Theory Filter

theory Filter
imports Infinite_Set
(*  Title:      HOL/NSA/Filter.thy
    Author:     Jacques D. Fleuriot, University of Cambridge
    Author:     Lawrence C Paulson
    Author:     Brian Huffman

header {* Filters and Ultrafilters *}

theory Filter
imports "~~/src/HOL/Library/Infinite_Set"

subsection {* Definitions and basic properties *}

subsubsection {* Filters *}

locale filter =
  fixes F :: "'a set set"
  assumes UNIV [iff]:  "UNIV ∈ F"
  assumes empty [iff]: "{} ∉ F"
  assumes Int:         "[|u ∈ F; v ∈ F|] ==> u ∩ v ∈ F"
  assumes subset:      "[|u ∈ F; u ⊆ v|] ==> v ∈ F"

lemma memD: "A ∈ F ==> - A ∉ F"
  assume "A ∈ F" and "- A ∈ F"
  hence "A ∩ (- A) ∈ F" by (rule Int)
  thus "False" by simp

lemma not_memI: "- A ∈ F ==> A ∉ F"
by (drule memD, simp)

lemma Int_iff: "(x ∩ y ∈ F) = (x ∈ F ∧ y ∈ F)"
by (auto elim: subset intro: Int)


subsubsection {* Ultrafilters *}

locale ultrafilter = filter +
  assumes ultra: "A ∈ F ∨ - A ∈ F"

lemma memI: "- A ∉ F ==> A ∈ F"
using ultra [of A] by simp

lemma not_memD: "A ∉ F ==> - A ∈ F"
by (rule memI, simp)

lemma not_mem_iff: "(A ∉ F) = (- A ∈ F)"
by (rule iffI [OF not_memD not_memI])

lemma Compl_iff: "(- A ∈ F) = (A ∉ F)"
by (rule iffI [OF not_memI not_memD])

lemma Un_iff: "(x ∪ y ∈ F) = (x ∈ F ∨ y ∈ F)"
 apply (rule iffI)
  apply (erule contrapos_pp)
  apply (simp add: Int_iff not_mem_iff)
 apply (auto elim: subset)


subsubsection {* Free Ultrafilters *}

locale freeultrafilter = ultrafilter +
  assumes infinite: "A ∈ F ==> infinite A"

lemma finite: "finite A ==> A ∉ F"
by (erule contrapos_pn, erule infinite)

lemma singleton: "{x} ∉ F"
by (rule finite, simp)

lemma insert_iff [simp]: "(insert x A ∈ F) = (A ∈ F)"
apply (subst insert_is_Un)
apply (subst Un_iff)
apply (simp add: singleton)

lemma filter: "filter F" ..

lemma ultrafilter: "ultrafilter F" ..


subsection {* Collect properties *}

lemma (in filter) Collect_ex:
  "({n. ∃x. P n x} ∈ F) = (∃X. {n. P n (X n)} ∈ F)"
  assume "{n. ∃x. P n x} ∈ F"
  hence "{n. P n (SOME x. P n x)} ∈ F"
    by (auto elim: someI subset)
  thus "∃X. {n. P n (X n)} ∈ F" by fast
  show "∃X. {n. P n (X n)} ∈ F ==> {n. ∃x. P n x} ∈ F"
    by (auto elim: subset)

lemma (in filter) Collect_conj:
  "({n. P n ∧ Q n} ∈ F) = ({n. P n} ∈ F ∧ {n. Q n} ∈ F)"
by (subst Collect_conj_eq, rule Int_iff)

lemma (in ultrafilter) Collect_not:
  "({n. ¬ P n} ∈ F) = ({n. P n} ∉ F)"
by (subst Collect_neg_eq, rule Compl_iff)

lemma (in ultrafilter) Collect_disj:
  "({n. P n ∨ Q n} ∈ F) = ({n. P n} ∈ F ∨ {n. Q n} ∈ F)"
by (subst Collect_disj_eq, rule Un_iff)

lemma (in ultrafilter) Collect_all:
  "({n. ∀x. P n x} ∈ F) = (∀X. {n. P n (X n)} ∈ F)"
apply (rule Not_eq_iff [THEN iffD1])
apply (simp add: Collect_not [symmetric])
apply (rule Collect_ex)

subsection {* Maximal filter = Ultrafilter *}

text {*
   A filter F is an ultrafilter iff it is a maximal filter,
   i.e. whenever G is a filter and @{term "F ⊆ G"} then @{term "F = G"}
text {*
  Lemmas that shows existence of an extension to what was assumed to
  be a maximal filter. Will be used to derive contradiction in proof of
  property of ultrafilter.

