--- a/src/HOL/Real/ContNotDenum.thy Thu Jun 21 15:42:14 2007 +0200
+++ b/src/HOL/Real/ContNotDenum.thy Thu Jun 21 15:42:15 2007 +0200
@@ -1,579 +1,579 @@
-(* Title : HOL/Real/ContNonDenum
- ID : $Id$
- Author : Benjamin Porter, Monash University, NICTA, 2005
-*)
-
-header {* Non-denumerability of the Continuum. *}
-
-theory ContNotDenum
-imports RComplete
-begin
-
-subsection {* Abstract *}
-
-text {* The following document presents a proof that the Continuum is
-uncountable. It is formalised in the Isabelle/Isar theorem proving
-system.
-
-{\em Theorem:} The Continuum @{text "\<real>"} is not denumerable. In other
-words, there does not exist a function f:@{text "\<nat>\<Rightarrow>\<real>"} such that f is
-surjective.
-
-{\em Outline:} An elegant informal proof of this result uses Cantor's
-Diagonalisation argument. The proof presented here is not this
-one. First we formalise some properties of closed intervals, then we
-prove the Nested Interval Property. This property relies on the
-completeness of the Real numbers and is the foundation for our
-argument. Informally it states that an intersection of countable
-closed intervals (where each successive interval is a subset of the
-last) is non-empty. We then assume a surjective function f:@{text
-"\<nat>\<Rightarrow>\<real>"} exists and find a real x such that x is not in the range of f
-by generating a sequence of closed intervals then using the NIP. *}
-
-subsection {* Closed Intervals *}
-
-text {* This section formalises some properties of closed intervals. *}
-
-subsubsection {* Definition *}
-
-definition
- closed_int :: "real \<Rightarrow> real \<Rightarrow> real set" where
- "closed_int x y = {z. x \<le> z \<and> z \<le> y}"
-
-subsubsection {* Properties *}
-
-lemma closed_int_subset:
- assumes xy: "x1 \<ge> x0" "y1 \<le> y0"
- shows "closed_int x1 y1 \<subseteq> closed_int x0 y0"
-proof -
- {
- fix x::real
- assume "x \<in> closed_int x1 y1"
- hence "x \<ge> x1 \<and> x \<le> y1" by (simp add: closed_int_def)
- with xy have "x \<ge> x0 \<and> x \<le> y0" by auto
- hence "x \<in> closed_int x0 y0" by (simp add: closed_int_def)
- }
- thus ?thesis by auto
-qed
-
-lemma closed_int_least:
- assumes a: "a \<le> b"
- shows "a \<in> closed_int a b \<and> (\<forall>x \<in> closed_int a b. a \<le> x)"
-proof
- from a have "a\<in>{x. a\<le>x \<and> x\<le>b}" by simp
- thus "a \<in> closed_int a b" by (unfold closed_int_def)
-next
- have "\<forall>x\<in>{x. a\<le>x \<and> x\<le>b}. a\<le>x" by simp
- thus "\<forall>x \<in> closed_int a b. a \<le> x" by (unfold closed_int_def)
-qed
-
-lemma closed_int_most:
- assumes a: "a \<le> b"
- shows "b \<in> closed_int a b \<and> (\<forall>x \<in> closed_int a b. x \<le> b)"
-proof
- from a have "b\<in>{x. a\<le>x \<and> x\<le>b}" by simp
- thus "b \<in> closed_int a b" by (unfold closed_int_def)
-next
- have "\<forall>x\<in>{x. a\<le>x \<and> x\<le>b}. x\<le>b" by simp
- thus "\<forall>x \<in> closed_int a b. x\<le>b" by (unfold closed_int_def)
-qed
-
-lemma closed_not_empty:
- shows "a \<le> b \<Longrightarrow> \<exists>x. x \<in> closed_int a b"
- by (auto dest: closed_int_least)
-
-lemma closed_mem:
- assumes "a \<le> c" and "c \<le> b"
- shows "c \<in> closed_int a b"
- using assms unfolding closed_int_def by auto
-
-lemma closed_subset:
- assumes ac: "a \<le> b" "c \<le> d"
- assumes closed: "closed_int a b \<subseteq> closed_int c d"
- shows "b \<ge> c"
-proof -
- from closed have "\<forall>x\<in>closed_int a b. x\<in>closed_int c d" by auto
- hence "\<forall>x. a\<le>x \<and> x\<le>b \<longrightarrow> c\<le>x \<and> x\<le>d" by (unfold closed_int_def, auto)
- with ac have "c\<le>b \<and> b\<le>d" by simp
- thus ?thesis by auto
-qed
-
-
-subsection {* Nested Interval Property *}
-
-theorem NIP:
- fixes f::"nat \<Rightarrow> real set"
- assumes subset: "\<forall>n. f (Suc n) \<subseteq> f n"
- and closed: "\<forall>n. \<exists>a b. f n = closed_int a b \<and> a \<le> b"
- shows "(\<Inter>n. f n) \<noteq> {}"
-proof -
- let ?g = "\<lambda>n. (SOME c. c\<in>(f n) \<and> (\<forall>x\<in>(f n). c \<le> x))"
- have ne: "\<forall>n. \<exists>x. x\<in>(f n)"
- proof
- fix n
- from closed have "\<exists>a b. f n = closed_int a b \<and> a \<le> b" by simp
- then obtain a and b where fn: "f n = closed_int a b \<and> a \<le> b" by auto
- hence "a \<le> b" ..
