1
$\begingroup$

The standard definition of a relation of an arbitrary set A is a subset of the set product of A, AxA. Is it okay to define relation R to be a subset of the set product AxA such that R has at least one property P (i.e. inequality, equality, difference, etc.), where any two element of a relation obeys P, and any two element of a set A either obeys P or not? The standard definition of a relation seems to me little ambiguous.

$\endgroup$
4
  • $\begingroup$ But all those properties $P$ are defined via relations. The properties that you seem to be describing are equivalence relations that, besides being relations, also have to be reflexive, symmetric, and transitive. $\endgroup$
    – Eduardo M.
    Jul 13 '16 at 2:25
  • $\begingroup$ Yes, I meant P to be any property that any two element of the set A can either satisfy or not. $\endgroup$
    – user205011
    Jul 13 '16 at 2:40
  • $\begingroup$ Yeah, but those are just specific kinds of relations; inequality and strict inequality are order relations, congruence and equality are equivalence relations. Basically you are asking if you can define a relation, and then induce some structure on that relation that is, itself, given by some specific kind of relation. $\endgroup$
    – Eduardo M.
    Jul 13 '16 at 2:46
  • $\begingroup$ What a great little exchange! Please let me add my misunderstanding. The questioner appeared to not appreciate that he has to define '<' beforehand. Having defined the operators, he can then feel perfectly 'safe' to use them in definitions. Of course, he can just borrow library functions, trusting the provenance of the library. Perhaps, the important thing is that ALL functional operators are defined - you have to begin, at the beginning! $\endgroup$ Feb 16 '19 at 10:43
1
$\begingroup$

Well I personally disagree with your sense of ambiguity that you feel. I think this definition perfectly encapsulates the intuitive notion of a relation. For example, for $a, b \in A$, the statement "$a$ is related to $b$" can be written in shorthand as $(a,b)$. A "relation" is then a specific way of grouping these related pairs together, so it seems intuitive to me to make $R \subseteq A \times A$ containing these $(a, b)$'s. Your addition specification of a property doesn't seem necessary to me.

$\endgroup$
5
  • $\begingroup$ Yes, but books use symbols like <, >, =, modulo notation, etc. to denote a relation. I just did not fee safe of using such symbols for notating the relation. Is it still okay to define relation < as R = {(a,b): <} $\endgroup$
    – user205011
    Jul 13 '16 at 2:44
  • $\begingroup$ @MathWanderer This is why the statement $(a,b) \in R$ is often written as $a \sim b$, especially when denoting an equivalence relation. $\endgroup$
    – JasonM
    Jul 13 '16 at 2:48
  • $\begingroup$ Of course, I am ware of it. I was wandering if it would be good to specify given equivalence relation in some arbitrary sets so I can specify the relation under interest. $\endgroup$
    – user205011
    Jul 13 '16 at 2:51
  • $\begingroup$ Given a set of integers Z, a~b can mean any equivalence relation possible for Z, and I would like to specify by giving property like = or congruence to a specific relation of interest rather than saying a~b. $\endgroup$
    – user205011
    Jul 13 '16 at 2:53
  • $\begingroup$ @MathWanderer The relation is typically defined in a specific enough way. I'm unclear on why this seems to be an insufficient definition for you. "...I would like to specify by giving property like = or congruence..." I'm not sure what you mean by that. = is a relation, and so is < or congruence, so we are allowed to say a=b as shorthand for $(a, b) \in R$. $\endgroup$
    – JasonM
    Jul 13 '16 at 2:56
1
$\begingroup$

$R$ is a relation over the set $A$, if and only if, $R$ is a subset of the Cartesian square of $A$.   $$R\subseteq A{\times}A$$

That is unambiguous.   All possible subsets of $A^2$ are each a relation over $A$.


Now we can describe some relations by set constructions when given some identified predicate, $P$. $$R=\{(a,b)\in A^2: P(a,b)\}$$

But there's a bit of chicken-and-egg redundancy there, and often there's no readily identifiable predicate other than asserting that the pair is in the given subset, $R$.

