As an exercise I'm trying to figure out what the index matrix of an order relation looks like.
Anyone who's willing to join the thread is welcome.
>>8641754
>index matrix
what?
>>8641754
Here, the matrix we use to represent [math]P[/math] depends on the linear order of [math]\mathbb{N}[/math] and on how [math]f[/math] is defined.
>>8641757
>what?
I don't know whether there is a name for it.
Since [math]\leq \subset P^{2}[/math] and finite, we can place its points on a grid.
I'm going to write down what I have left on paper now.
I realize now that since [math]f[/math] is a bijection maybe we can do without proving much.
Either way, given a partial ordering on [math]\{1,...,n\}^{2}[/math], we have an ordering on [math]P^{2}[/math].
>>8641811
>Either way, given a partial ordering on {1,...,n}2, we have an ordering on P2.
This but without the squares.
Example:
[math]n=4[/math] so [math](1,1),(2,2),(3,3),(4,4)\in R[/math]
If [math](1,3)\in R[/math] then [math](3,1)\not \in R[/math].
[If [math](3,1)\in R[/math] then [math]1=3[/math] which is false.]
Of the form [math](3,j)[/math] we have only [math](3,2)[/math] and [math](3,4)[/math] left.
[math](3,2)[/math] lends to pick [math](1,2)[/math]. [math](3,4)[/math], [math](1,4)[/math].