#### Howdy, Stranger!

It looks like you're new here. If you want to get involved, click one of these buttons!

Options

# 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.

## Comments

• Options
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: $$\begin{eqnarray} (S_1, S_1) \\\\ (S_1, S_2) \\\\ (S_1, S_3) \\\\ (S_1, S_4) \\\\ (S_1, S_5) \\\\ (S_2, S_2) \\\\ (S_2, S_5) \\\\ (S_3, S_3) \\\\ (S_3, S_5) \\\\ (S_4, S_4) \\\\ (S_4, S_5) \\\\ (S_5, S_5) \\\\ \end{eqnarray}$$
• Options
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 \$$
• Options
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.
Sign In or Register to comment.