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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 ;)

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  • $\begingroup$ Why include $6$? $\endgroup$ Jan 15 at 0:22
  • $\begingroup$ Why are there pairs with $6$ if $6$ is not in $A$? $\endgroup$
    – jjagmath
    Jan 15 at 0:22
  • $\begingroup$ What do you mean by "make it"? What more is there to do than to simply define the relevant inequalities? $\endgroup$
    – lulu
    Jan 15 at 0:23
  • $\begingroup$ So sorry it was my mistake, I think now it's ok $\endgroup$
    – sankiago
    Jan 15 at 0:24
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    $\begingroup$ I don't understand. If you were to write, say, $3<1<2<4<5<6$, you are done, right? That completely defines the order and it is clear that $3$ is minimal. What else is there to be said? $\endgroup$
    – lulu
    Jan 15 at 0:27

1 Answer 1

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If you insist in writing the relation in comprehension notation, you could write $R=\{(a,b)\in A\times A : a = 3 \text{ and } b \in A \text{ or } a = 1 \text{ and } b \in \{1,2,4,5,6\} \text{ or } a=2 \text{ and } b \in \{2,4,5,6\} \text{ or } a=4 \text{ and } b \in \{4,5,6\} \text{ or } a=5 \text{ and } b \in \{5,6\} \text{ or } a=b=6\}$

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