- with closed_not_empty have "\<exists>x. x\<in>closed_int a b" by simp
- with fn show "\<exists>x. x\<in>(f n)" by simp
- qed
-
- have gdef: "\<forall>n. (?g n)\<in>(f n) \<and> (\<forall>x\<in>(f n). (?g n)\<le>x)"
- proof
- fix n
- from closed have "\<exists>a b. f n = closed_int a b \<and> a \<le> b" ..
- then obtain a and b where ff: "f n = closed_int a b" and "a \<le> b" by auto
- hence "a \<le> b" by simp
- hence "a\<in>closed_int a b \<and> (\<forall>x\<in>closed_int a b. a \<le> x)" by (rule closed_int_least)
- with ff have "a\<in>(f n) \<and> (\<forall>x\<in>(f n). a \<le> x)" by simp
- hence "\<exists>c. c\<in>(f n) \<and> (\<forall>x\<in>(f n). c \<le> x)" ..
- thus "(?g n)\<in>(f n) \<and> (\<forall>x\<in>(f n). (?g n)\<le>x)" by (rule someI_ex)
- qed
-
- -- "A denotes the set of all left-most points of all the intervals ..."
- moreover obtain A where Adef: "A = ?g ` \<nat>" by simp
- ultimately have "\<exists>x. x\<in>A"
- proof -
- have "(0::nat) \<in> \<nat>" by simp
- moreover have "?g 0 = ?g 0" by simp
- ultimately have "?g 0 \<in> ?g ` \<nat>" by (rule rev_image_eqI)
- with Adef have "?g 0 \<in> A" by simp
- thus ?thesis ..
- qed
-
- -- "Now show that A is bounded above ..."
- moreover have "\<exists>y. isUb (UNIV::real set) A y"
- proof -
- {
- fix n
- from ne have ex: "\<exists>x. x\<in>(f n)" ..
- from gdef have "(?g n)\<in>(f n) \<and> (\<forall>x\<in>(f n). (?g n)\<le>x)" by simp
- moreover
- from closed have "\<exists>a b. f n = closed_int a b \<and> a \<le> b" ..
- then obtain a and b where "f n = closed_int a b \<and> a \<le> b" by auto
- hence "b\<in>(f n) \<and> (\<forall>x\<in>(f n). x \<le> b)" using closed_int_most by blast
- ultimately have "\<forall>x\<in>(f n). (?g n) \<le> b" by simp
- with ex have "(?g n) \<le> b" by auto
- hence "\<exists>b. (?g n) \<le> b" by auto
- }
- hence aux: "\<forall>n. \<exists>b. (?g n) \<le> b" ..
-
- have fs: "\<forall>n::nat. f n \<subseteq> f 0"
- proof (rule allI, induct_tac n)
- show "f 0 \<subseteq> f 0" by simp
- next
- fix n
- assume "f n \<subseteq> f 0"
- moreover from subset have "f (Suc n) \<subseteq> f n" ..
- ultimately show "f (Suc n) \<subseteq> f 0" by simp
- qed
- have "\<forall>n. (?g n)\<in>(f 0)"
- proof
- fix n
- from gdef have "(?g n)\<in>(f n) \<and> (\<forall>x\<in>(f n). (?g n)\<le>x)" by simp
- hence "?g n \<in> f n" ..
- with fs show "?g n \<in> f 0" by auto
- qed
- moreover from closed
- obtain a and b where "f 0 = closed_int a b" and alb: "a \<le> b" by blast
- ultimately have "\<forall>n. ?g n \<in> closed_int a b" by auto
- with alb have "\<forall>n. ?g n \<le> b" using closed_int_most by blast
- with Adef have "\<forall>y\<in>A. y\<le>b" by auto
- hence "A *<= b" by (unfold setle_def)
- moreover have "b \<in> (UNIV::real set)" by simp
- ultimately have "A *<= b \<and> b \<in> (UNIV::real set)" by simp
- hence "isUb (UNIV::real set) A b" by (unfold isUb_def)
- thus ?thesis by auto
- qed
- -- "by the Axiom Of Completeness, A has a least upper bound ..."
- ultimately have "\<exists>t. isLub UNIV A t" by (rule reals_complete)
-
- -- "denote this least upper bound as t ..."
- then obtain t where tdef: "isLub UNIV A t" ..
-
- -- "and finally show that this least upper bound is in all the intervals..."
