Proof. Assume that f is onto and Ke

Proof. Assume that f is onto and Ker f = {0}. Let A be a non-empty subset
of L. We have always f(A∗) ⊆ {f(A)}∗. Let x ∈ {f(A)}∗ ⊆ L′. Since f is
onto, there exists y ∈ L such that f(y) = x. By knowing that f(y) ∈ {f(A)}∗.
Then f(y) ∧ m = 0 for all m ∈ f(A). Let a ∈ A. Then f(y) ∧ f(a) = 0′. That is
f(y ∧ a) = 0′ which means y ∧ a ∈ Ker f = {0}. Then y ∧ a = 0. Hence y ∈ A∗.
Therefore {f(A)}∗ ⊆ f(A∗). Thus {f(A)}∗ = f(A∗). Therefore f is annihilator
preserving.

ผลลัพธ์ (อังกฤษ) 2:

Proof. Assume that f is onto and Ker f = {0}. Let A be a non-Empty Subset
of L. We always have F (A *) ⊆ {F (A)} *. Let x ∈ {f (A)} * ⊆ L '. Since F is
onto, there exists Y ∈ L such that F (Y) = x. By knowing that F (Y) ∈ {F (A)} *.
Then F (Y) = 0 for all M ∈ M ∧ F (A). Let a ∈ A. Then f (y) ∧ f (a) = 0 '. That is
F (a ∧ Y) = 0 'which means a ∧ Y = {0} ∈ Ker F. Then Y = 0. Hence a ∧ Y ∈ A *.
Therefore {F (A) ⊆ F} * (A *). Thus {f (A)} * = f (A *). Therefore F is Annihilator
preserving.