--- Título: s ⊆ f⁻¹[f[s]] Autor: José A. Alonso --- [mathjax] Demostrar que si \\(s\\) es un subconjunto del dominio de la función \\(f\\), entonces \\(s\\) está contenido en la imagen inversa de la imagen de \\(s\\) por \\(f\\); es decir, \\[ s ⊆ f⁻¹[f[s]] \\] Para ello, completar la siguiente teoría de Lean4:
import Mathlib.Data.Set.Function
open Set
variable {α β : Type _}
variable (f : α → β)
variable (s : Set α)
example : s ⊆ f ⁻¹' (f '' s) :=
by sorry
import Mathlib.Data.Set.Function
open Set
variable {α β : Type _}
variable (f : α → β)
variable (s : Set α)
-- 1ª demostración
-- ===============
example : s ⊆ f ⁻¹' (f '' s) :=
by
intros x xs
-- x : α
-- xs : x ∈ s
-- ⊢ x ∈ f ⁻¹' (f '' s)
have h1 : f x ∈ f '' s := mem_image_of_mem f xs
show x ∈ f ⁻¹' (f '' s)
exact mem_preimage.mp h1
-- 2ª demostración
-- ===============
example : s ⊆ f ⁻¹' (f '' s) :=
by
intros x xs
-- x : α
-- xs : x ∈ s
-- ⊢ x ∈ f ⁻¹' (f '' s)
apply mem_preimage.mpr
-- ⊢ f x ∈ f '' s
apply mem_image_of_mem
-- ⊢ x ∈ s
exact xs
-- 3ª demostración
-- ===============
example : s ⊆ f ⁻¹' (f '' s) :=
by
intros x xs
-- x : α
-- xs : x ∈ s
-- ⊢ x ∈ f ⁻¹' (f '' s)
apply mem_image_of_mem
-- ⊢ x ∈ s
exact xs
-- 4ª demostración
-- ===============
example : s ⊆ f ⁻¹' (f '' s) :=
fun _ ↦ mem_image_of_mem f
-- 5ª demostración
-- ===============
example : s ⊆ f ⁻¹' (f '' s) :=
by
intros x xs
-- x : α
-- xs : x ∈ s
-- ⊢ x ∈ f ⁻¹' (f '' s)
show f x ∈ f '' s
use x, xs
-- 6ª demostración
-- ===============
example : s ⊆ f ⁻¹' (f '' s) :=
by
intros x xs
-- x : α
-- xs : x ∈ s
-- ⊢ x ∈ f ⁻¹' (f '' s)
use x, xs
-- 7ª demostración
-- ===============
example : s ⊆ f ⁻¹' (f '' s) :=
subset_preimage_image f s
-- Lemas usados
-- ============
-- variable (x : α)
-- variable (t : Set β)
-- #check (mem_preimage : x ∈ f ⁻¹' t ↔ f x ∈ t)
-- #check (mem_image_of_mem f : x ∈ s → f x ∈ f '' s)
-- #check (subset_preimage_image f s : s ⊆ f ⁻¹' (f '' s))
Se puede interactuar con las demostraciones anteriores en Lean 4 Web.
theory Imagen_inversa_de_la_imagen
imports Main
begin
(* 1ª demostración *)
lemma "s ⊆ f -` (f ` s)"
proof (rule subsetI)
fix x
assume "x ∈ s"
then have "f x ∈ f ` s"
by (simp only: imageI)
then show "x ∈ f -` (f ` s)"
by (simp only: vimageI)
qed
(* 2ª demostración *)
lemma "s ⊆ f -` (f ` s)"
proof
fix x
assume "x ∈ s"
then have "f x ∈ f ` s" by simp
then show "x ∈ f -` (f ` s)" by simp
qed
(* 3ª demostración *)
lemma "s ⊆ f -` (f ` s)"
by auto
end