0

I understand it in theory,but not in practice So here is the exampl

  1. LetA={a,b,c,d,e,f}
  2. G,H be equivalence relation in A

3.G=$1_a$ $\cup$ {(a,b),(b,a),(b,c),(c,b),(a,c), (c,a),(d,e),(e,d)}

H=$1_a$ $\cup$ {(b,c),(c,b)}

Find A/G

Here is my guess

A/G=G

Any help would be mighty appreciated to figure out how to do this

1 Answers1

1

$a$ is related to $b$ under the equivalence relation given by $G$, $b$ is related to $c$, so $a$ is related to $b$ and by transitivity to $c$, hence the equivalence class of $a$ is the subset $\{a,b,c\}$ of $A$. Then $d$ is related to $e$ and $f$ is related to $f$ itself. Hence the set of distinct equivalence classes of $A$ modulo $G$ is given by $\{a,b,c\}, \{d,e\}$, and $\{f\}$.

Maryam
  • 1,889
  • 1
    Hence $A/G={{a, b, c}, {d, e}, {f} }$ – Gribouillis Jun 19 '20 at 17:18
  • yes, I am editing, thanks – Maryam Jun 19 '20 at 17:18
  • So in A/G you are checking transitivity in G? So for –  Jun 19 '20 at 17:41
  • You don't need to check transitivity, by assumption you have that $G$ is an equivalence relation, that is, reflexive, symmetric and transitive. I am using transitivity to put elements of $A$ into classes of $A/G$. – Maryam Jun 19 '20 at 17:44
  • Ok So what do I have to do in general? Use any of the given properties. So A/H={b,c} ? b is related to c under the equiv. relation of under H and c to b. –  Jun 19 '20 at 17:53
  • $G/H={ {a}, {b,c}, {d}, {e}, {f}}$ – Maryam Jun 19 '20 at 17:56
  • Thanks. Can u explain how you got it ? Looking at H, by assumption bHc-> cHb and all the other are singletons . Thus you get your set? –  Jun 19 '20 at 18:14
  • Don't forget reflexivity: for each $a\in A$, $a$ is in the class $[a]$ represented by $a$, both with respect to $H$ and to $G$. – Maryam Jun 19 '20 at 18:17
  • Thanks for the help –  Jun 19 '20 at 20:09