- have "\<forall>n. t \<in> f n"
- proof
- fix n::nat
- from closed obtain a and b where
- int: "f n = closed_int a b" and alb: "a \<le> b" by blast
-
- have "t \<ge> a"
- proof -
- have "a \<in> A"
- proof -
- (* by construction *)
- from alb int have ain: "a\<in>f n \<and> (\<forall>x\<in>f n. a \<le> x)"
- using closed_int_least by blast
- moreover have "\<forall>e. e\<in>f n \<and> (\<forall>x\<in>f n. e \<le> x) \<longrightarrow> e = a"
- proof clarsimp
- fix e
- assume ein: "e \<in> f n" and lt: "\<forall>x\<in>f n. e \<le> x"
- from lt ain have aux: "\<forall>x\<in>f n. a \<le> x \<and> e \<le> x" by auto
-
- from ein aux have "a \<le> e \<and> e \<le> e" by auto
- moreover from ain aux have "a \<le> a \<and> e \<le> a" by auto
- ultimately show "e = a" by simp
- qed
- hence "\<And>e. e\<in>f n \<and> (\<forall>x\<in>f n. e \<le> x) \<Longrightarrow> e = a" by simp
- ultimately have "(?g n) = a" by (rule some_equality)
- moreover
- {
- have "n = of_nat n" by simp
- moreover have "of_nat n \<in> \<nat>" by simp
- ultimately have "n \<in> \<nat>"
- apply -
- apply (subst(asm) eq_sym_conv)
- apply (erule subst)
- .
- }
- with Adef have "(?g n) \<in> A" by auto
- ultimately show ?thesis by simp
- qed
- with tdef show "a \<le> t" by (rule isLubD2)
- qed
- moreover have "t \<le> b"
- proof -
- have "isUb UNIV A b"
- proof -
- {
- from alb int have
- ain: "b\<in>f n \<and> (\<forall>x\<in>f n. x \<le> b)" using closed_int_most by blast
-
- have subsetd: "\<forall>m. \<forall>n. f (n + m) \<subseteq> f n"
- proof (rule allI, induct_tac m)
- show "\<forall>n. f (n + 0) \<subseteq> f n" by simp
- next
- fix m n
- assume pp: "\<forall>p. f (p + n) \<subseteq> f p"
- {
- fix p
- from pp have "f (p + n) \<subseteq> f p" by simp
- moreover from subset have "f (Suc (p + n)) \<subseteq> f (p + n)" by auto
- hence "f (p + (Suc n)) \<subseteq> f (p + n)" by simp
- ultimately have "f (p + (Suc n)) \<subseteq> f p" by simp
- }
- thus "\<forall>p. f (p + Suc n) \<subseteq> f p" ..
- qed
- have subsetm: "\<forall>\<alpha> \<beta>. \<alpha> \<ge> \<beta> \<longrightarrow> (f \<alpha>) \<subseteq> (f \<beta>)"
- proof ((rule allI)+, rule impI)
- fix \<alpha>::nat and \<beta>::nat
- assume "\<beta> \<le> \<alpha>"
- hence "\<exists>k. \<alpha> = \<beta> + k" by (simp only: le_iff_add)
- then obtain k where "\<alpha> = \<beta> + k" ..
- moreover
- from subsetd have "f (\<beta> + k) \<subseteq> f \<beta>" by simp
- ultimately show "f \<alpha> \<subseteq> f \<beta>" by auto
- qed
-
- fix m
- {
- assume "m \<ge> n"
- with subsetm have "f m \<subseteq> f n" by simp
- with ain have "\<forall>x\<in>f m. x \<le> b" by auto
- moreover
- from gdef have "?g m \<in> f m \<and> (\<forall>x\<in>f m. ?g m \<le> x)" by simp
- ultimately have "?g m \<le> b" by auto
- }
- moreover
- {
- assume "\<not>(m \<ge> n)"
- hence "m < n" by simp
- with subsetm have sub: "(f n) \<subseteq> (f m)" by simp
- from closed obtain ma and mb where
- "f m = closed_int ma mb \<and> ma \<le> mb" by blast
- hence one: "ma \<le> mb" and fm: "f m = closed_int ma mb" by auto
- from one alb sub fm int have "ma \<le> b" using closed_subset by blast
- moreover have "(?g m) = ma"
- proof -
- from gdef have "?g m \<in> f m \<and> (\<forall>x\<in>f m. ?g m \<le> x)" ..
