LARA

Differences

This shows you the differences between two versions of the page.

Link to this comparison view

Both sides previous revision Previous revision
Next revision
Previous revision
Next revision Both sides next revision
partial_order [2008/04/30 10:00]
vkuncak
partial_order [2008/04/30 15:21]
vkuncak
Line 12: Line 12:
 Given a partial ordering relation $\le$, the corresponding **strict ordering relation** $x < y$ is defined by $x \le y \land x \neq y$ and can be viewed as a shorthand for this conjunction. Given a partial ordering relation $\le$, the corresponding **strict ordering relation** $x < y$ is defined by $x \le y \land x \neq y$ and can be viewed as a shorthand for this conjunction.
  
-We can view partial order $(A,r)$ as a first-order interpretation $I=(A,​\alpha)$ of language ${\cal L}={\le\}$ where $\alpha({\le\})=r$.+We can view partial order $(A,r)$ as a first-order interpretation $I=(A,​\alpha)$ of language ${\cal L}=\{\le\}$ where $\alpha({\le})=r$.
  
 ===== Example Partial Orders ===== ===== Example Partial Orders =====
Line 31: Line 31:
  
 Given a partial order $(A,\le)$ and a set $S \subseteq A$, we call an element $a \in A$ Given a partial order $(A,\le)$ and a set $S \subseteq A$, we call an element $a \in A$
-  * **upper bound** of $S$ if for all $a' \in S$ we have $a \le a'$+  * **upper bound** of $S$ if for all $a' \in S$ we have $a' ​\le a$
   * **lower bound** of $S$ if for all $a' \in S$ we have $a \le a'$   * **lower bound** of $S$ if for all $a' \in S$ we have $a \le a'$
   * **minimal element** of $S$ if $a \in S$ and there is no element $a' \in S$ such that $a' < a$   * **minimal element** of $S$ if $a \in S$ and there is no element $a' \in S$ such that $a' < a$