1

Yea, I hate myself when I ask this question. But I got a question below:

Consider the following relations on the set $\{a, b, c\}$. Which of the relations are equivalence relations?

It's from a multiple choice. One of the correct answers is:

$\{(a, a),(b, b),(c, c)\}$

I understand all the other answers but I just have to convince my self about this is true. Here is my ideas:

  • It's reflective because $aRa$.

Let $a=b$ then

  • It's s symmetric because $aRb$ and $bRa$ where $a=b$.

Let $a=b=c$ then

  • It's transitive because $aRb$ and $bRc$ then $aRc$ where $a=b=c$.

Am I correct? If not, how can I convince my self it's true?

Gaurang Tandon
  • 6,449
Amir Hassan
  • 425

3 Answers3

0

Your step-by-step method and the logic for each step is entirely correct. This correct method should convince you that it is true. That's really all that is there to this.

Gaurang Tandon
  • 6,449
  • Thanks man, I will try to convince my self.

    Btw how about this: ${(a, a),(a, b),(b, a),(b, b),(c, c)}$

    I can see it's reflective and symmetric. It should be transitive, but how? I see that $aRb$ and $bRa$, but I can't find $bRc$ so if $bRc$ does not exist, then I don't have to worry about transitive there? I can see it's simmilar to the first I posted, but am I right?

     – Amir Hassan Jan 15 '18 at 15:31
  • 1
    @KasperMedK Yeah, you don't need to worry about $bRc$. In transitivity, remember that you only need to check $aRc$ *IFF* $aRb$ AND $bRc$. In the above set, there do not exist any two pairs $bRx$ and $xRc$ ($x\in{a,b,c}$), so you needn't worry about whether $bRc$ exists or not. – Gaurang Tandon Jan 15 '18 at 15:35
  • Ohhh! I see, thank you very much! – Amir Hassan Jan 15 '18 at 15:38
  • @KasperMedK If your query is solved, please consider accepting the answer. Thank you! – Gaurang Tandon Jan 15 '18 at 15:56
0

I'm going to have to disagree with the first answer about this solution being correct. Your relation is explicitly on the set $\{a,b,c\}$. That is, $a,b,c$ are fixed objects. You can't "let $a = b$" in your reasoning, any more than you can "let $2=5$" if you were investigating a relation on the integers.

It is reflexive because every member of the set is related to itself.

The arguments for symmetry and reflexivity are essentially vacuous and degenerate cases of those properties. Your arguments do have the right idea behind them, but you need to express these ideas with different symbols (IE, let $xRy$, then $x=y$. Then...). You can't declare explicit members of your set to be equal.

Notice that this is the identity relation on your set. Equivalence relations are essentially a generalization and abstraction of this idea in the same way that a ring is a generalization of the integers or a metric is a generalization of distance.

Jonathan Hebert
  • 23
0

The axioms defining an equivalence relation $R$ over a set $A$ use universal quantification, and those defining symmetry and transitivity also use implication – and this is important.

The reflexivity axiom states that

For every $a \in A$ we have that $aRa$

Clearly, if $a = \{ a,b,c \}$ this holds for our choice of $R$.

The symmetry axiom states that

For every $a,b \in R$, if $aRb$ then also $bRa$

To check that this axiom holds for our choice of $R$, we need to check that for any choice of $a$ and $b$, then whenever $aRb$ then also $bRa$. We only have that $aRa$, $bRb$ and $cRc$, and clearly if $aRa$, then $aRa$ -- the cases for $bRb$ and $cRc$ are similar. All other cases are vacuously true; for instance if $aRc$ (which is not the case) then also $cRa$.

The case of transitivity is similar; the axiom here states that

For all $a,b,c \in A$, if $aRb$ and $bRc$ then also $aRc$.

Here, the only cases that must be checked are that

  • if $aRa$ and $aRa$ then also $aRa$
  • if $bRb$ and $bRb* *then* also $bRb$ - *if* $cRc$ and $cRc$ *then* also $cRc$
Hans Hüttel
  • 4,271