Why must a preorder decompose into disjoint ordered chains? If I have a partial preference w1<w3 and another partial preference w2<w3 how do those induce disjoint ordered chains where worlds between chains are incomparable? Perhaps you are asking us to assume that the preorder decomposes into disjoint ordered chains?
How do cycles vanish in ¯¯¯¯¯¯W? Can you work through the example where the partial preference expressed by the human is w1<w2<w3<w1?
we can extend this to ¯¯¯¯¯¯W by setting U(w)=U(p(w)).
Why must a preorder decompose into disjoint ordered chains?
They don’t have to; I’m saying that sensible partial preferences (eg P) should do so. I then see how I’d deal with sensible preorders, then generalise to all preorders in the next section.
How do cycles vanish in ¯¯¯¯¯¯W? Can you work through the example where the partial preference expressed by the human is w1<w2<w3<w1?
Note that what you’ve written is impossible as w<w′ means w≤w′ but not w′≤w. A preorder is transitive, so the best you can get is w1≤w2≤w3≤w1.
Then projecting down (via p) to ¯¯¯¯¯¯W will project all these wi down to the same element. That’s why there are no cycles, because all cycles go to points.
Then we need to check some math. Define ≤ on ¯¯¯¯¯¯W by p(w)≤p(w′) iff w≤w′.
This is well defined (independently of which w and w′ we use to represent p(w) and p(w′)), because if p(w′′)=p(w), then w′′≤w, so, by transitivity, w′′≤w′. The same argument works for w′.
We now want to show the ≤ is a partial order on ¯¯¯¯¯¯W. It’s transitive, because if p(w)≤p(w′) and p(w′)≤p(w′′), then w≤w′≤w′′, and the transitivity in W implies w≤w′′ and hence p(w)≤p(w′′).
That shows it’s a preorder. To show partial order, we need to show there are no cycles. So, if p(w)≤p(w′) and p(w′)≤p(w), then w≤w′ and w′≤w, hence, by definition of p, p(w)=p(w′). So it’s a partial order.
For cycles, it looks like the projection to ¯¯¯¯¯¯W is akin to taking all the worlds that form a given cycle, and compressing them into a single world.
In your example, it’s true wi<wj and wj<wi when i≠j. That’s the condition for equivalence in the project, so you have that w1=w2=w3. If you’re thinking about the ordering as a directed graph, you can collapse those worlds to a single point and not mess up the ordering.
For the disjoint chain part, first imagine all the worlds in the from (n,v) and split them into equivalence classes based on equality of v. The preference P can only compare two worlds that are in the same equivalence class, so each equivalence class will be totally ordered, but no class will be comparable to any other (hence decomposition into disjoint chains).
I thought the (n, v) thing was just an example. If all the worlds are meant to be represented via (n, v), then it seems like you are only ever allowed to give a partial preference based on whatever feature n represents, and you can never talk about any of the other features in v. This seems bad.
(I do agree that if you only give partial preferences on (n, v) worlds in the way you describe, then you get a decomposition into disjoint chains.)
I do believe the OP is talking about partial pref on (n,v) form worlds. Yeah, this seems bad in the “How do I act when looking at different v?” sense, but I get the sense that it’s not supposed to answer that question. Or at least Stuart plans to build a lot from here before it will answer that sort of question.
I am very confused by the math in this post:
Why must a preorder decompose into disjoint ordered chains? If I have a partial preference w1<w3 and another partial preference w2<w3 how do those induce disjoint ordered chains where worlds between chains are incomparable? Perhaps you are asking us to assume that the preorder decomposes into disjoint ordered chains?
How do cycles vanish in ¯¯¯¯¯¯W? Can you work through the example where the partial preference expressed by the human is w1<w2<w3<w1?
I think this is extending to W?
Should that be ||U(w′)−U(w)||=2?
Thanks, corrected a few typos.
They don’t have to; I’m saying that sensible partial preferences (eg P) should do so. I then see how I’d deal with sensible preorders, then generalise to all preorders in the next section.
Note that what you’ve written is impossible as w<w′ means w≤w′ but not w′≤w. A preorder is transitive, so the best you can get is w1≤w2≤w3≤w1.
Then projecting down (via p) to ¯¯¯¯¯¯W will project all these wi down to the same element. That’s why there are no cycles, because all cycles go to points.
Then we need to check some math. Define ≤ on ¯¯¯¯¯¯W by p(w)≤p(w′) iff w≤w′.
This is well defined (independently of which w and w′ we use to represent p(w) and p(w′)), because if p(w′′)=p(w), then w′′≤w, so, by transitivity, w′′≤w′. The same argument works for w′.
We now want to show the ≤ is a partial order on ¯¯¯¯¯¯W. It’s transitive, because if p(w)≤p(w′) and p(w′)≤p(w′′), then w≤w′≤w′′, and the transitivity in W implies w≤w′′ and hence p(w)≤p(w′′).
That shows it’s a preorder. To show partial order, we need to show there are no cycles. So, if p(w)≤p(w′) and p(w′)≤p(w), then w≤w′ and w′≤w, hence, by definition of p, p(w)=p(w′). So it’s a partial order.
Thanks!
For cycles, it looks like the projection to ¯¯¯¯¯¯W is akin to taking all the worlds that form a given cycle, and compressing them into a single world.
In your example, it’s true wi<wj and wj<wi when i≠j. That’s the condition for equivalence in the project, so you have that w1=w2=w3. If you’re thinking about the ordering as a directed graph, you can collapse those worlds to a single point and not mess up the ordering.
Ah yes, that makes sense, thanks! I didn’t realize what ¯¯¯¯¯¯W was the set of equivalence classes of W
For the disjoint chain part, first imagine all the worlds in the from (n,v) and split them into equivalence classes based on equality of v. The preference P can only compare two worlds that are in the same equivalence class, so each equivalence class will be totally ordered, but no class will be comparable to any other (hence decomposition into disjoint chains).
I thought the (n, v) thing was just an example. If all the worlds are meant to be represented via (n, v), then it seems like you are only ever allowed to give a partial preference based on whatever feature n represents, and you can never talk about any of the other features in v. This seems bad.
(I do agree that if you only give partial preferences on (n, v) worlds in the way you describe, then you get a decomposition into disjoint chains.)
I do believe the OP is talking about partial pref on (n,v) form worlds. Yeah, this seems bad in the “How do I act when looking at different v?” sense, but I get the sense that it’s not supposed to answer that question. Or at least Stuart plans to build a lot from here before it will answer that sort of question.
That makes the math make sense, but I really object to calling this the “sensible” case.