theorems on equivalence classes

Question

I have a few short proofs that I wish to be checked regarding equivalent classes.

Suppose that R is an equivalence relation on set X. If $a, b \in X$, then

1. $a\in [a]$
2. $[a] = [b] \iff (a, b) \in R$
3. $[a]\cap[b]=\emptyset\iff(a, b)\notin R$

For number 1 I understand that I have to make the claim "every element in X must be in one equivalence class." But how may I write out a rigid proof?

For number 2/3 I have minimal idea on how to begin. Would you shed some light to these short proofs please? Thank you.

Edit: Thanks Author. Equivalent relation is defined here as a relation that is reflexive, symmetric and transitive.

Answer 1

For $(1)$, since $a\in X$, $R$ is an equivalence relation, we have that $aRa$. Therefore, $a\in [a]$. Because $[a] = \{x: xRa\}$.

Now for $(2)$, Assume $[a] = [b]$. You have deduced that $a\in [a]$. Since $[a]=[b]$ you can deduce that $a\in [b]$ because sets are equal if and only if they have the same elements. Since $a\in [b]$, $aRb \longrightarrow (a,b) \in R$. Now assume $(a,b)\in R$. Let $x\in [a]$. We desire to show that $x\in [b]$, (so that $[a]\subseteq [b]$). Since $x\in [a]$, we have $xRa$. Since $R$ is an equivalence relation, $R$ is transitive. By hypothesis, $aRb$. Therefore $xRa \land aRb \longrightarrow xRb$ so that $x\in [b]$. Now let $y\in [b]$. We desire to show that $y\in [a]$. By hypothesis, $aRb$. Since $aRb$ and $R$ is an equivalence relation, $R$ is symmetric and therefore $bRa$. Applying transitivity, we have $yRb \land bRa \longrightarrow yRa$, so $y\in [a]$. Thus we have $[a] \subseteq [b] \land [b] \subseteq [a] \longrightarrow [a] = [b]$.

Now for $(3)$, assume $[a] \cap [b] = \emptyset$. Later assume for sake of contradiction $(a,b) \in R$. $(a,b) \in R \longrightarrow a\in [b]$. We already have deduced that $a\in [a]$. So $\{a\} \subseteq [a]\cap [b] = \emptyset$, a contradiction. Now assume that $(a,b) \notin R$. We desire to show that $[a]\cap [b] = \emptyset$. Assume for sake of contradiction that $[a]\cap [b] \ne \emptyset$. Then there is some $x$ such that $xRa$ and $xRb$. Since $xRa$, and $R$ is symmetric, $aRx$. Now since $aRx \land xRb$, by transitivity of $R$, we have that $aRb$. Therefore, $(a,b) \in R$, a contradiction.

Remark: Most of these proofs are about unraveling definitions, applying them when needed, and following your nose. Hope this helps.

Answer 2

I guess $[a] := \{b\in X : (a,b)\in R\}$, is that right?

1. You know $(a,a)\in R$ for any $a$, so ...

2. $a\in [a] = [b]$ tells you that $(a,b)\in R$. That's one direction. For the other, say $(a,b)\in R$. Then which of the following is true: $a\in [b]$ or $a\notin [b]$?

3. Say you can choose some $x\in [a]\cap [b]$. Then $(a,x)\in R$ and $(b,x)\in R$. Now transitivity of $R$ gives ...

Hope that helps!