[bubblyworld]

Looks that way to me, yeah, though this is obviously a super simple case. It's clearly scale invariant and there are only two partitions, which your algorithm hits, so it's rich. Completeness is trivially satisfied in both cases too.

[n0bra1n]

i think i found the issue: the paper says distance function is 0 IFF elements are equal. so for this example, you can not define d(1,2) as equal to 0. so it is not rich, as this is the only way to get the partition {{1,2}}.

[bubblyworld]

Oh, I see, it's not a true metric. That's fair enough, though I wonder if the result depends critically on that assumption.

(you can pass to equivalence classes to recover a true metric, and I didn't see anything obviously incompatible with that in the paper, but I admit I didn't look very deeply)