# Why can the author just switch the order of the inequality without any reprecussions?

Note: This example is from Discrete Mathematics and Its Applications [7th ed, example 2, page 598].

I understand the idea of a symmetric closure. You add all pairs of (b, a) where (a,b) already exists in the relation. And now you have a symmetric relation that is a subset of all other symmetric relations that contain R.
My question is in this problem, why can the author introduce the inverse of R? The original relation R states the ordered pairs that are in this relation are the ones in which the first integer is greater than the second integer.(Order Matters!).
I understand what the author did, it was i mentioned earlier "You add all pairs of (b, a) where (a,b) already exists in the relation.". But shouldn't you only be able to add ones that can be in the relation, the ones that satisfy the requirement? In that case, it be none?
Does anyone agree with me that the symmetric closure of this relation doesn't exist because the fact that the ordered pairs you would be adding would violate the initial relation condition?

This was an example that the author used earlier
The relation {(1,1), (1,2), (2,2), (2,3), (3,1), (3,2)} on {1, 2, 3} is not symmetric. To produce a symmetric relation, we need only to add (2,1) and (1,3). And that would be the symmetric closure. To me those points can be added because there is no rule no place on what coordinates can be added or not added to the relation.

• Well, look at it this way: If we "added in" stuff that was already in the relation, we wouldn't be adding anything. The idea of symmetric closure is to find the smallest symmetric relation containing R. So taking the union with R-inverse is just the right thing to do. Mar 5 '15 at 21:15
• But if $R$ is the rlation $R= \{(a,b) \mid a > b \}$ the symmetric relation $R \cup R^{-1}$, as you can see, does not satisfy the condition $a > b$ any more, but the condition $a \ne b$. Mar 5 '15 at 21:19
• That's why i was saying there would be no symmetric closure then. If you add (6,4) to this relation, it would break the initial rule set in place of (a, b), a > b. In the second example, the addition of (2,1) and (1,3) aren't breaking any rules. Mar 5 '15 at 21:19
• @MauroALLEGRANZA But to add elements, don't you need to meet the initial requirements? It doesn't make sense to add a odd number to a list of evens for example. Mar 5 '15 at 21:21
• If we start with the relation "father of", its symmetric closure does not contains only couples (father,son) but also couples (son, father) and thus it is nor more the relation "father of"... Mar 5 '15 at 21:22

## 1 Answer

The author begins with the relation: $$R = \{ (a,b) : a < b\} \subseteq \mathbb{N}\times\mathbb{N}$$

The author then makes a new relation $$S = \{ (a,b) : a < b \text{ or } b < a \} \subseteq \mathbb{N}\times\mathbb{N}$$

The author actually constructs $S$ as $R \cup R^{-1}$. But $S$ is not $R$: $S$ is an entirely different relation. There is no reason that $S$ has to satisfy the condition that was used to define $R$. The relation $R$ is unchanged.

This is one way that the term "adding" can be misleading. One cannot really "add" elements to a set. You can take a set, and a new element, and form a new set consisting of the old set and the new element. But the old set still exists, just as it did before - you have not changed it in any way by making a new set.