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May 13, 2019 at 12:59 vote accept oNgStrIng
May 10, 2019 at 16:48 history edited Andrés E. Caicedo
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May 10, 2019 at 14:10 answer added Pat Devlin timeline score: 4
May 10, 2019 at 13:50 history edited oNgStrIng CC BY-SA 4.0
clarification
May 10, 2019 at 13:48 comment added oNgStrIng Hi @Devlin Each, partition will perform on the particular subset containing $s_1,s_2$, since if $s_1,s_2$ are separated, the partition process will stop. So I think (a) is not. The block, i.e. the subset, is the very one, not random. So (b) is not. It seems that (c) is also not. I will re-edit and clarify the process.
May 10, 2019 at 13:41 comment added Pat Devlin Could you elaborate on the random process? Here are a few options about what you could mean. (a) At each step, refine the partition by picking a block and splitting it into two blocks. Among all such partitions, pick uniformly at random [so big blocks are more likely to be split up]. Or perhaps you mean (b) do (a), but first pick the block to split up with each block being equally likely. Or perhaps you mean the very different (c) pick sets S_1, S_2, S_3, ... independently uniformly at random [with/without replacement], and consider minimal non-empty sets in the topology generated by them.
May 10, 2019 at 12:25 review First posts
May 10, 2019 at 12:39
May 10, 2019 at 12:23 history asked oNgStrIng CC BY-SA 4.0