Equivalence Relation Statements Proof

Question

Let $\sim$ be an equivalence relation on a class $X$. The following are equivalent for $x,y \in X$.

1) $[x]=[y]$

2) $x \sim y$

3) $x \in [y]$

4) $y \in [x]$

5) $[x] \bigcap [y] \neq \emptyset$

I have a feeling this should be a very basic proof but I don't understand how to go about proving the equivalence of all these statements.

Answer 1

This is a classic problem where you should try something like this: $$1\Rightarrow2\Rightarrow3\Rightarrow4\Rightarrow5\Rightarrow1 $$

So for example let's do the first step $1\Rightarrow2$:

We assume $[x]=[y]$ and we want to show, that $x\sim y$.

First recall the definition of an equivalence class:

$[x]=\{ x'\in X \mid x\sim x' \}$

Since $[x]=[y]$, we have $[x]=\{ x'\in X \mid x\sim x' \}=\{ y'\in X \mid y\sim y' \}=[y]$.

Hence there must be an element, say $a\in [x]=[y]$ satisfying: $x\sim a \sim y$ and by transitivity of the equivalance relation we get $2$.

Now it's your turn :)

On your comment:

$2\Rightarrow 3$:

You assume $2$ is true and you want to show that $3$ is true:

$$x\sim y \Rightarrow x\in [y]$$

You don't have to show that $\sim $ is an equivalence relation, in fact you should use that. It's a prerequisite stated in the problem, so you can use it whenever you like (or when it makes sense). Just try to proof one step at a time.

Answer 2

Hint $\ $ Show that the equivalence classes form a partition of $X,$ from which the proofs of the equivalences follows straightforwardly.

Answer 3

Class of equivalence is defined $[x]=\{z:x\sim z\}$. Therefore, $$y\in [x]\Leftrightarrow x\sim y.$$ Since $\sim$ is equivallence relation, it is symmetric, so we have $x\sim y\Rightarrow y\sim x$. From all we wrote, we have: $$y\in [x]\Leftrightarrow x\sim y \Rightarrow y\sim x\Leftrightarrow x\in [y].$$ In the same way, it can be shown that $x\in [y]\Rightarrow y\in [x]$.

Therefore, $$2)\Leftrightarrow 3) \Leftrightarrow 4).$$ Now we will show that $1)\Rightarrow 2)$:

Since $\sim$ is reflexive, we know that $x\sim x$ and $x\in [x]=[y]$. Then, from the proven equivalence $2)\Leftrightarrow 3) \Leftrightarrow 4)$, we have $x\in [y]\Leftrightarrow x\sim y$.

Proof of $2)\Rightarrow 1)$:

Let $z\in [x]$ and $x\sim y$. Then, $z\sim x$ (since $\sim$ is symmetric and $z\in [x]\Leftrightarrow x\sim z$) and $x\sim y$, from where, by transitivity, we have $z\sim y$ and $z\in [y]$. So, we have shown $[x]\subset [y]$. Other inclusion $[y]\subset [x]$ can be shown in the same way.

It remains to show $1)\Leftrightarrow 5)$. $1)\Rightarrow 5)$ is trivial, so let us prove $5)\Rightarrow 1)$:

$[x]\cap [y]\neq\emptyset$. Let $z\in [x]\cap [y]$. Then, $z\in [x]\Leftrightarrow x\sim z$, and by symmetry $z\sim x$. Since we also have $z\in [y]\Leftrightarrow y\sim z$, by transitivity we can conclude $y\sim x$ and $[y]\subset[x]$. Similarly, $[x]\subset[y]$. Finally, $[x]=[y]$ and $$1)\Leftrightarrow 2)\Leftrightarrow 3)\Leftrightarrow 4)\Leftrightarrow 5).$$