- moreover from one have
- "ma \<in> closed_int ma mb \<and> (\<forall>x\<in>closed_int ma mb. ma \<le> x)"
- by (rule closed_int_least)
- with fm have "ma\<in>f m \<and> (\<forall>x\<in>f m. ma \<le> x)" by simp
- ultimately have "ma \<le> ?g m \<and> ?g m \<le> ma" by auto
- thus "?g m = ma" by auto
- qed
- ultimately have "?g m \<le> b" by simp
- }
- ultimately have "?g m \<le> b" by (rule case_split)
- }
- with Adef have "\<forall>y\<in>A. y\<le>b" by auto
- hence "A *<= b" by (unfold setle_def)
- moreover have "b \<in> (UNIV::real set)" by simp
- ultimately have "A *<= b \<and> b \<in> (UNIV::real set)" by simp
- thus "isUb (UNIV::real set) A b" by (unfold isUb_def)
- qed
- with tdef show "t \<le> b" by (rule isLub_le_isUb)
- qed
- ultimately have "t \<in> closed_int a b" by (rule closed_mem)
- with int show "t \<in> f n" by simp
- qed
- hence "t \<in> (\<Inter>n. f n)" by auto
- thus ?thesis by auto
-qed
-
-subsection {* Generating the intervals *}
-
-subsubsection {* Existence of non-singleton closed intervals *}
-
-text {* This lemma asserts that given any non-singleton closed
-interval (a,b) and any element c, there exists a closed interval that
-is a subset of (a,b) and that does not contain c and is a
-non-singleton itself. *}
-
-lemma closed_subset_ex:
- fixes c::real
- assumes alb: "a < b"
- shows
- "\<exists>ka kb. ka < kb \<and> closed_int ka kb \<subseteq> closed_int a b \<and> c \<notin> (closed_int ka kb)"
-proof -
- {
- assume clb: "c < b"
- {
- assume cla: "c < a"
- from alb cla clb have "c \<notin> closed_int a b" by (unfold closed_int_def, auto)
- with alb have
- "a < b \<and> closed_int a b \<subseteq> closed_int a b \<and> c \<notin> closed_int a b"
- by auto
- hence
- "\<exists>ka kb. ka < kb \<and> closed_int ka kb \<subseteq> closed_int a b \<and> c \<notin> (closed_int ka kb)"
- by auto
- }
- moreover
- {
- assume ncla: "\<not>(c < a)"
- with clb have cdef: "a \<le> c \<and> c < b" by simp
- obtain ka where kadef: "ka = (c + b)/2" by blast
-
- from kadef clb have kalb: "ka < b" by auto
- moreover from kadef cdef have kagc: "ka > c" by simp
- ultimately have "c\<notin>(closed_int ka b)" by (unfold closed_int_def, auto)
- moreover from cdef kagc have "ka \<ge> a" by simp
- hence "closed_int ka b \<subseteq> closed_int a b" by (unfold closed_int_def, auto)
- ultimately have
- "ka < b \<and> closed_int ka b \<subseteq> closed_int a b \<and> c \<notin> closed_int ka b"
- using kalb by auto
- hence
- "\<exists>ka kb. ka < kb \<and> closed_int ka kb \<subseteq> closed_int a b \<and> c \<notin> (closed_int ka kb)"
- by auto
-
- }
- ultimately have
- "\<exists>ka kb. ka < kb \<and> closed_int ka kb \<subseteq> closed_int a b \<and> c \<notin> (closed_int ka kb)"
- by (rule case_split)
- }
- moreover
- {
- assume "\<not> (c < b)"
- hence cgeb: "c \<ge> b" by simp
-
- obtain kb where kbdef: "kb = (a + b)/2" by blast
- with alb have kblb: "kb < b" by auto
- with kbdef cgeb have "a < kb \<and> kb < c" by auto
- moreover hence "c \<notin> (closed_int a kb)" by (unfold closed_int_def, auto)
- moreover from kblb have
- "closed_int a kb \<subseteq> closed_int a b" by (unfold closed_int_def, auto)
- ultimately have
- "a < kb \<and> closed_int a kb \<subseteq> closed_int a b \<and> c\<notin>closed_int a kb"
- by simp
- hence
- "\<exists>ka kb. ka < kb \<and> closed_int ka kb \<subseteq> closed_int a b \<and> c \<notin> (closed_int ka kb)"
- by auto
- }
- ultimately show ?thesis by (rule case_split)
-qed
-
-subsection {* newInt: Interval generation *}
-
-text {* Given a function f:@{text "\<nat>\<Rightarrow>\<real>"}, newInt (Suc n) f returns a
-closed interval such that @{text "newInt (Suc n) f \<subseteq> newInt n f"} and
-does not contain @{text "f (Suc n)"}. With the base case defined such
-that @{text "(f 0)\<notin>newInt 0 f"}. *}
-
-subsubsection {* Definition *}
-
-consts newInt :: "nat \<Rightarrow> (nat \<Rightarrow> real) \<Rightarrow> (real set)"
-primrec
-"newInt 0 f = closed_int (f 0 + 1) (f 0 + 2)"
-"newInt (Suc n) f =
- (SOME e. (\<exists>e1 e2.
- e1 < e2 \<and>
- e = closed_int e1 e2 \<and>
- e \<subseteq> (newInt n f) \<and>
- (f (Suc n)) \<notin> e)
- )"
-
-subsubsection {* Properties *}
-
-text {* We now show that every application of newInt returns an
-appropriate interval. *}
-
-lemma newInt_ex:
- "\<exists>a b. a < b \<and>
- newInt (Suc n) f = closed_int a b \<and>
- newInt (Suc n) f \<subseteq> newInt n f \<and>
- f (Suc n) \<notin> newInt (Suc n) f"
-proof (induct n)
- case 0
-
- let ?e = "SOME e. \<exists>e1 e2.
- e1 < e2 \<and>
- e = closed_int e1 e2 \<and>
- e \<subseteq> closed_int (f 0 + 1) (f 0 + 2) \<and>
- f (Suc 0) \<notin> e"
-
- have "newInt (Suc 0) f = ?e" by auto
- moreover
- have "f 0 + 1 < f 0 + 2" by simp
- with closed_subset_ex have
- "\<exists>ka kb. ka < kb \<and> closed_int ka kb \<subseteq> closed_int (f 0 + 1) (f 0 + 2) \<and>
- f (Suc 0) \<notin> (closed_int ka kb)" .
