Suppose a function $f:A→B$ is given. Define a relation ~ on $A$ as follows:
$a_1$~$a_2$ ⟺ $f(a_1)$ = $f(a_2)$
Since ~ is an equivalence relation, it induces a partition of $A$ into equivalence classes. Describe these equivalence classes in each of the following cases. ($R$ is the set of real numbers).
(a) $A = B = R, f(x)=x^2$
(b) $A = B = R, f(x)=|x|$
(c) $A = R×R , B = R, f(x,y)=x+y$
my approach:
(a): the equivalence class contains no elements because the square of different numbers is different.
(b) : the equivalence class contains the set of real numbers.
(c): don't know about c
Is my approach anywhere near to correct? I am not so sure about the answers. please help!
Best Answer
You are starting your approaches with "the equivalence class contains …" as if there was only one equivalence class in each example. This is wrong. For each $x\in A$ there is an equivalence class $[x]=\{y\in A\mid x\sim y\}$ and this is never empty as it always contains $x$ itself.
In the first example consider $x=2$. We have $x\sim y$ if and only if $x^2=y^2$ so $y^2=4$, which has the solutions $-2$ and $2$. This means that $[2]=\{-2,2\}=[-2]$. The same argument shows that $[x]=\{-x,x\}$ for all $x\in\mathbb R$ with $x\neq 0$. Finally, since $y^2=0^2$ has the unique solution $y=0$, we have $[0]=\{0\}$ as the only equivalence class with just one element, while all other classes have two elements.
With this input I'll leave it to you to try the other examples again.