Show that $x$ is an element of its own equivalence class?

Question

If $R$ is an equivalence relation on $X$ and $x$ is an element of $X$, the equivalence class is defined as $[x]_R = [y ∈ X : xRy]$.

Since $x$ is equivalent to itself, doesn't that automatically make it in its own equivalence class? It seems pretty self explanatory, but I don't know how to show this more formally.

Answer 1

This might help you see how to come up with a precise proof: consider the relation $R$ on the set $\{1, 2\}$ given by $1R1, 2R1, 1R2$. $R$ is not an equivalence relation (why?), but we can still (in analogy with equivalence classes) define the "$R$-class" of an element $x$ to be the set of $y$ which are $R$-related to $x$.

Now, is $2$ in the $R$-class of $2$? Well, for that to be the case, we'd have to have $2R2$, but we don't; so $2$ is not in its own $R$-class!

If you think about this for a bit, I think you'll see how to write a precise proof that any element is an element of its own equivalence class with respect to an equivalence relation . . .

Answer 2

Yep, $x$ is in its own equivalence class because it is equivalent to itself. If you wanted to write this a bit more formally, you could say that $xRx$ by reflexivity of $R$, and hence $x\in[x]_R$ by the definition of $[x]_R$ (if you substitute $y=x$ into the condition defining what it means to be an element of $[x]_R$, you just get $xRx$).

Answer 3

$x$ doesn't become an equivalence class, but it is (of course) a member of the equivalence class $[x]_R$.

Answer 4

An equivalence class is a set of all elements in an equivalence relation with a specified element.   That is if we have a set $S$ and an equivalence relation $R$ on that set, then we define the ($R$-)equivalence class for element $a$ of $S$ as: $${[a]}_R~:=~\big\{x{\in}S: (x,a){\in} R\big\}$$

An equivalence relation is a relation that is reflexive, symmetric, and transitive.   One of these properties tells us that $~a\in[a]_R~$, can you tell me which one?

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Basically, yes, an element is included in its own equivalence class because it is "equivalent to" itself, due to the fact that any equivalence relation is reflexive.

Answer 5

By the definition

$$y\in[x]_R\iff y\in X\land xRy.$$

Then

$$x\in X\land xRx\implies x\in[x]_R.$$