- hence
- "\<exists>e. \<exists>ka kb. ka < kb \<and> e = closed_int ka kb \<and>
- e \<subseteq> closed_int (f 0 + 1) (f 0 + 2) \<and> f (Suc 0) \<notin> e" by simp
- hence
- "\<exists>ka kb. ka < kb \<and> ?e = closed_int ka kb \<and>
- ?e \<subseteq> closed_int (f 0 + 1) (f 0 + 2) \<and> f (Suc 0) \<notin> ?e"
- by (rule someI_ex)
- ultimately have "\<exists>e1 e2. e1 < e2 \<and>
- newInt (Suc 0) f = closed_int e1 e2 \<and>
- newInt (Suc 0) f \<subseteq> closed_int (f 0 + 1) (f 0 + 2) \<and>
- f (Suc 0) \<notin> newInt (Suc 0) f" by simp
- thus
- "\<exists>a b. a < b \<and> newInt (Suc 0) f = closed_int a b \<and>
- newInt (Suc 0) f \<subseteq> newInt 0 f \<and> f (Suc 0) \<notin> newInt (Suc 0) f"
- by simp
-next
- case (Suc n)
- hence "\<exists>a b.
- a < b \<and>
- newInt (Suc n) f = closed_int a b \<and>
- newInt (Suc n) f \<subseteq> newInt n f \<and>
- f (Suc n) \<notin> newInt (Suc n) f" by simp
- then obtain a and b where ab: "a < b \<and>
- newInt (Suc n) f = closed_int a b \<and>
- newInt (Suc n) f \<subseteq> newInt n f \<and>
- f (Suc n) \<notin> newInt (Suc n) f" by auto
- hence cab: "closed_int a b = newInt (Suc n) f" by simp
-
- let ?e = "SOME e. \<exists>e1 e2.
- e1 < e2 \<and>
- e = closed_int e1 e2 \<and>
- e \<subseteq> closed_int a b \<and>
- f (Suc (Suc n)) \<notin> e"
- from cab have ni: "newInt (Suc (Suc n)) f = ?e" by auto
-
- from ab have "a < b" by simp
- with closed_subset_ex have
- "\<exists>ka kb. ka < kb \<and> closed_int ka kb \<subseteq> closed_int a b \<and>
- f (Suc (Suc n)) \<notin> closed_int ka kb" .
- hence
- "\<exists>e. \<exists>ka kb. ka < kb \<and> e = closed_int ka kb \<and>
- closed_int ka kb \<subseteq> closed_int a b \<and> f (Suc (Suc n)) \<notin> closed_int ka kb"
- by simp
- hence
- "\<exists>e. \<exists>ka kb. ka < kb \<and> e = closed_int ka kb \<and>
- e \<subseteq> closed_int a b \<and> f (Suc (Suc n)) \<notin> e" by simp
- hence
- "\<exists>ka kb. ka < kb \<and> ?e = closed_int ka kb \<and>
- ?e \<subseteq> closed_int a b \<and> f (Suc (Suc n)) \<notin> ?e" by (rule someI_ex)
- with ab ni show
- "\<exists>ka kb. ka < kb \<and>
- newInt (Suc (Suc n)) f = closed_int ka kb \<and>
- newInt (Suc (Suc n)) f \<subseteq> newInt (Suc n) f \<and>
- f (Suc (Suc n)) \<notin> newInt (Suc (Suc n)) f" by auto
-qed
-
-lemma newInt_subset:
- "newInt (Suc n) f \<subseteq> newInt n f"
- using newInt_ex by auto
-
-
-text {* Another fundamental property is that no element in the range
-of f is in the intersection of all closed intervals generated by
-newInt. *}
-
-lemma newInt_inter:
- "\<forall>n. f n \<notin> (\<Inter>n. newInt n f)"
-proof
- fix n::nat
- {
- assume n0: "n = 0"
- moreover have "newInt 0 f = closed_int (f 0 + 1) (f 0 + 2)" by simp
- ultimately have "f n \<notin> newInt n f" by (unfold closed_int_def, simp)
- }
- moreover
- {
- assume "\<not> n = 0"
- hence "n > 0" by simp
- then obtain m where ndef: "n = Suc m" by (auto simp add: gr0_conv_Suc)
-
- from newInt_ex have
- "\<exists>a b. a < b \<and> (newInt (Suc m) f) = closed_int a b \<and>
- newInt (Suc m) f \<subseteq> newInt m f \<and> f (Suc m) \<notin> newInt (Suc m) f" .
