Show that if $a \sim b$, then $C(a) = C(b)$ where $C(x)$ is the equivalence class containing $x$.

Question

I'm working through a textbook on my own, so I don't want the full answer. I'm only looking for a hint on this problem.

Show that if $a \sim b$, then $C(a) = C(b)$ where $C(x)$ is the equivalence class containing $x$.

I'm reading through Paul Sally's Tools of the Trade, and he gives this problem as an example in a theorem. An equivalence class is defined like this: If $a \in X$, then we write $C(a) = \lbrace b \in X \mid b \sim a \rbrace$. The proof is as "Transitivity" and that's it, but I'm not sure how to use transitivity to prove it. Here's what I have so far:

Let $a, b \in X$ be arbitrary elements of $X$. If $a \sim b$, then $a \in C(b)$ by definition. But, if $a \sim b$, then $b \sim a$ by symmetry, so $b \in C(a)$. This is where I get stuck. Since $a, b$ are arbitrary elements in $X$, can I just immediately conclude that $C(a) = C(b)$ for any $a,b \in X$ where $a \sim b$. I'm not sure how to use transitivity here.

Answer 1

Let us introduce some more variety in the letters used. For $x \in X$, define $C(x)$ by:

$$C(x) = \{y \in X\mid y \sim x\}$$

Hence:

$$C(a) = \{y \in X \mid y \sim a\}, \quad C(b) =\{y \in X\mid y \sim b\}$$

What does it mean for these to be equal? How does transitivity come into play when proving this?

Answer 2

Hint:

Prove $C(a)\subseteq C(b)$, and $C(b)\subseteq C(a)$.

Answer 3

HINT: If $x\in C(a)$, then $x\sim a\sim b$, so ...