The intersection $\bigcap_{i \in I}G_i$ is an order relation

Question

Let $\{G_i\}_{i \in I}$ be an indexed family of order relations in a set $A$. Show that $\bigcap_{i \in I}G_i$ is an order relation in $A$.

My attempt:

Since $G_i$ is reflexive for all $i \in I$,

$\forall x \in A, (x,x) \in G_i \forall i \in I$

$\implies \forall x \in A, (x,x) \in \bigcap_{i \in I}G_i$

Is this the right way of proving the reflexive part? And could anyone help me with the antisymmetric and transitive part. Any help would be appreciated.

Answer 1

Let $G = \displaystyle \bigcap_{i \in I} G_i$.

Reflexivity

$$\begin{array}{rcl} \forall x \in A: && \forall i \in I: (x,x) \in G_i \\ &\implies& (x,x) \in G \end{array}$$

Antisymmetry

$$\begin{array}{rcl} \forall x,y \in A: && (x,y) \in G \land (y,x) \in G \\ &\implies& \forall i \in I: (x,y) \in G_i \land (y,x) \in G_i \\ &\implies& x = y \end{array}$$

Transitivity

$$\begin{array}{rcl} \forall x,y,z \in A: && (x,y) \in G \land (y,z) \in G \\ &\implies& \forall i \in I: (x,y) \in G_i \land (y,z) \in G_i \\ &\implies& \forall i \in I: (x,z) \in G_i \\ &\implies& (x,z) \in G \end{array}$$

Answer 2

For each i : I we have an order relation Gᵢ, and their intersection G ≔ ∩ᵢ Gᵢ is also an order relation.

Indeed, it is reflexive: For any element x we have,

   x G x
 =⟨ definition of G and intersection membership ⟩
   ∀ i : I  • x Gᵢ x       -- the bullet is a syntactic separator only
 =⟨ each Gᵢ is reflexive ⟩
   ∀ i : I • true
 =⟨ quantifiers ⟩
   true

Likewise, it is antisymmetric: For any element x and element y,

   x G y  ∧  y G x
 =⟨ definition of G, intersection membership, quantifiers ⟩
   ∀ i,j : I •  x Gᵢ y ∧ y Gⱼ x
 ⇒⟨ instantiating j ≔ i ⟩
   ∀ i : I •  x Gᵢ y ∧ y Gᵢ x
 =⟨ each Gᵢ is antisymmetric ⟩
   ∀ i : I • x ≈ y
 =⟨ quantifiers, assuming I non-empty ⟩
   x ≈ y

Transitivity is similar :-)