the set of objects satsfying an identically false condition is empty, so $\uparrow\!\!\emptyset=\emptyset$. ∎

###### Theorem 2.

$A\subseteq\uparrow\!\!A$

###### Proof.

This follows from reflexivity  — for every $a\in A$, one has $a\leq a$, hence $a\in\uparrow\!\!A$. ∎

###### Theorem 3.

$\uparrow\uparrow\!\!A=\uparrow\!\!A$

###### Proof.

By the previous result, $\uparrow\!\!A\subseteq\uparrow\uparrow\!\!A$. Hence, it only remains to show that $\uparrow\uparrow\!\!A\subseteq\uparrow\!\!A$. This follows from transitivity. In order for some $a$ to be an element of $\uparrow\uparrow\!\!A$, there must exist $b$ and $c$ such that $a\geq b\geq C$ and $C\in A$. By transitivity, $A\geq C$, so $a\in\uparrow\!\!A$, hence $\uparrow\uparrow\!\!A\subseteq\uparrow\!\!A$ as well. ∎

###### Theorem 4.

If $A$ and $B$ are subsets of a partially ordered set  , then

 $\uparrow\!\!(A\cup B)=(\uparrow\!\!A)\cup(\uparrow\!\!B)$
###### Proof.

On the one hand, if $a\in\uparrow\!\!(A\cup B)$, then $a\geq b$ for some $b\in A\cup B$. It then follows that either $b\in A$ or $b\in B$. In the former case, $a\in\uparrow\!\!A$, in the latter case, $a\in\uparrow\!\!B$ so, either way $a\in(\uparrow\!\!A)\cup(\uparrow\!\!B)$. Hence $\uparrow\!\!(A\cup B)\subseteq(\uparrow\!\!A)\cup(\uparrow\!\!B)$.

On the other hand, if $a\in(\uparrow\!\!A)\cup(\uparrow\!\!B)$, then either $a\in(\uparrow\!\!A)$ or $a\in(\uparrow\!\!B)$. In the former case, there exists $b$ such that $a\geq b$ and $b\in A$. Since $A\subseteq A\cup B$, we also have $b\in A\cup B$, hence $a\in\uparrow\!\!(A\cup B)$. Likewise, in the second case, we also conclude that $a\in\uparrow\!\!(A\cup B)$. Therefore, we have $(\uparrow\!\!A)\cup(\uparrow\!\!B)\subseteq\uparrow\!\!(A\cup B)$. ∎

###### Theorem 5.

$\uparrow\!\!P=P$

###### Theorem 6.

$A\subseteq B$, $\uparrow\!\!A\subseteq\uparrow\!\!B$

Title upper set operation is a closure operator UpperSetOperationIsAClosureOperator 2013-03-22 16:41:43 2013-03-22 16:41:43 rspuzio (6075) rspuzio (6075) 8 rspuzio (6075) Theorem  msc 06A06