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.

How do you formally define equivalence classes ? – Mar 21 '16 at 07:25 — , Mar 21 '16 at 07:25
Just updated the question to include it. – Chris Mar 21 '16 at 07:31 — Chris, Mar 21 '16 at 07:31

score 1 · Answer 1 · answered Mar 21 '16 at 07:52

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?

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.

score 0 · Answer 2 · answered Mar 21 '16 at 07:32

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 . . .

score 0 · Answer 3 · answered Mar 21 '16 at 07:33

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 $xRx$).

score 0 · Answer 4 · answered Mar 21 '16 at 07:33

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

answered Mar 21 '16 at 07:33

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score 0 · Answer 5 · answered Mar 21 '16 at 07:55

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By the definition

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

Then

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

answered Mar 21 '16 at 07:55

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

5 Answers5