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I'm trying to do some chapter problems on equivalence relations. I'm stuck in the second section "properties of relations."

Question: Let $A=\{a,b,c,d\}$. Give an example of a relation $R$ on $A$ that is neither reflexive, symmetric ,or transitive.

What I tried doing was writing out all the pairs and then canceling out the ones that matched with the laws. I was left with $\{(d,b) (d,c)\}$. Does that mean $R=\{(d,b) (d,c)\}$ is not reflexive, transitive, or symmetric ?

Sloan
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Ali Raza
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    Let's see. Is R reflexive? Is $xRx$ for all $x$. No. In fact $x\not Rx$ for all $x$ so not reflexive. Is it symmetric? $dRb$ but $b\not R d$ so not symmettric. Is it transitive. If $x R y$ and $y R z$ is it $x R z$. Well if $x R y$ then $x = d$ and $y \ne d$ so there is no $y R z$. So as there is no $x R y $ and $y R z$ is is "vacuously true" that it is transitive. Sorry. – fleablood Sep 26 '16 at 20:53
  • What set are you basing your answer off? How do I know I am writing the correct R based on given A to apply the rules? – Ali Raza Sep 26 '16 at 20:58
  • I'm basing my answer on your R. You wrote a relationship that is not symmetric or reflexive and only vacuously transitive. – fleablood Sep 26 '16 at 21:04
  • So you're saying it is correct to write all ordered pairs and only use the pairs that are not part of the reflexive, transitive, symmetric justification and say that these pairs don't satisfy the properties of relations. Your answer is the definition of the properties and I understand that part. – Ali Raza Sep 26 '16 at 21:09
  • Yes, but that seems a little haphazzardous and random. There are easier ways to do it. – fleablood Sep 26 '16 at 21:19

3 Answers3

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Well, here's a way to do it. Maybe over kill.

Let $S = A \times A$. That is the universal relationship.

As $(x,x) \in S$ for all $x$ it is reflexive. Let's destroy that be removing $(a,a)$.

$T = A \times A \setminus \{(a,a)\}$.

$T$ is not reflexive because $a \not T a$.

But if $x T y$ then $y T x$. So T is symmetric. Let's kill that. Let's remove $(a,b)$.

$W = A \times A \setminus \{(a,a),(a,b)\}$

Not $W$ is not symetric because $b W a$ but $a \not W b$.

Is $W$ transitive? To be honest, I'm not sure. But $a R c$ and $c W d$ and $a W d$. Let's remove $(a,d)$.

Let $R = A \times A \setminus \{(a,a),(a,b),(a,d)\}$

$R$ is not transitive because $a R c$ and $c R d$ but $a \not R d$.

And it's not symmetric as $a \not R a$ nor reflexive as $b R a$ but $a \not R b$.

Or we could have built from scratch;

$S = \{(a,b)\}$

Not reflexive: $a \not S a$.

Not symmetric: $a S b$ but $b \not S a$.

Transitive? Vacuously so. There are no $x S y$ and $y S z$. So for all zero of those it is vacuously true $x S y; y S z \implies x S z$.

But let's kill this by adding $R = \{(a,b),(b,c)\}$

Then we have $a R b$ and $b R c$ but $a \not R c$ so it is not transitive.

fleablood
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Question: Let $A=\{a,b,c,d\}$. Give an example of a relation $R$ on $A$ that is neither reflexive, symmetric ,or transitive.

I built that from knowing to exclude reflexivity; there's no symmetry, and no transitivity.

Let $$R =\{(a,b ), (b, c), (c, d)\}$$

amWhy
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  • wow. So, did you list your pairs or just think about it intuitively? I'm just trying to understand how you got your answer. Why were my pairs excluded from your answer? – Ali Raza Sep 27 '16 at 03:39
  • https://www.facebook.com/MNdabbi – Mikasa Sep 29 '16 at 07:45
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Well, the relation you give is indeed neither reflexive nor transitive, but there are many ways to get such a relation. "Cancelling out pairs that don't match the laws" is not a well defined procedure. (Well, it is defined for reflexivity, but it does not do what you want: You would be left with $\{(a,a),(b,b),(c,c),(d,d)\}$ which is reflexive).

I think the problem would like you to find a relation that involves all four of the members of the set. You could, for example, add in $(a,b)$. Or even $(a,a)$; just one element being equivalent to itself does not make the relation reflexive if others are not.

Mark Fischler
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  • Actually it is vacuously transitive. For all $x R y$ and $y R z$ it is true that $x R z$ because there are no $x R y$ and $y R z$. – fleablood Sep 26 '16 at 20:54
  • The questions wants a set that is not symmetric, transitive, or reflexive. In the form R={...}. I assume my problem is devising the set. How do I define the set and then apply the rules. What if the question asked "is the set reflexive" and I started my answer by writing a random combo of ordered pairs. – Ali Raza Sep 26 '16 at 21:05
  • Reflexive means $x R x$ for all $x \in A$. Simply leave out a $(x,x)$, say omit (a,a), and it won't be reflexive. Or simply never include any in the first place. It seems too hard and arbitrary to try every subset and to see if it satisfies things or not. It'd be easier to deliberate make your set so that the satisfy or not. There are $2^{16} $ relations. It's not practical the check every one. For $A = \mathbb R$ there are uncountable relations. Forget randomly checking.... – fleablood Sep 26 '16 at 21:23
  • Wow, I didn't now there were that many pairs. I just listed the pairs based some sort of combinatorics for {a,b,c,d}. Using the definitions of reflexive, transitive, and symmetric directly seems like a better idea. I guess I need to be a little more intuitive in this case. I don't think the old work by hand can help here. – Ali Raza Sep 27 '16 at 03:28