= 27d567bc2d48d9f40e9ccc20ea370b15)
shminux comments on A Pure Math Argument for Total Utilitarianism - Less Wrong
You are viewing a comment permalink. View the original post to see all comments and the full post content.
= 783df68a0f980790206b9ea87794c5b6)
Loading…= b715d02e85f7b3b4ae65ca3d3e450868)
Subscribe to RSS Feed
Powered by Reddit
= f037147d6e6c911a85753b9abdedda8d)
You are viewing a comment permalink. View the original post to see all comments and the full post content.
Comments (36)
First, I think that what you call lattice order is more like partial order, unless you can also show that a join always exists. The pictures have it, but I am not convinced that they constitute a proof.
It looks like all you have "shown" is that if you embed some partial order into a total order, then you can map this total ordering into integers. I am not a mathematician, but this seems rather trivial.
I agree, I didn't show this. It's not hard, but it's a bit of writing to prove that (x1x2 \/ y1y2)=(x1\/y1)(x2\/y2) which inductively shows that this is an l-group.
It's not a total order, nor is it true that all totally ordered groups can be embedded into Z (consider R^2, lexically ordered, for example. Heck, even R itself can't be mapped to Z since it's uncountable!). So not only would this be a non-trivial proof, it would be an impossible one :-)
Not all, just countable...
Z^2 lexically ordered is countable but can't be embedded in Z.
It seems like your intuition is shared by a lot of LW though - people seem to think it's "obvious" that these restrictions result in total utilitarianism, even though it's actually pretty tricky.