**Proof of: B ∩ (∪**_{i∈I} A_{i}) ⊆ x∈ ∪_{i∈I} (B ∩ A_{i})

**Let x be an arbitrary element, suppose that x∈B∩(∪**_{i∈I}A_{i}). So that x∈B and x∈ (∪_{i∈I} A_{i}), therefore, we can choose j∈I such that x∈A_{j}. Since x∈B and x∈ A_{j}, x∈B∩ A_{j}, so that x∈ ∪_{i∈I} (B ∩ A_{i}). Therfore, if x∈B ∩ (∪_{i∈I} A_{i}), then x∈ ∪_{i∈I} (B ∩ A_{i}). And since the element x was arbitrary, B ∩ (∪_{i∈I} A_{i}) ⊆ x∈ ∪_{i∈I} (B ∩ A_{i}).