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# Exercise 34 - Chapter 1

edited June 2018

Let $$X$$ be the set of partitions of {•, ◦, ∗}; it has five elements and an order by coarseness, as shown in the Hasse diagram Eq. (1.2) [below]. Write down every pair $$(x, y)$$ of elements in $$X$$ such that $$x \le y$$. There should be 12.

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1.
edited April 2018

Let $$S_1, \cdots, S_5$$ denote the five partitions:

$$\begin{eqnarray} S_1 &=& \left\{\{\bullet\}, \{\circ\}, \{\star\}\right\}\\ S_2 &=& \left\{\{\bullet\}, \{\circ, \star\}\right\}\\ S_3 &=& \left\{\{\circ\}, \{\bullet, \star\}\right\}\\ S_4 &=& \left\{\{\star\}, \{\bullet, \circ\}\right\}\\ S_5 &=& \left\{\{\bullet, \circ, \star\}\right\}\\ \end{eqnarray}$$ Then the twelve pairs are:

$$\begin{eqnarray} (S_1, S_1) \\ (S_1, S_2) \\ (S_1, S_3) \\ (S_1, S_4) \\ (S_1, S_5) \\ (S_2, S_2) \\ (S_2, S_5) \\ (S_3, S_3) \\ (S_3, S_5) \\ (S_4, S_4) \\ (S_4, S_5) \\ (S_5, S_5) \\ \end{eqnarray}$$

Comment Source:Let \$$S_1, \cdots, S_5\$$ denote the five partitions: $$\begin{eqnarray} S_1 &=& \left\\{\\{\bullet\\}, \\{\circ\\}, \\{\star\\}\right\\}\\\\ S_2 &=& \left\\{\\{\bullet\\}, \\{\circ, \star\\}\right\\}\\\\ S_3 &=& \left\\{\\{\circ\\}, \\{\bullet, \star\\}\right\\}\\\\ S_4 &=& \left\\{\\{\star\\}, \\{\bullet, \circ\\}\right\\}\\\\ S_5 &=& \left\\{\\{\bullet, \circ, \star\\}\right\\}\\\\ \end{eqnarray}$$ Then the twelve pairs are: $$\begin{eqnarray} (S_1, S_1) \\\\ (S_1, S_2) \\\\ (S_1, S_3) \\\\ (S_1, S_4) \\\\ (S_1, S_5) \\\\ (S_2, S_2) \\\\ (S_2, S_5) \\\\ (S_3, S_3) \\\\ (S_3, S_5) \\\\ (S_4, S_4) \\\\ (S_4, S_5) \\\\ (S_5, S_5) \\\\ \end{eqnarray}$$
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2.
edited April 2018

To summarize: there are 5 identity arrows; 6 drawn arrows, and; 1 composed arrow [3 equivalent composed arrows] from $$S_1 \dots S_5$$. $$5 + 6 + 1 = 12$$

Comment Source:To summarize: there are 5 identity arrows; 6 drawn arrows, and; 1 composed arrow [3 equivalent composed arrows] from \$$S_1 \dots S_5 \$$. \$$5 + 6 + 1 = 12 \$$
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3.
edited April 2018

Just a minor additional explanation point that might help others. Coarser partition (as per John's lecture 10), means that more points are connected (or equivalent). Which is why in Dan's answer $$S_5$$ always appears on the right, because it is the most coarse --that is, in that partition most points are equivalent. May be another way to think about 'coarsest' partition, is that it will contain the 'least' number of blocks.

Comment Source:Just a minor additional explanation point that might help others. Coarser partition (as per John's lecture 10), means that more points are connected (or equivalent). Which is why in Dan's answer \$$S_5\$$ always appears on the right, because it is the most coarse --that is, in that partition most points are equivalent. May be another way to think about 'coarsest' partition, is that it will contain the 'least' number of blocks.