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I've looked all around this stackexchange before asking this question. Just don't want to get penalized for a repeating question in case there is one.

I'm having trouble finding relations that are

  1. transitive but not reflexive or symmetric
  2. reflexive and transitive but not symmetric (not quite sure that this is possible)
  3. symmetric and transitive but not reflexive

To clarify, I'm looking for three different relations.

Thanks in advance!

Martin Sleziak
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Anon E. Muss
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    Have you checked http://math.stackexchange.com/questions/1592652/example-of-a-relation-that-is-symmetric-and-transitive-but-not-reflexive ? –  Dec 17 '16 at 15:04
  • Yeah, I did. I'm pretty sure that #2 is not possible, but not 100% sure. I should probably add that to my question. – Anon E. Muss Dec 17 '16 at 15:06
  • For 2, take the base set to be {a, b, c} and the relation {(a,a), (b, b), (c, c), (a, b), (b, c), (a, c)}. It is both "reflexive" and "transitive" but not "symmetric". – user247327 Dec 17 '16 at 15:09
  • See also [this question about the empty relation, which is bot transitive and symmetric, but not reflexive](http://math.stackexchange.com/questions/1081333/prove-that-the-empty-relation-is-transitive-symmetric-but-not-reflexive?rq=1 – amWhy Dec 17 '16 at 15:49

2 Answers2

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Let $A = \{1,2,3\}$, and $R$ be a relation on $A$.

  1. $R = \{(1,3),(3,2),(1,2)\}$

  2. $R = \{(1,1),(2,2),(3,3),(1,2)\}$

  3. $R = \{(1,1),(2,2),(2,1),(1,2)\}$

amWhy
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Kanwaljit Singh
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  • You're welcome! But you are the one who should be thanked for providing a sound answer! – amWhy Dec 17 '16 at 16:05
  • I am learning things and people like you are truly inspiration for me and other new joining learners. So its really proud when someone like you appreciate us. Again thank u. – Kanwaljit Singh Dec 17 '16 at 16:11
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HINTS:
$(1)$ A relation $R$ in $\mathbb R$ with $R=\{(a,b): a<b\}$.
$(2)$ A relation $R$ in $\mathbb R$ with $R=\{(a,b): a^3\geq b^3\}$.
$(3)$ Mentioned in @Fib1123's comment. EDIT: If you want another example, take a set $A=\{a,b\} $ where $a,b $ are distinct. Now take a relation $R $ in $A $ with $R=\{(a,a)\} $. Sorry for the wrong example.

  • You last example in $(3)$ is not transitive nor is it symmetric, transitivity because R lacks $(5, -5)$ (necessary to be symmetric), and it lacks $(-5, -5)$, which we need for transitivity.($(-5, -6 ) \in R$ and $(-6, -5))\in R$, but $(-5, -5) \notin R$. (Similarly we have $(-6, -5), (-5, -6) \in R$, but $(-6, -6) \notin R$. So if we were to require both symmetry and transitivity on A, then the relation would also need to be reflexive. If you meant for the last ordered pair of $R$ in (3) to be $(-5, -5),$ then the relation would be symmetric, but not transitive, since it lacks $(-6,-6)$. – amWhy Dec 17 '16 at 15:38
  • But, while the relation is symmetric now, after your edit $\sim$ 40 minues after your post, and 10 minutes of my comment, the relation you give for $R$ on $A$ still fails to be transitive: $(-6, -5)\in R,$ and $(-5, -6) \in R$, but $(-6, -6)\notin R$. and so, like I said, the relation at this point in time is not transitive. To make it transitive, you'd need to add $(-6,-6)$ to $R$, after which, the relation also becomes reflexive. Sorry! – amWhy Dec 17 '16 at 15:54
  • Bottom line, your answer was wrong on two accounts, at the point of its acceptance. You made one edit since that point to make the relation symmetric. The next edit you make (since the relation you chose for $(3)$ will fail), please indicate it is an edit to the post. E.g. below the body of your incorrect post wrt $(3)$, use: EDIT to be the header to your third attempt at creating a relation that satisfies $(3)$. – amWhy Dec 17 '16 at 16:01
  • For the set $A = {(-6, -5}$ you could simply have chosen the relation $R = {(-5, -5)}$, and as a relation on A, it is transitive, and it is symmetric, but it fails to be reflexive. – amWhy Dec 17 '16 at 16:12