Proof Verification: $aRb$ iff $\exists c \in A: a,b \in [c]$

Question

In an exercise I'm asked to prove the following:

Let $R$ be an equivalence relation in a set $A$, and let $a,b \in A$, then:

$$aRb \ \text{ iff } \ \exists c \in A: a,b \in [c]$$ Where $[c]$ is the equivalence class of $c$.

I proved that $\exists c \in A: a,b \in [c] \Rightarrow aRb$, using the fact that $R$ is transitive, but I'm not sure whether my proof for $aRb \Rightarrow \exists c \in A: a,b \in [c]$ is valid. This is what I did:

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My proof:

Let's assume that $aRb$. I'll do this part by contradiction. Let's assume that $\nexists c \in A: a,b \in [c]$. This means that , one of the following is true for all $c \in A$:

• $c R a \wedge c\not R b$ $\ \ \ $ (1)

• $c \not R a \wedge c R b$ $\ \ \ $(2)

• $c \not R a \wedge c\not R b$ $\ \ \ $(3)

If (1):

$c R a \wedge a R b$, so by transitivity $c R b \to$ contradiction

If (2):

$c R b \wedge b R a$, so by transitivity $ c R a \to $ contraditcion

If (3):

$\forall c \in A \setminus \{a,b\}, c \notin [a] \wedge c \notin [b]$. This means that $[a] = [b] = \{a,b\}$, so if we set $c = a$, we have $c R a \wedge c R b \to$ contradiction.

So assuming that $\nexists c \in A: a,b \in [c]$ leads to a contradiction.

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I think that (1) and (2) are correct but I'm a little sceptical about (3). Did I made some mistake or is this correct?

Answer 1

Your proof is incorrect. In point (3) you stated that "$c \in A \setminus \{a,b\}$" but then you arrived at the conclusion that $c = a$.