Currently I am trying to understand the equivalence class from the Herstein's book. And I had understood almost the complete proof of the theorem given on the page number 8. Book name is Topics in algebra, second edition. But struggling to give justification for one simple line.
I am struggling at the last line which says that "it is an exercise to prove that this is an equivalence relation on A and the distinct equivalent classes are the Aα's. The former one is quite simple but for the later one what I had thought is that we have to show:
For any $a\in A$, $cl(a)=A_α$ for some $α$
For this: Let any $b \in cl(a)$, which is true iff b~a iff b and a are in same $A_α$. Or $b,a \in A_α$ for some $α\in T$
I don't know whether this is enough or not? Can someone please give valid steps justifying the same?
I think you are correct in that you need to show the equivalence classes are precisely the $A_\alpha$. I assume you have already shown that $\sim$ is an equivalence relation and will use this fact in my solution. Here is how I would proceed:
Let $a\in A$ and consider $cl(a)$. As already shown, $a\sim a$ and so there exists an $\alpha \in T$ such that $a\in A_\alpha$. Now we want to show that this $A_\alpha = cl (a)$. Take $b\in cl(a)$. You basically have this already but for completeness sake, I'll repeat it. By definition, $a \sim b$ and, again, by definition, $b\in A_\alpha$ (since we already have that $a \in A_\alpha$). So we have just shown that $cl(a)\subseteq A_\alpha$. For the other direction, take $b\in A_\alpha$. But since $a,b\in A_\alpha$, $a\sim b$ and so $b\in cl(a)$. Thus we have shown that $A_\alpha\subseteq cl(a)$. So, $cl(a)=A_\alpha$. Therefore, for any $a \in A$, there is some $\alpha$ such that $cl(a)=A_\alpha$.
We can also show that, for $\alpha\not=\beta\in T$, we have that, $cl(a)\not=cl(b)$, if $a\in A_\alpha$ and $b\in A_\beta$.