There's a set D in R¹²
{x1, x2, x3, ... , x12 | (x1)²+(x2)²+(x3)²+...+(x12)²⩽1}
A relation is defined on R¹² such that a and b both belong to D or a and b both do not belong to D at the same time.
Prove that this is an equivalence relation.
How i proved it which i feel is incorrect is by showing that if a and b both belong to D then using the same language i proved reflexivity symmetry and transitivity. For example i proved transitivity by saying if a ∈ D and b ∈ D, and b ∈ D and c ∈ D then a ∈ D and c ∈ D
Is there any other way of proving this? Is this correct?
What $D$ is actually irrelevant. It happens that $D$ is a unit sphere, but it doesn't matter $D$ could be any set at all.
Now you claimed that the relation was defined on $D$. I assume you are mistaken. If not then this relationship is that $a$ is related to $b$ no matter what $a$ and $b$ are the only elements we are considering are points in $D$. .... so this would be a relationship where $a$ relates to $a$ and $a$ relates to $b$ means $b$ relates to $a$ and $a$ relates to $b$ and $b$ relates to $c$ means $a$ relates to $c$ because EVERYTHING relates to EVERYTHING.
I think you mean the is a relation on $\mathbb R^{12}$ where $a R b$ means either $a,b\in D$ or $a,b \not \in D$.
This is pretty trivial. The universe is devided into two equivalence classes: $D$ and $D^c$ where every element in $D$ is related to every element in $D$ and to no elements in $D^c$ and every element of $D^c$ is related to every element of $D^c$ and to no element of $D$.
To prove this is an equivalence relationship:
Reflexive:
For any $x \in R^{12}$ either $x \in D$ or $x \not \in D$. If $x \in D$ then $x \in D$ and $x R x$, and if $x \not \in D$ then $x \not \in D$ and $x\not \in D$ so $x R x$.
Symmetric:
If $a R b $ then either 1) $a \in D$ and $b \in D$ or 2) $a \not \in D$ and $b \not \in D$. If 1) then $b\in D$ and $a \in D$ so $b R a$. And if 2) then $b\not \in D$ and $a \not \in D$ so $b R a$.
Transitive:
If $a R b$ and $b R c$ then either 1) $a\in D$ and $b \in D$ and as $b R c$, $c\in D$. Or 2) $a \not \in D $ and $b \not \in D$ and as $b R c$ then $c \not \in D$. In 1) we have $a\in D$ and $c\in D$ so $a R c$ and in 2) we have $a\not \in D$ and $c\not \in D$ so $a R c$.