- then have "f (Suc m) \<notin> newInt (Suc m) f" by auto
- with ndef have "f n \<notin> newInt n f" by simp
- }
- ultimately have "f n \<notin> newInt n f" by (rule case_split)
- thus "f n \<notin> (\<Inter>n. newInt n f)" by auto
-qed
-
-
-lemma newInt_notempty:
- "(\<Inter>n. newInt n f) \<noteq> {}"
-proof -
- let ?g = "\<lambda>n. newInt n f"
- have "\<forall>n. ?g (Suc n) \<subseteq> ?g n"
- proof
- fix n
- show "?g (Suc n) \<subseteq> ?g n" by (rule newInt_subset)
- qed
- moreover have "\<forall>n. \<exists>a b. ?g n = closed_int a b \<and> a \<le> b"
- proof
- fix n::nat
- {
- assume "n = 0"
- then have
- "?g n = closed_int (f 0 + 1) (f 0 + 2) \<and> (f 0 + 1 \<le> f 0 + 2)"
- by simp
- hence "\<exists>a b. ?g n = closed_int a b \<and> a \<le> b" by blast
- }
- moreover
- {
- assume "\<not> n = 0"
- then have "n > 0" by simp
- then obtain m where nd: "n = Suc m" by (auto simp add: gr0_conv_Suc)
-
- have
- "\<exists>a b. a < b \<and> (newInt (Suc m) f) = closed_int a b \<and>
- (newInt (Suc m) f) \<subseteq> (newInt m f) \<and> (f (Suc m)) \<notin> (newInt (Suc m) f)"
- by (rule newInt_ex)
- then obtain a and b where
- "a < b \<and> (newInt (Suc m) f) = closed_int a b" by auto
- with nd have "?g n = closed_int a b \<and> a \<le> b" by auto
- hence "\<exists>a b. ?g n = closed_int a b \<and> a \<le> b" by blast
- }
- ultimately show "\<exists>a b. ?g n = closed_int a b \<and> a \<le> b" by (rule case_split)
- qed
- ultimately show ?thesis by (rule NIP)
-qed
-
-
-subsection {* Final Theorem *}
-
-theorem real_non_denum:
- shows "\<not> (\<exists>f::nat\<Rightarrow>real. surj f)"
-proof -- "by contradiction"
- assume "\<exists>f::nat\<Rightarrow>real. surj f"
- then obtain f::"nat\<Rightarrow>real" where "surj f" by auto
- hence rangeF: "range f = UNIV" by (rule surj_range)
- -- "We now produce a real number x that is not in the range of f, using the properties of newInt. "
- have "\<exists>x. x \<in> (\<Inter>n. newInt n f)" using newInt_notempty by blast
- moreover have "\<forall>n. f n \<notin> (\<Inter>n. newInt n f)" by (rule newInt_inter)
- ultimately obtain x where "x \<in> (\<Inter>n. newInt n f)" and "\<forall>n. f n \<noteq> x" by blast
- moreover from rangeF have "x \<in> range f" by simp
- ultimately show False by blast
-qed
-
-end
+(* Title : HOL/Real/ContNonDenum
+ ID : $Id$
+ Author : Benjamin Porter, Monash University, NICTA, 2005
+*)
+
+header {* Non-denumerability of the Continuum. *}
+
+theory ContNotDenum
+imports RComplete
+begin
+
+subsection {* Abstract *}
+
+text {* The following document presents a proof that the Continuum is
+uncountable. It is formalised in the Isabelle/Isar theorem proving
+system.
+
+{\em Theorem:} The Continuum @{text "\<real>"} is not denumerable. In other
+words, there does not exist a function f:@{text "\<nat>\<Rightarrow>\<real>"} such that f is
+surjective.
+
+{\em Outline:} An elegant informal proof of this result uses Cantor's
+Diagonalisation argument. The proof presented here is not this
+one. First we formalise some properties of closed intervals, then we
+prove the Nested Interval Property. This property relies on the
+completeness of the Real numbers and is the foundation for our
+argument. Informally it states that an intersection of countable
+closed intervals (where each successive interval is a subset of the
+last) is non-empty. We then assume a surjective function f:@{text
+"\<nat>\<Rightarrow>\<real>"} exists and find a real x such that x is not in the range of f
+by generating a sequence of closed intervals then using the NIP. *}
+
+subsection {* Closed Intervals *}
+
+text {* This section formalises some properties of closed intervals. *}
+
+subsubsection {* Definition *}
+
+definition
+ closed_int :: "real \<Rightarrow> real \<Rightarrow> real set" where
+ "closed_int x y = {z. x \<le> z \<and> z \<le> y}"
+
+subsubsection {* Properties *}
+
+lemma closed_int_subset:
+ assumes xy: "x1 \<ge> x0" "y1 \<le> y0"
+ shows "closed_int x1 y1 \<subseteq> closed_int x0 y0"