lemma extend_lemma1: "UNIV ∈ F ==> A ∈ {X. ∃f∈F. A ∩ f ⊆ X}"
by blast

lemma extend_lemma2: "F ⊆ {X. ∃f∈F. A ∩ f ⊆ X}"
by blast

lemma (in filter) extend_filter:
assumes A: "- A ∉ F"
shows "filter {X. ∃f∈F. A ∩ f ⊆ X}" (is "filter ?X")
proof (rule filter.intro)
  show "UNIV ∈ ?X" by blast
  show "{} ∉ ?X"
  proof (clarify)
    fix f assume f: "f ∈ F" and Af: "A ∩ f ⊆ {}"
    from Af have fA: "f ⊆ - A" by blast
    from f fA have "- A ∈ F" by (rule subset)
    with A show "False" by simp
  fix u and v
  assume u: "u ∈ ?X" and v: "v ∈ ?X"
  from u obtain f where f: "f ∈ F" and Af: "A ∩ f ⊆ u" by blast
  from v obtain g where g: "g ∈ F" and Ag: "A ∩ g ⊆ v" by blast
  from f g have fg: "f ∩ g ∈ F" by (rule Int)
  from Af Ag have Afg: "A ∩ (f ∩ g) ⊆ u ∩ v" by blast
  from fg Afg show "u ∩ v ∈ ?X" by blast
  fix u and v
  assume uv: "u ⊆ v" and u: "u ∈ ?X"
  from u obtain f where f: "f ∈ F" and Afu: "A ∩ f ⊆ u" by blast
  from Afu uv have Afv: "A ∩ f ⊆ v" by blast
  from f Afv have "∃f∈F. A ∩ f ⊆ v" by blast
  thus "v ∈ ?X" by simp

lemma (in filter) max_filter_ultrafilter:
assumes max: "!!G. [|filter G; F ⊆ G|] ==> F = G"
shows "ultrafilter_axioms F"
proof (rule ultrafilter_axioms.intro)
  fix A show "A ∈ F ∨ - A ∈ F"
  proof (rule disjCI)
    let ?X = "{X. ∃f∈F. A ∩ f ⊆ X}"
    assume AF: "- A ∉ F"
    from AF have X: "filter ?X" by (rule extend_filter)
    from UNIV have AX: "A ∈ ?X" by (rule extend_lemma1)
    have FX: "F ⊆ ?X" by (rule extend_lemma2)
    from X FX have "F = ?X" by (rule max)
    with AX show "A ∈ F" by simp

lemma (in ultrafilter) max_filter:
assumes G: "filter G" and sub: "F ⊆ G" shows "F = G"
  show "F ⊆ G" using sub .
  show "G ⊆ F"
    fix A assume A: "A ∈ G"
    from G A have "- A ∉ G" by (rule filter.memD)
    with sub have B: "- A ∉ F" by blast
    thus "A ∈ F" by (rule memI)

subsection {* Ultrafilter Theorem *}

text "A local context makes proof of ultrafilter Theorem more modular"
  fixes   frechet :: "'a set set"
  and     superfrechet :: "'a set set set"

  assumes infinite_UNIV: "infinite (UNIV :: 'a set)"

  defines frechet_def: "frechet ≡ {A. finite (- A)}"
  and     superfrechet_def: "superfrechet ≡ {G. filter G ∧ frechet ⊆ G}"

lemma superfrechetI:
  "[|filter G; frechet ⊆ G|] ==> G ∈ superfrechet"
by (simp add: superfrechet_def)

lemma superfrechetD1:
  "G ∈ superfrechet ==> filter G"
by (simp add: superfrechet_def)

lemma superfrechetD2:
  "G ∈ superfrechet ==> frechet ⊆ G"
by (simp add: superfrechet_def)

