First suppose that 𝑝 and 𝑞 are equivalent, i.e. there exist 𝑏 ≥𝑎 >0 such that 𝑎𝑞(𝑥) ≤𝑝(𝑥) ≤𝑏𝑞(𝑥) for all 𝑥 ∈𝑋. Now suppose 𝑈 ⊆𝑋 is open under 𝑝. Then for every 𝑥 ∈𝑈 there exists some 𝛿 >0 such that 𝑝(𝑦 −𝑥) <𝛿 ⟹ 𝑦 ∈𝑈. But 𝑝(𝑦 −𝑥) ≤𝑏𝑞(𝑦 −𝑥), and thus for every 𝑥 ∈𝑈 there exists 𝛿𝑏 >0 such that 𝑞(𝑦 −𝑥) <𝛿𝑏 ⟹ 𝑝(𝑦 −𝑥) <𝛿 ⟹ 𝑦 ∈𝑈. Hence 𝑈 is open under 𝑞. Therefore T𝑝 ⊆T𝑞 Since equivalence of norms is symmetric, by the same argument T𝑞 ⊆T𝑝, and thus T𝑝 =T𝑞.