[keithalewis]

Let S = {1, 2}. Every distance function is determined by d(1,2) = a, a >= 0. Define f(d) = {{1,2}} if a = 0 and f(d) = {{1},{2}} otherwise. Isn't this a clustering algorithm that is scale invariant, rich, and consistent?

[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}}.