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I have a question regarding Relations on Sets. Here is the problem:

Let $S=\left\{ a, b, c\right\}$. Then $R=\left\{ (a,a), (a,b), (a,c)\right\}$. Which of the properties (reflexive, symmetric, or transitive) does the relation, $R$, possess?

Here is what I was thinking:

Not reflexive. Since $(b,b)$ and $(c,c)$ are not in $S$, we have that $(b \not R b) , (c \not R c)$-- therefore $R$ is not reflexive.

Not symmetric. Given $(a,b)$, then we would also need $(b,a)$ to show that $aRb$ and $bRa$.

Not transitive. To show the transitive property, we need to show if $aRb$ and $bRc$, then $aRc$. This would require $(a,b), (b,c)$ -- which we do not have.

So, to me it would appear as though this relation, $R$, is not reflexive, symmetric, nor transitive. Does this seem like I have the right mindset?

Thanks for looking!

Mia

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    $\begingroup$ I think it might be vacuously true that it is transitive. $\endgroup$ – Cameron Williams Jul 17 '15 at 23:31
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All are right, but the transitivity, we do have transitivity.

Here we have $3$ elements, we can check all $3\times 3$ possibilities, like taking $(a,a),(a,b)$ then we do have $(a,b)$, or taking $(a,b),(a,c)$, then we do not have to think about if $b\neq a$, but if $b=a$, then we have $(a,c)$, and so on you can check all $9$ possibilities.

To visuallise transitivity, we can make them as a directed graph, where $(x,y)$ means a directed arc, and transitivity means every walk of length two can be shortcut by some arc.

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  • $\begingroup$ So, I was thinking that it could be vacuously transitive.... but I was not certain. $\endgroup$ – Mr. Meeseeks Jul 17 '15 at 23:32
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    $\begingroup$ @mia.avery I like to think about the transitivity condition as a promise. If I say that $R$ is transitive, I mean that every time you can find $x, y, z$ such that $xRy$ and $yRz$ both hold, then I can promise you that $xRz$ also holds. If there is no triple in $R$ such that the first condition holds, then it will be very easy to keep my promise, won't it? $\endgroup$ – Arthur Jul 17 '15 at 23:44
  • $\begingroup$ @Arthur Thanks for your help. Your analogy was just what I needed. Thanks! $\endgroup$ – Mr. Meeseeks Jul 17 '15 at 23:48
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when p is false $$ p \Rightarrow q$$ is true so the relation is transitive because you need $$ (a,b) \wedge (b,c) \Rightarrow (a,c)$$
let's check $$ (a,a) \wedge (a,b) \Rightarrow (a,b) \space \surd \\ (a,a) \wedge (a,c) \Rightarrow (a,c) \space \surd $$ and we have not any thing to put for $ \Box \space\Rightarrow \Diamond $

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