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.
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!