---- > [!definition] Definition. ([[dictionary order relation]]) > Suppose that $A$ and $B$ are two sets with order relations lt;_{A}$ and lt;_{B}$. Define an [[strict order relation]] lt;$ on $A \times B$ by defining $(a_{1},b_{1}) < (a_{2} , b_{2})$ if $(a_{1} <_{A} a_{2})$ or if $(a_{1}=a_{2}$ and $b_{1}<_{B}b_{2})$. It is called the [[dictionary order relation]] on $A \times B$. > [!intuition] > The terminology arises as follows— the rule defining lt;$ is that used to order dictionary words: first compare first letters, break ties with second letters, break ties with third letters,... > [!basicexample] > Consider the **dictionary order** on the plane $\rr \times \rr$. In this order, a point $p$ is less than every point above it on a vertical line through $p$, and less than every point to the right of this vertical line. > [!justification] > We show by cases that this indeed defines an [[strict order relation]] on $A \times B$. >1. (Comparability) Let $(a_{1},b_{1})$, $(a_{2},b_{2}) \in A \times B$ such that $(a_{1},b_{1}) \neq (a_{2},b_{2})$. If $a_{1} <_{A} a_{2}$ then $(a_{1},b_{1}) < (a_{2},b_{2})$; since lt;_{A}$ is an [[strict order relation]] we cannot have $(a_{1},b_{1}) > (a_{2},b_{2})$ in this case. If $a_{1}=a_{2}$ then we compare the $b$s; since lt;_{B}$ is an [[strict order relation]] we're good. >2. (Reflexivity) obvious >3. (Transitivity) easy ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```