text {* A few properties of free filters *}

lemma filter_cofinite:
assumes inf: "infinite (UNIV :: 'a set)"
shows "filter {A:: 'a set. finite (- A)}" (is "filter ?F")
proof (rule filter.intro)
  show "UNIV ∈ ?F" by simp
  show "{} ∉ ?F" using inf by simp
  fix u v assume "u ∈ ?F" and "v ∈ ?F"
  thus "u ∩ v ∈ ?F" by simp
  fix u v assume uv: "u ⊆ v" and u: "u ∈ ?F"
  from uv have vu: "- v ⊆ - u" by simp
  from u show "v ∈ ?F"
    by (simp add: finite_subset [OF vu])

text {*
   We prove: 1. Existence of maximal filter i.e. ultrafilter;
             2. Freeness property i.e ultrafilter is free.
             Use a locale to prove various lemmas and then 
             export main result: The ultrafilter Theorem

lemma filter_frechet: "filter frechet"
by (unfold frechet_def, rule filter_cofinite [OF infinite_UNIV])

lemma frechet_in_superfrechet: "frechet ∈ superfrechet"
by (rule superfrechetI [OF filter_frechet subset_refl])

lemma lemma_mem_chain_filter:
  "[|c ∈ chains superfrechet; x ∈ c|] ==> filter x"
by (unfold chains_def superfrechet_def, blast)

subsubsection {* Unions of chains of superfrechets *}

text "In this section we prove that superfrechet is closed
with respect to unions of non-empty chains. We must show
  1) Union of a chain is a filter,
  2) Union of a chain contains frechet.

Number 2 is trivial, but 1 requires us to prove all the filter rules."

lemma Union_chain_UNIV:
  "[|c ∈ chains superfrechet; c ≠ {}|] ==> UNIV ∈ \<Union>c"
proof -
  assume 1: "c ∈ chains superfrechet" and 2: "c ≠ {}"
  from 2 obtain x where 3: "x ∈ c" by blast
  from 1 3 have "filter x" by (rule lemma_mem_chain_filter)
  hence "UNIV ∈ x" by (rule filter.UNIV)
  with 3 show "UNIV ∈ \<Union>c" by blast

lemma Union_chain_empty:
  "c ∈ chains superfrechet ==> {} ∉ \<Union>c"
  assume 1: "c ∈ chains superfrechet" and 2: "{} ∈ \<Union>c"
  from 2 obtain x where 3: "x ∈ c" and 4: "{} ∈ x" ..
  from 1 3 have "filter x" by (rule lemma_mem_chain_filter)
  hence "{} ∉ x" by (rule filter.empty)
  with 4 show "False" by simp

lemma Union_chain_Int:
  "[|c ∈ chains superfrechet; u ∈ \<Union>c; v ∈ \<Union>c|] ==> u ∩ v ∈ \<Union>c"
proof -
  assume c: "c ∈ chains superfrechet"
  assume "u ∈ \<Union>c"
    then obtain x where ux: "u ∈ x" and xc: "x ∈ c" ..
  assume "v ∈ \<Union>c"
    then obtain y where vy: "v ∈ y" and yc: "y ∈ c" ..
  from c xc yc have "x ⊆ y ∨ y ⊆ x" using c unfolding chains_def chain_subset_def by auto
  with xc yc have xyc: "x ∪ y ∈ c"
    by (auto simp add: Un_absorb1 Un_absorb2)
  with c have fxy: "filter (x ∪ y)" by (rule lemma_mem_chain_filter)
  from ux have uxy: "u ∈ x ∪ y" by simp
  from vy have vxy: "v ∈ x ∪ y" by simp
  from fxy uxy vxy have "u ∩ v ∈ x ∪ y" by (rule filter.Int)
  with xyc show "u ∩ v ∈ \<Union>c" ..

lemma Union_chain_subset:
  "[|c ∈ chains superfrechet; u ∈ \<Union>c; u ⊆ v|] ==> v ∈ \<Union>c"
proof -
  assume c: "c ∈ chains superfrechet"
     and u: "u ∈ \<Union>c" and uv: "u ⊆ v"
  from u obtain x where ux: "u ∈ x" and xc: "x ∈ c" ..
  from c xc have fx: "filter x" by (rule lemma_mem_chain_filter)
  from fx ux uv have vx: "v ∈ x" by (rule filter.subset)
  with xc show "v ∈ \<Union>c" ..

