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shminux comments on A Pure Math Argument for Total Utilitarianism - Less Wrong Discussion

-5 Post author: Xodarap 27 October 2013 05:05PM

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Comment author: Xodarap 27 October 2013 07:17:36PM *  1 point [-]

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.

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 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.

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 :-)

Comment author: shminux 27 October 2013 07:46:35PM -1 points [-]

Not all, just countable...

Comment author: Xodarap 27 October 2013 10:08:41PM *  2 points [-]

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.

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