On page 217 of Velleman's "How to Prove It", Velleman states a lemma. It goes like this (note that $[x]_R = \{y \in A : y \, R \, x \})$:
Lemma 4.6.5. Suppose R is an equivalence relation on A. Then:
- For every $x \in A$, $x \in [x]_R$
- For every $x \in$ and $y \in A$, $y \in [x]_R$ iff $[y]_R=[x]_R.$
I don't understand why (1) should be true. After all, an equivalence class is a set $[a]_R = \{ x \in A : x \ R\ a \}$, where $R$ is an equivalence relation - i.e., a relation that is reflexive, symmetric and transitive.
As I didn't understand it, I turned to the commentary on the page, and found the author had written:
According to the definition of equivalence classes, $x \in [x]_R$ means $x R \ x$. This is what leads us to apply the fact that R is reflexive.
...so? Yes, $x \in [x]_R$ means that $x R \ x$, but that isn't the only hard requirement for it to be in $[x]_R$ - we just established that an eq. class is one defined using an equivalence relation, that is to say, a relation that is not only reflexive but also symmetric and transitive! How can we say that $x \in [x]_R$ by just looking at one of 3 requirements?
So I turned to another book. Cunningham's "A Logical Introduction to Proof", pg. 216. I found an almost identical list on that page, and the first pointer was yet again
- For every $x \in A$, $x \in [x]_R$
So I turned to the proof, and in the proof the author goes
To prove (1), let $x \in A$. Clearly, $x \in [x]_R$, as $R$ is reflexive.
What? How can you conclude that $x \in [x]_R$ by just considering one of three requirements of the eq. relation R?
The same thing in Lay's "Analysis with an Introduction to Proof":
Given an equivalence relation $R$ on a set $S$ [...] we define the equivalence class (with respect to $R$) of $x \in S$ to be the set: $$E_x = \{y \in S : y \, R \,x \}$$ Since $R$ is reflexive, each element of $S$ is in some equivalence class.
Why! Why would $R$ being reflexive have anything to do with each element of $S$ being in some equivalence class! To be in an eq. class, they're supposed to be reflexive, symmetric and transitive - why does everybody keep repeating they only need to be reflexive!
I'm tall, heavy, and old. To answer the question "Can you reach the top shelf?", I need only consider my property of tallness. Tallness is sufficient to establish whether I can reach the top shelf. My heaviness and my oldness do not enter into resolving the question.
A relation has three properties: reflexivity, symmetry and trainsitivity. To answer the question, "Is $x$ related to $x$?" The relation's property of reflexivity is sufficient to establish whether $x$ is related to $x$. Symmetry and transitivity do not enter into resolving the question.
You are having a useful thought. We could imagine the class of "sameness"s. These are weaker than equivalence relations -- they only satisfy reflexivity. Note that every equivalence relation is a sameness relation because every equivalence relation satisfies the one requirement to be a sameness (and also some other requirements which sameness doesn't require). The thing you are asking about $x \in [x]_R \iff xRx$ is satisfied by every sameness, via reflexivity, so it is satisfied by every equivalence relation as well. In short, the other two additional properties of an equivalence relation cannot prevent reflexivity giving $x \in [x]_R \iff xRx$.
Edit following comment...
An equivalence relation is defined to be a relation having the properties of reflexivity, transitivity, and symmetry. If we establish by hypothesis that a thing is an equivalence relation, it automatically has all three properties (from the definition of "equivalence relation"). If we establish by hypothesis that a thing is only a relation, then it does not automatically have these three properties.
Your lemma 4.6.5 starts "Let $R$ be an equivalence relation on $A$." So we have hypothesized that $R$ is an equivalence relation. This gives us four facts:
If we need any of these four facts to get to the conclusion of our proof, they are available to us (as many times as we need) throughout.
The first thing you want to show is "For every $x \in A$, $x \in [x]_R$.". So we proceed : \begin{align*} 1 &: x \in A && \text{[hypothesis]} \\ 2 &: R \text{ is an equivalence relation on } A && \text{[given]} \\ 3 &: R \text{ is a reflexive relation on $A$} && \text{[2, definition of equivalence relation]} \\ 4 &: x \, R \, x && \text{[1, 3, definition of reflexivity]} \\ 5 &: x \in \{y \in A : y \, R \, x \} && \text{[1, 4, definition of set membership]} \\ 6 &: x \in [x]_R && \text{[5, definition of equivalence class]} \\ \end{align*}
The only constructive fact that we use in this proof is that $R$ is reflexive. You do not need the other two properties of an equivalence relation to get this conclusion. Thus, it is fair for the writers to write "because $R$ is reflexive, $x \in [x]_R$". By definition, "$R$ is reflexive" is equivalent to "$x \, R \, x$", so from "$x \, R \, x$", the proof above allows us to conclude "$x \in [x]_R$", which you have also quoted...