Reflexivity, symmetry, and transitivity of difference operation between two binary relations

Question

This is exercise $4.3.15$ from the book How to Prove it by Velleman $($$2^{nd}$ edition$)$:

Suppose $R_1$ and $R_2$ are relations on $A$. For each part, give either a proof or a counterexample to justify your answer.

$(a)$ If $R_1$ and $R_2$ are reflexive, must $R_1\setminus R_2$ be reflexive$?$

$(b)$ If $R_1$ and $R_2$ are symmetric, must $R_1\setminus R_2$ be symmetric$?$

$(a)$ If $R_1$ and $R_2$ are transitive, must $R_1\setminus R_2$ be transitive$?$

Here are my answers:

Part $(a)$: Counterexample:

Suppose $A=\{1\}$, $R_1=\{(1,1)\}$, and $R_2=\{(1,1)\}$. Then $R_1\setminus R_2=\{\}$. Thus $R_1$ and $R_2$ are both transitive but $R_1\setminus R_2$ is not transitive and hence a counterexample.

Part $(b)$: Proof:

Suppose $R_1$ and $R_2$ are symmetric. Let $a$ and $b$ be arbitrary elements of $A$ such that $(a,b)\in R_1\setminus R_2$ which means $(a,b)\in R_1$ and $(a,b)\notin R_2$. Since $R_1$ is symmetric, $(b,a)\in R_1$ and since $R_2$ is symmetric, $(b,a)\notin R_2$. Ergo $(b,a)\in R_1\setminus R_2$. Since $a$ and $b$ are arbitrary, $R_1\setminus R_2$ is symmetric. Therefore if $R_1$ and $R_2$ are symmetric then $R_1\setminus R_2$ is symmetric. $Q.E.D.$

Part $(c)$: Counterexample:

Suppose $A=\{1,2,3\}$, $R_1=\{(1,3),(3,2),(1,2)\}$, and $R_2=\{(1,2)\}$. Then $R_1\setminus R_2=\{(1,3),(3,2)\}$. Thus $R_1$ and $R_2$ are both transitive but $R_1\setminus R_2$ is not transitive and hence a counterexample.

Here are my questions:

$1.$ Are my answers to parts $(a)$, $(b)$, and $(c)$ correct$?$

$2.$ I was certain about my answer to part $(c)$ until I saw this post. Is my counterexample a valid one$?$ If yes, then why it seems that there is a discrepancy between my answer and the above linked post$?$

Thanks for your attention.

Edit:

Question $2$ is irrelevant and I see it now. I mixed up the counterexample $(a)$ and counterexample $(c)$ in my mind. Thanks.

Regarding question $2$, how is the linked post relevant? It concerns the empty relation, which your counterexample is not. — PrincessEev, Aug 12 '20 at 07:23
@EeveeTrainer You are absolutely right. Thank you. In my mind I mixed up counterexample $(a)$ with counterexample $(c)$. I will edit the post accordingly. — Khashayar Baghizadeh, Aug 12 '20 at 07:25

score 2 · Answer 1 · answered Aug 12 '20 at 07:24

2

Everythink is perfectly correct. The link you posted is about the empy relation $R$ while in c) neither $R_1$, $R_2$ nor $R_1\backslash R_2$ are empty so it does not apply to your example c).

answered Aug 12 '20 at 07:24

crush3dice

317

score 2 · Accepted Answer · answered Aug 12 '20 at 07:25

2

(a) is fine, minimal counterexample.

To (b) I'd write "Since $R_1$ is symmetric, $(b,a) \in R_1$ and also $(b,a) \notin R_2$ or else $(a,b) \in R_2$, as $R_2$ is symmetric, and we knew $(a,b) \notin R_2$. So $(b,a) \in R_1 \setminus R_2$". (making the small argument using the symmetry of $R_2$ explicit).

To (c): I'd say the example looks good. But part of having an example is showing that actually $R_1,R_2$ are transitive (tedious case checking). It's also part of the solution. So it's somewhat incomplete as it stands now. I don't see the relevance of the linked post, though.

answered Aug 12 '20 at 07:25

Henno Brandsma

250,824

Thank you very much. In the book that I am studying (Velleman), the author recommends that we do not make our thought processes of the reasoning behind proof statements explicit. That's why I didn't mention that reasoning explicitly. Does not mentioning such a reasoning makes my proof invalid? – Khashayar Baghizadeh Aug 12 '20 at 07:35
I also don't understand why the counterexample $(c)$ is incomplete. Could you please explain more? – Khashayar Baghizadeh Aug 12 '20 at 07:37
@KhashayarBaghizadeh You have to prove that $R_1$ is transitive and $R_2$ too. You just claim it. – Henno Brandsma Aug 12 '20 at 07:39
I do not understand it. They are mere examples. Why should we prove them? Sorry I am self-studying the material. – Khashayar Baghizadeh Aug 12 '20 at 07:41
@KhashayarBaghizadeh An example is also a proof, but of a different type. – Henno Brandsma Aug 12 '20 at 07:42
So how does one prove that $R_2={(1,2)}$ is transitive? Isn't it trivial? – Khashayar Baghizadeh Aug 12 '20 at 07:44
1

@KhashayarBaghizadeh Yes, but you have to say why. $R_1$ has some more cases to consider. $R_2$ just one: Suppose $(x,y),(y,z) \in R_2$. Then $x=1$,$y=2$ as $(x,y) \in R_2$ implies $(x,y)=(1,2)$. But then $y=1,z=2$ too, contradiction. So there are no pairs $(x,y), (y,z) \in R_2$ and so $R_2$ is transitive voidly. – Henno Brandsma Aug 12 '20 at 07:48
Thank you very much. :$)$ – Khashayar Baghizadeh Aug 12 '20 at 07:49

Reflexivity, symmetry, and transitivity of difference operation between two binary relations

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