+proof -
+ {
+ fix x::real
+ assume "x \<in> closed_int x1 y1"
+ hence "x \<ge> x1 \<and> x \<le> y1" by (simp add: closed_int_def)
+ with xy have "x \<ge> x0 \<and> x \<le> y0" by auto
+ hence "x \<in> closed_int x0 y0" by (simp add: closed_int_def)
+ }
+ thus ?thesis by auto
+qed
+
+lemma closed_int_least:
+ assumes a: "a \<le> b"
+ shows "a \<in> closed_int a b \<and> (\<forall>x \<in> closed_int a b. a \<le> x)"
+proof
+ from a have "a\<in>{x. a\<le>x \<and> x\<le>b}" by simp
+ thus "a \<in> closed_int a b" by (unfold closed_int_def)
+next
+ have "\<forall>x\<in>{x. a\<le>x \<and> x\<le>b}. a\<le>x" by simp
+ thus "\<forall>x \<in> closed_int a b. a \<le> x" by (unfold closed_int_def)
+qed
+
+lemma closed_int_most:
+ assumes a: "a \<le> b"
+ shows "b \<in> closed_int a b \<and> (\<forall>x \<in> closed_int a b. x \<le> b)"
+proof
+ from a have "b\<in>{x. a\<le>x \<and> x\<le>b}" by simp
+ thus "b \<in> closed_int a b" by (unfold closed_int_def)
+next
+ have "\<forall>x\<in>{x. a\<le>x \<and> x\<le>b}. x\<le>b" by simp
+ thus "\<forall>x \<in> closed_int a b. x\<le>b" by (unfold closed_int_def)
+qed
+
+lemma closed_not_empty:
+ shows "a \<le> b \<Longrightarrow> \<exists>x. x \<in> closed_int a b"
+ by (auto dest: closed_int_least)
+
+lemma closed_mem:
+ assumes "a \<le> c" and "c \<le> b"
+ shows "c \<in> closed_int a b"
+ using assms unfolding closed_int_def by auto
+
+lemma closed_subset:
+ assumes ac: "a \<le> b" "c \<le> d"
+ assumes closed: "closed_int a b \<subseteq> closed_int c d"
+ shows "b \<ge> c"
+proof -
+ from closed have "\<forall>x\<in>closed_int a b. x\<in>closed_int c d" by auto
+ hence "\<forall>x. a\<le>x \<and> x\<le>b \<longrightarrow> c\<le>x \<and> x\<le>d" by (unfold closed_int_def, auto)
+ with ac have "c\<le>b \<and> b\<le>d" by simp
+ thus ?thesis by auto
+qed
+
+
+subsection {* Nested Interval Property *}
+
+theorem NIP:
+ fixes f::"nat \<Rightarrow> real set"
+ assumes subset: "\<forall>n. f (Suc n) \<subseteq> f n"
+ and closed: "\<forall>n. \<exists>a b. f n = closed_int a b \<and> a \<le> b"
+ shows "(\<Inter>n. f n) \<noteq> {}"
+proof -
+ let ?g = "\<lambda>n. (SOME c. c\<in>(f n) \<and> (\<forall>x\<in>(f n). c \<le> x))"
+ have ne: "\<forall>n. \<exists>x. x\<in>(f n)"
+ proof
+ fix n
+ from closed have "\<exists>a b. f n = closed_int a b \<and> a \<le> b" by simp
+ then obtain a and b where fn: "f n = closed_int a b \<and> a \<le> b" by auto
+ hence "a \<le> b" ..
+ with closed_not_empty have "\<exists>x. x\<in>closed_int a b" by simp
+ with fn show "\<exists>x. x\<in>(f n)" by simp
+ qed
+
+ have gdef: "\<forall>n. (?g n)\<in>(f n) \<and> (\<forall>x\<in>(f n). (?g n)\<le>x)"
+ proof
+ fix n
+ from closed have "\<exists>a b. f n = closed_int a b \<and> a \<le> b" ..
+ then obtain a and b where ff: "f n = closed_int a b" and "a \<le> b" by auto
+ hence "a \<le> b" by simp
+ hence "a\<in>closed_int a b \<and> (\<forall>x\<in>closed_int a b. a \<le> x)" by (rule closed_int_least)
+ with ff have "a\<in>(f n) \<and> (\<forall>x\<in>(f n). a \<le> x)" by simp
+ hence "\<exists>c. c\<in>(f n) \<and> (\<forall>x\<in>(f n). c \<le> x)" ..
+ thus "(?g n)\<in>(f n) \<and> (\<forall>x\<in>(f n). (?g n)\<le>x)" by (rule someI_ex)
+ qed
+
+ -- "A denotes the set of all left-most points of all the intervals ..."
+ moreover obtain A where Adef: "A = ?g ` \<nat>" by simp
+ ultimately have "\<exists>x. x\<in>A"
+ proof -
+ have "(0::nat) \<in> \<nat>" by simp
+ moreover have "?g 0 = ?g 0" by simp
+ ultimately have "?g 0 \<in> ?g ` \<nat>" by (rule rev_image_eqI)
+ with Adef have "?g 0 \<in> A" by simp
+ thus ?thesis ..
+ qed
+
+ -- "Now show that A is bounded above ..."
+ moreover have "\<exists>y. isUb (UNIV::real set) A y"
+ proof -
+ {
+ fix n
+ from ne have ex: "\<exists>x. x\<in>(f n)" ..
+ from gdef have "(?g n)\<in>(f n) \<and> (\<forall>x\<in>(f n). (?g n)\<le>x)" by simp
+ moreover
+ from closed have "\<exists>a b. f n = closed_int a b \<and> a \<le> b" ..
