Help needed with an equivalence relation task on natural numbers

Question

I'm having a bit difficulties understanding and solving this task. I would appreciate any help on how you can solve tasks like this.

Here is the task:

Let ~ be an equivalence relation on the natural numbers, and let $E$ be $[0]$, ie equivalence class to $0$.

1. Prove that E is not equal to the empty set
2. Prove that all natural numbers $x$ and $y$ that are elements of $E$ are such that $x$ ~ $y$.
3. If $[x] = [y]$, then $x$ ~ $y$? If yes, give a proof for it, if no, give a counterexample.

Thanks a lot for any help.

Answer 1

For the first one, consider what three properties an equivalence relation has, and note that $$E=\{z\in\Bbb N:0\sim z\}.$$ One of the three properties of an equivalence relation will allow you to prove the claim.

For the third one, we will use the same property of an equivalence relation. You want to show that if $$\{z\in\Bbb N:x\sim z\}=\{z\in\Bbb N:y\sim z\},$$ then $x\sim y.$

I believe that the second one should read "Prove that all natural numbers $x,y$ which are elements of $E$ are such that $x\sim y.$" That is, we must show that if $x,y\in E,$ then $x\sim y.$ This time, we will be using the other two properties of equivalence relations to prove it.

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Edit: As you know, a binary relation $R$ is said to be an equivalence relation on a set $A$ if and only if it is reflexive, symmetric, and transitive on $A$. In particular, this means:

• For all $x\in A,$ we have $x\:R\:x.$ ($R$ is reflexive on $A$.)
• For all $x,y\in A,$ if $x\:R\:y,$ then $y\:R\:x.$ ($R$ is symmetric on $A$.)
• For all $x,y,z\in A,$ if $x\:R\:y$ and $y\:R\:z,$ then $x\:R\:z.$ ($R$ is transitive on $A$.)

Now, suppose that $R$ is any binary relation, $A$ any set. For each $x\in A,$ define:

$$[x]_{A,R}:=\{z\in A:x\:R\:z\}$$

Let's use this notation to rewrite the definitions of reflexivity, symmetry, and transitivity of $R$ on $A$:

• For all $x\in A,$ we have $x\in[x]_{A,R}.$ ($R$ is reflexive on $A$.)
• For all $x,y\in A,$ if $y\in[x]_{A,R},$ then $x\in[y]_{A,R}.$ ($R$ is symmetric on $A$.)
• For all $x,y,z\in A,$ if $y\in[x]_{A,R}$ and $z\in[y]_{A,R},$ then $z\in[x]_{A,R}.$ ($R$ is transitive on $A$.)

Observe then that for any $x\in\Bbb N,$ we have $[x]:=[x]_{\Bbb N,\sim}.$ Do you see, then, which of the three properties immediately tells us that $[0]\ne\emptyset$? We will use this property again to show that if $[x]=[y],$ then $y\in[x]$ (why is this true?), and so $x\sim y$ by definition.

For $(2),$ we will use the other two properties. In particular, we want to show that if $0\sim x$ and $0\sim y,$ then $x\sim y.$