$\endgroup$
2
  • $\begingroup$ I assume P(a,b) can be things like <, >, etc.? $\endgroup$
    – user205011
    Jul 13 '16 at 2:46
  • $\begingroup$ @MathWanderer Yes, Including but not limited to those comparison operators. $\endgroup$ Jul 13 '16 at 3:26
0
$\begingroup$

A relation is designed to be as generic a thing as possible, hence why it's just a set of pairs of elements. It's possible to create a rule, or identify a property, that defines the relation, but that's just a shorthand so you don't always have to write out all the pairs manually (which is particularly a pain when the relation isn't finite). It's like how a set is just a collection of objects (with a few caveats), so that $\{0, 1, 2, \mbox{cat}, \mbox{dog}, \mbox{eggplant emoji}, \emptyset, \pi, \{3+2i\}, \mbox{a lingering sense of dread}\}$ is as valid a set as $\mathbb{C}$ or $\mathbb{N}$ or $\{x \in \mathbb{R} | \sin{(x)} = 0\}$.

$\endgroup$
1
  • $\begingroup$ Yes, I noticed that the notation of relation can be things like <, >, =, congruence notation, etc. I just did not like to use < as a notation. $\endgroup$
    – user205011
    Jul 13 '16 at 2:42
0
$\begingroup$

I think what you're trying to do could be illustrated in the following example, here's a way to define a relation:

Let $ \sim $ be a relation over $\mathbb{Z}$ defined by $$ x \sim y \iff (x=y) \lor (x+y=3).$$

In this case, $\sim$ represents $R$ and, by giving the appropriate definition, you can specify which elements are in the relation. In my example $(x,y) \in R$ (or $x \sim y$) if and only if $x$ and $y$ meet the chosen definition. Of course '$\sim$' could be any symbol you choose.

$\endgroup$
0
$\begingroup$

Your intuition is correct, but it's not necessary to add another part to it since what you're suggesting is already implied by the definition of a Relation of a Set.

A relation describes what elements in a Cartesian product are related to each other. Say we have set $A$ and set $B$, then $A \times B$ gives us all the possible ordered pairs resulting from the cross product.

Within the set of ordered pairs in $A \times B$, lies a subset that consists of elements which are $\mathbf{related\ in\ some\ way\ by\ a\ certain\ condition}$ (this is our relation $R$ from $A$ to $B$).

It makes sense that $R$ is a subset of $A \times B$, because all the ordered pairs that fall within $R$ are the ordered pairs from $A \times B$ that either meet or don't meet the condition given by $R$.

I believe the definition that you are trying to add is already implied by the definition of a relation. The only way to $\mathbf{distinguish}$ whether an ordered pair is an element of $R$, is to test that pair against a condition given by $R$.

Here's a quick example:

Let, $A = \{2 , 6, 5 \}$ and $B = \{3, 7, 8 \}$ and $R$ be a relation from $A$ to $B$, and given any ordered pair $(a , b) \in A \times B$

Our condition is $( a , b ) \in R$ iff $a \lt b$

Ok, so we have a condition that an ordered pair must meet to fall into the relation: $a \lt b$ (This is what you were adding in your definition)

First, here's our $A \times B$:

$ A \times B = \{ (2,3) , (2,7) , (2,8), (6,3), (6,7), (6,8) , (5,3) , (5,7) , (5,8) \}$

Here's a few examples showing which ordered pairs meets the relation $R$ condition:

$2 \lt 3$ , so $(2 , 3) \in R$

$6 \not\lt 3$, so $( 6 , 3 ) \notin R$

... and so on giving us:

$R = \{ (2,3) , (2,7) , (2,8) , (6,7), (6,8), (5,7) , (5,8) \}$

Here's the key: all the elements that fall within our relation $R$ are $\mathbf{contained\ in}$ $A \times B$ and are the ordered pairs that $\mathbf{meet\ a\ certain\ condition}$.

$\endgroup$
0
$\begingroup$

A relation on $A$ is as such defined to be a subset of $A\times A$.

Defining a relation on $A$ is to explicitly enumerate which elements of $A\times A$ that is included in the relation or determine the relation implicitly as those elements in $A\times A$ which are related by some predicate $P(x,y)$.

I think you mix up the definitions of the term relation with the definition of a specific relation.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy