Relations (discrete math)

Question

I am stuck with this problem, any help will be appreciated.

Let $S$ be the set of all the planes in 3D, and $\alpha$ the relation on $S$ defined so that for $\Sigma_1, \Sigma_2 \in S$
$\Sigma_1 \alpha \Sigma_2$ if and only if there is a line $p$ in the space such that $p \perp \Sigma_1, p \perp \Sigma_2$.

Prove that $\alpha$ is an equivalence relation.

Answer 1

We can say that α = {(Σ1,Σ2)| Σ1 || Σ2} and S consists every Σ in 3D. The definition for equivalence relation says that α will be equivalence if it is transitive, reflective, and symmetric. So we can prove that these three are true.

1. Transitive (∀Σ1, Σ2, Σ3 ∈ S)(Σ1, Σ2)∈α ^ (Σ2, Σ3)∈α -> (Σ1, Σ3)∈α In this case we have to prove that if Σ1||Σ2 and Σ2||Σ3 then Σ1||Σ3. Two planes are parallel if there is a line p such that p⊥Σ1, p⊥Σ2. So, if Σ1||Σ2 it means there is a line p, p⊥Σ1 and p⊥Σ2 and because Σ2|| Σ3 it proves that Σ1||Σ3. So, α is transitive.

2. Reflective (∀Σ ∈ S)(Σ, Σ)∈α It is true because every plane is parallel with itself.

3. Symmetric (∀Σ1, Σ2 ∈ S)(Σ1, Σ2)∈α -> (Σ2, Σ1)∈α Which is also true because if Σ1||Σ2 it also means that Σ2||Σ1

Therefore, it is proven that α is an equivalence relation.