let $A:=\{1,2,3,4,5,6\}$
Define a total order relation in A where 3 is the minimum
I'd like to make a total order relation where the order is $3<1<2<4<5<6$. How could I make it?
I guess it's a real numbers usual order modification. I think my relation should take these ordered pairs $$ R = \{ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6), \\ (1,1),(1,2),(1,4),(1,5), 1,6), \\ (2,2),(2,4),(2,5),(2,6), \\ (4,4),(4,5),(4,6),\\ (5,5),(5,6),\\ (6,6) \\ \} $$
Obviously it is a $A\times A$ subset so
$$R = \{(a,b) \in A \times A : \text{I don't know what should I write here}\}$$
I can't figure out how to do it
Maybe it could be $(x=3 \rightarrow \text{something})$ otherwise usual order
Thanks in advance ;)
If you insist in writing the relation in comprehension notation, you could write $R=\{(a,b)\in A\times A : a = 3 \text{ and } b \in A \text{ or } a = 1 \text{ and } b \in \{1,2,4,5,6\} \text{ or } a=2 \text{ and } b \in \{2,4,5,6\} \text{ or } a=4 \text{ and } b \in \{4,5,6\} \text{ or } a=5 \text{ and } b \in \{5,6\} \text{ or } a=b=6\}$