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Exercise 47 - Chapter 1
Given preorders \(A\) and \(B\), let their product \(A \times B\) be the cartesian product of sets \(A\) and \(B\) made into a preorder as follows:
$$ (a,b) \le (a',b') \textrm{ if and only if } a \le a' \textrm{ and } b \le b' .$$ Draw the product of the two preorders drawn below:
Comments
<center>  </center>This is pretty fascinating to me. You can imagine acquiring the product by extruding \(A\) from \(B\). But it also works just as well by imagining extruding \(B\) from \(A\)!
It's like we're replacing every element in one preorder with a copy of the other preorder, and distributing edges of the original preorder through the copies.
Product constructions aren't particularly new to me, but... There's something incredibly pleasant about a geometric explanation.
This is pretty fascinating to me. You can imagine acquiring the product by extruding \\(A\\) from \\(B\\). But it also works just as well by imagining extruding \\(B\\) from \\(A\\)!  It's like we're replacing every element in one preorder with a copy of the other preorder, and distributing edges of the original preorder through the copies. Product constructions aren't particularly _new_ to me, but... There's something incredibly pleasant about a geometric explanation.Here's another image showing what I mean by "replacing every element". The outer edge "distributes" through the larger cells onto every analogous pair.
Here's another image showing what I mean by "replacing every element". The outer edge "distributes" through the larger cells onto every analogous pair. Nice illustrations @JonathanCastello . https://forum.azimuthproject.org/discussion/comment/17639/#Comment_17639 in particular shows commutivity of paths.
Nice illustrations @JonathanCastello . https://forum.azimuthproject.org/discussion/comment/17639/#Comment_17639 in particular shows commutivity of paths.