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This was in the book I was studying from so I went to prove it just for the sake of further understanding it, although it wasn't an exercise so I can't really go anywhere to check if my proof is correct. This is my first time working with equivalent relations so I'd really like to get some insight.

Two elements of the quotient $S/{\sim}$ are equal if and only if the corresponding elements in $S$ are related by $\sim$.

Let $A,B,a,b \in S$, and \begin{align*} S/{\sim} \ &:= \{[A]_\sim , \ [B]_\sim, \ \dots \},\\ [A]_\sim \ &:=\{ a \in S \mid a \sim A\},\\ [B]_\sim \ &:=\{ b \in S \mid b \sim B\}. \end{align*}

If $a \sim A$, $b \sim B$, and $a \sim b$, then because $\sim$ is transitive and symmetric, $a \sim b \to a \sim B$ so $a \in [B]_\sim$ and $b \sim a \to b \sim A$ so $b \in [A]_\sim$.

What do you guys think?

2 Answers 2

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Well, you are heading the right way. This is how I will do it, first let's translate into maths your statement:

$[A]_\sim=[B]_\sim$ if and only if $A\sim B$

Now we can just work with your definitions.

1) Assume $[A]_\sim=[B]_\sim$ then, $A\in [A]_\sim$ implies $A\in [B]_\sim$, which then implies $A\sim B$.

2) Assume $A\sim B$ then you have to prove an equality of sets, so one inclusion and then the other.

Take $a\in[A]_\sim$, that means $a\sim A$ and now you can use transitivity and $A\sim B$, to get $a\sim B$ which means $a\in[B]_\sim$

So you have $[A]_\sim\subseteq[B]_\sim$ and the other inclusion is similar.

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    Oh, thats much better! I was also unsure of how to show the converse, so thank you for indicating your method :) – 2017-01-14
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    If you want to be really strict to prove that other inclusion you will need the symmetry apart from the transitivity. – 2017-01-14
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    I was actually writing down the proof in my notebook and when I was showing $b \sim B$ implies $b \sim A$ I made sure to indicate usage of the symmetry property! Thanks again! – 2017-01-14
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You are only showing the "if" part of the equivalence; so you need to write the "only if" part for it to be complete. Also, your notation $a \sim A$ should be $a \in [A]_\sim$, since equivalence classes are sets. I would prove this result as follows:

For any $x \in S$, we have $x \sim x$ (reflexivity), so $\{ a \in S \mid a \sim x \}$ is not empty (it contains at least $x$). So we can take a shorter notation for equivalence classes, and write for example $\dot{x}$ to denote the equivalence class of the element $x$.

Taking $y \in \dot{x}$, we have by definition of $\dot{x}$, $x \sim y$. For any $z \in S$ such that $y \sim z$, the transitivity of $\sim$ allows us to conclude that $x \sim z$. From this we conclude that $\dot{y} \subset \dot{x}$. We can switch the roles of $x$ and $y$ in the above to get the reverse inclusion $\dot{x} \subset \dot{y}$. We have shown that $\forall x,y \in S,\; x \sim y \implies \dot{x} = \dot{y}$

For the converse, take $x$ and $y$ such that $\dot{x} = \dot{y}$. Since $\dot{x}$ is not empty, take an element $z \in \dot{x}$. Such a $z$ verifies $x \sim z$. Since $\dot{x} = \dot{y}$, $z$ is also an element of $\dot{y}$, and so $y \sim z$. The symmetry of $\sim$ shows that $z \sim y$, and then the transitivity of $\sim$ allows us to conclude that $x \sim y$. We have shown that $\forall \dot{x}, \dot{y} \in S/\sim,\; \dot{x} = \dot{y} \implies x \sim y$