This is another question based on "An Invitation to Applied Category Theory", this time based on exercise 1.12 (p. 9). I was able to instinctively answer the question but I don't know why the answer is correct and I was hoping someone could explain.
Here we have the set $ \{ 11 , 12 , 13 , 21 , 22 , 23 \} $ partitioned into $ ( 11 , 12 ) , ( 13 ), (21), (22, 23)$ and the exercise is this:
For any two elements $a,b \in \{11, 12, 13, 21, 22, 23\}$ ... [we] write $a \sim b$ if $a$ and $b$ are both in the same part. Write down every pair of elements $(a,b)$ that are in the same part. There should be 10.
So the correct answer is: $ (11,11 ), ( 11,12 ), ( 12,11 ) , ( 12, 12 ) , ( 13, 13 ), ( 21, 21 ), ( 22 , 22 ) , ( 22 , 23 ) , (23 , 22), ( 23, 23 )$.
And my question is this: order appears to matter here, so that $(22, 23)$ is seen as distinguishable from $( 23, 22 )$ yet there is only one, e.g., $( 22, 22 )$. Why is one type of pair distinguishable by order in this way and another not? A pair seems distinguished from a set here, as order would not matter in a set?
The reason why you only need one $(22,22)$ is that taking the pair the other way round gives you $(22,22)$ again. Order still matters so you still need "both ways round", but because both ways round are the same ordered pair you only need to write it down once.
This exercise is basically exploring the relationship between partitions and equivalence relations. A relation like $\sim$, $=$, $\leq$ or $|$ ("divides") could be thought of as a function that takes two inputs and returns a truth value: i.e. "Does $a \sim b$, yes/no?". But it can also be thought of as a set of ordered pairs: these are the set of pairs $(a,b)$ such that $a \sim b$.
This shows why we use ordered pairs for relations, since e.g. $(3,4)$ is going to be a pair in the relation $\leq$, while $(4,3)$ is not. We should also make the point here than when viewing a relation as a set of pairs, we need to keep all the pairs in that relation. If you leave some of the pairs out you'll end up with a different relation, not the one you started with.
An equivalence relation is one that is reflexive, symmetric and transitive. It turns out that equivalence relations on a set give rise to partitions and vice-versa (the proof is probably coming soon in your book). This exercise gave you a partition of your set into subsets (and they should really have been written with curly brackets e.g. $\{11, 12 \}$ to show this is a subset, not an ordered pair) and asks you to construct the relation $a \sim b$ if they are in the same subset. This relation is a set of ordered pairs, so the exercise asks you to give every ordered pair in that subset.
If you didn't include both ways round $(22,23)$ and $(23,22)$ you would end up with a different relation than the one it was asking for. And this relation wouldn't be symmetric, so it wouldn't be the equivalence relation that gives rise to this partition.