Equivalence classes notation meaning

Question

What does it mean for some element $x$ to be an element of the equivalence class $[y]$.

If I have some set $A$, and an equivalence relation, $R$, and the equivalence class is not explicitly defined, I know $A \times A$ gives me the subset $R$, does that mean for every $s \in A$, $[y] = \{ s \in A \mid (s,y)\}.$

So, isn't it already implied that $x$ is an element of $[y]$ because within $A \times A$ we have $(x,y)$.

Answer 1

$[y] := \{s \in A \mid (s,y) \in R\}$. If $A = \{1,2\}$, one possible equivalence relation is $R = \{(1,1), (2,2)\}$. Then $1 \not \in [2]$, since $(1,2) \not \in R$.

Answer 2

Two elements are in the same equivalence class if they're in the same partition given by the fundamental theorem of equivalence relations. We can also say that two things in the same equivalence class are equal under the equivilence relation, so in your example this would mean $x \equiv y$ because $x \in [y]$.

Answer 3

$R \subseteq A\times A$.

$a R b$ is defined as $(a,b) \in R$.

And $[b] = \{x \in A| xRb\} = \{x \in A| (x,b) \in R\}$.

It is absolutely true that for any $x\in A$ we have $(x, b) \in A\times A$ but that in NO way, implies $(x,b)\in R$.

.....

Consider $A = \mathbb R$ and we say $a R b$ if $a-b \in \mathbb Z$.

That means $R = \{(x,y) \in \mathbb R^2| y-x \in \mathbb Z\} = \{x,y)\in \mathbb R^2| y = x + k$ for some $k \in \mathbb Z\}$.

Let $b = 2.5689$ What is $[b]=[2.5689]$? Well. if $a R b$ then $a - 2.5689= k$ for some integer. Now $b= 2 + 0.5689$ so $a - (2+0.5689) = k$ so $a = (k+2) + 0.5689$ and $k+2$ is some integer so $a$ is ... $0.5689$ plus some integer.

In other words:

$[2.5689] = \{x \in \mathbb R| x R 2.5689\} = \{x\in \mathbb R| (x,2.5689) \in R\}=$

$\{x \in \mathbb R| x - 2.5689\in \mathbb Z\} =$

$\{x \in \mathbb R| x= (k+2) + 0.5689$ where $k+2$ is some integer$\}=$

$\{m + 0.5689| m = k+2=$ some integer $\}$.

$= \{....., -2.4311, -1.4311, -0.4311, 0.5689, 1.5689, 2.5689, 3.5689,.....\}$.

......

You will note that $\pi \ne k + 0.5689$ for any integer $k$ and that $k - 2.5689\in \mathbb Z$ so $\pi \not R b$ and $(\pi, 2.5689) \not \in R$ and $\pi \not \in [2.5689]$.

But you claimed that $(\pi, 2.5689) \in A\times A$. So how can $(\pi, 2.5689) \in A\times A$ but $(\pi, 2.5689) \not \in R$.

Well, $A\times A \not \subset R$.

It's true that $R \subset A \times A$. But it need not be true that $A\times A \subset R$.