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# Exercise 47 - Chapter 1

edited June 2018

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:

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edited April 2018

Comment Source:<center> ![Product](https://docs.google.com/drawings/d/e/2PACX-1vSDPTMFV6q0bYV4ee-Khkg5_1f_5Kfv3dBj7Pc2NhqiT3b7PyGQexE87KbIqRfAdUf8N0iyM9hT9PE6/pub?w=282&h=280) </center>
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edited April 2018

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.

Comment Source: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\$$! ![](https://i.imgur.com/iP02W2C.png) 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.
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Here's another image showing what I mean by "replacing every element". The outer edge "distributes" through the larger cells onto every analogous pair.

Comment Source:Here's another image showing what I mean by "replacing every element". The outer edge "distributes" through the larger cells onto every analogous pair. ![](https://i.imgur.com/j9oGOC0.png)
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Nice illustrations @JonathanCastello . https://forum.azimuthproject.org/discussion/comment/17639/#Comment_17639 in particular shows commutivity of paths.

Comment Source:Nice illustrations @JonathanCastello . https://forum.azimuthproject.org/discussion/comment/17639/#Comment_17639 in particular shows commutivity of paths.