lemma Union_chain_filter:
assumes chain: "c ∈ chains superfrechet" and nonempty: "c ≠ {}"
shows "filter (\<Union>c)" 
proof (rule filter.intro)
  show "UNIV ∈ \<Union>c" using chain nonempty by (rule Union_chain_UNIV)
  show "{} ∉ \<Union>c" using chain by (rule Union_chain_empty)
  fix u v assume "u ∈ \<Union>c" and "v ∈ \<Union>c"
  with chain show "u ∩ v ∈ \<Union>c" by (rule Union_chain_Int)
  fix u v assume "u ∈ \<Union>c" and "u ⊆ v"
  with chain show "v ∈ \<Union>c" by (rule Union_chain_subset)

lemma lemma_mem_chain_frechet_subset:
  "[|c ∈ chains superfrechet; x ∈ c|] ==> frechet ⊆ x"
by (unfold superfrechet_def chains_def, blast)

lemma Union_chain_superfrechet:
  "[|c ≠ {}; c ∈ chains superfrechet|] ==> \<Union>c ∈ superfrechet"
proof (rule superfrechetI)
  assume 1: "c ∈ chains superfrechet" and 2: "c ≠ {}"
  thus "filter (\<Union>c)" by (rule Union_chain_filter)
  from 2 obtain x where 3: "x ∈ c" by blast
  from 1 3 have "frechet ⊆ x" by (rule lemma_mem_chain_frechet_subset)
  also from 3 have "x ⊆ \<Union>c" by blast
  finally show "frechet ⊆ \<Union>c" .

subsubsection {* Existence of free ultrafilter *}

lemma max_cofinite_filter_Ex:
  "∃U∈superfrechet. ∀G∈superfrechet. U ⊆ G --> G = U" 
proof (rule Zorn_Lemma2, safe)
  fix c assume c: "c ∈ chains superfrechet"
  show "∃U∈superfrechet. ∀G∈c. G ⊆ U" (is "?U")
  proof (cases)
    assume "c = {}"
    with frechet_in_superfrechet show "?U" by blast
    assume A: "c ≠ {}"
    from A c have "\<Union>c ∈ superfrechet"
      by (rule Union_chain_superfrechet)
    thus "?U" by blast

lemma mem_superfrechet_all_infinite:
  "[|U ∈ superfrechet; A ∈ U|] ==> infinite A"
  assume U: "U ∈ superfrechet" and A: "A ∈ U" and fin: "finite A"
  from U have fil: "filter U" and fre: "frechet ⊆ U"
    by (simp_all add: superfrechet_def)
  from fin have "- A ∈ frechet" by (simp add: frechet_def)
  with fre have cA: "- A ∈ U" by (rule subsetD)
  from fil A cA have "A ∩ - A ∈ U" by (rule filter.Int)
  with fil show "False" by (simp add: filter.empty)

text {* There exists a free ultrafilter on any infinite set *}

lemma freeultrafilter_Ex:
  "∃U::'a set set. freeultrafilter U"
proof -
  from max_cofinite_filter_Ex obtain U
    where U: "U ∈ superfrechet"
      and max [rule_format]: "∀G∈superfrechet. U ⊆ G --> G = U" ..
  from U have fil: "filter U" by (rule superfrechetD1)
  from U have fre: "frechet ⊆ U" by (rule superfrechetD2)
  have ultra: "ultrafilter_axioms U"
  proof (rule filter.max_filter_ultrafilter [OF fil])
    fix G assume G: "filter G" and UG: "U ⊆ G"
    from fre UG have "frechet ⊆ G" by simp
    with G have "G ∈ superfrechet" by (rule superfrechetI)
    from this UG show "U = G" by (rule max[symmetric])
  have free: "freeultrafilter_axioms U"
  proof (rule freeultrafilter_axioms.intro)
    fix A assume "A ∈ U"
    with U show "infinite A" by (rule mem_superfrechet_all_infinite)
  from fil ultra free have "freeultrafilter U"
    by (rule freeultrafilter.intro [OF ultrafilter.intro])
    (* FIXME: unfold_locales should use chained facts *)
  then show ?thesis ..


hide_const (open) filter