+ then obtain a and b where "f n = closed_int a b \<and> a \<le> b" by auto
+ hence "b\<in>(f n) \<and> (\<forall>x\<in>(f n). x \<le> b)" using closed_int_most by blast
+ ultimately have "\<forall>x\<in>(f n). (?g n) \<le> b" by simp
+ with ex have "(?g n) \<le> b" by auto
+ hence "\<exists>b. (?g n) \<le> b" by auto
+ }
+ hence aux: "\<forall>n. \<exists>b. (?g n) \<le> b" ..
+
+ have fs: "\<forall>n::nat. f n \<subseteq> f 0"
+ proof (rule allI, induct_tac n)
+ show "f 0 \<subseteq> f 0" by simp
+ next
+ fix n
+ assume "f n \<subseteq> f 0"
+ moreover from subset have "f (Suc n) \<subseteq> f n" ..
+ ultimately show "f (Suc n) \<subseteq> f 0" by simp
+ qed
+ have "\<forall>n. (?g n)\<in>(f 0)"
+ proof
+ fix n
+ from gdef have "(?g n)\<in>(f n) \<and> (\<forall>x\<in>(f n). (?g n)\<le>x)" by simp
+ hence "?g n \<in> f n" ..
+ with fs show "?g n \<in> f 0" by auto
+ qed
+ moreover from closed
+ obtain a and b where "f 0 = closed_int a b" and alb: "a \<le> b" by blast
+ ultimately have "\<forall>n. ?g n \<in> closed_int a b" by auto
+ with alb have "\<forall>n. ?g n \<le> b" using closed_int_most by blast
+ with Adef have "\<forall>y\<in>A. y\<le>b" by auto
+ hence "A *<= b" by (unfold setle_def)
+ moreover have "b \<in> (UNIV::real set)" by simp
+ ultimately have "A *<= b \<and> b \<in> (UNIV::real set)" by simp
+ hence "isUb (UNIV::real set) A b" by (unfold isUb_def)
+ thus ?thesis by auto
+ qed
+ -- "by the Axiom Of Completeness, A has a least upper bound ..."
+ ultimately have "\<exists>t. isLub UNIV A t" by (rule reals_complete)
+
+ -- "denote this least upper bound as t ..."
+ then obtain t where tdef: "isLub UNIV A t" ..
+
+ -- "and finally show that this least upper bound is in all the intervals..."
+ have "\<forall>n. t \<in> f n"
+ proof
+ fix n::nat
+ from closed obtain a and b where
+ int: "f n = closed_int a b" and alb: "a \<le> b" by blast
+
+ have "t \<ge> a"
+ proof -
+ have "a \<in> A"
+ proof -
+ (* by construction *)
+ from alb int have ain: "a\<in>f n \<and> (\<forall>x\<in>f n. a \<le> x)"
+ using closed_int_least by blast
+ moreover have "\<forall>e. e\<in>f n \<and> (\<forall>x\<in>f n. e \<le> x) \<longrightarrow> e = a"
+ proof clarsimp
+ fix e
+ assume ein: "e \<in> f n" and lt: "\<forall>x\<in>f n. e \<le> x"
+ from lt ain have aux: "\<forall>x\<in>f n. a \<le> x \<and> e \<le> x" by auto
+
+ from ein aux have "a \<le> e \<and> e \<le> e" by auto
+ moreover from ain aux have "a \<le> a \<and> e \<le> a" by auto
+ ultimately show "e = a" by simp
+ qed
+ hence "\<And>e. e\<in>f n \<and> (\<forall>x\<in>f n. e \<le> x) \<Longrightarrow> e = a" by simp
+ ultimately have "(?g n) = a" by (rule some_equality)
+ moreover
+ {
+ have "n = of_nat n" by simp
+ moreover have "of_nat n \<in> \<nat>" by simp
+ ultimately have "n \<in> \<nat>"
+ apply -
+ apply (subst(asm) eq_sym_conv)
+ apply (erule subst)
+ .
+ }
+ with Adef have "(?g n) \<in> A" by auto
+ ultimately show ?thesis by simp
+ qed
+ with tdef show "a \<le> t" by (rule isLubD2)
+ qed
+ moreover have "t \<le> b"
+ proof -
+ have "isUb UNIV A b"
+ proof -
+ {
+ from alb int have
+ ain: "b\<in>f n \<and> (\<forall>x\<in>f n. x \<le> b)" using closed_int_most by blast
+
+ have subsetd: "\<forall>m. \<forall>n. f (n + m) \<subseteq> f n"
+ proof (rule allI, induct_tac m)
+ show "\<forall>n. f (n + 0) \<subseteq> f n" by simp
+ next
+ fix m n
+ assume pp: "\<forall>p. f (p + n) \<subseteq> f p"
+ {
+ fix p
+ from pp have "f (p + n) \<subseteq> f p" by simp
+ moreover from subset have "f (Suc (p + n)) \<subseteq> f (p + n)" by auto
+ hence "f (p + (Suc n)) \<subseteq> f (p + n)" by simp
+ ultimately have "f (p + (Suc n)) \<subseteq> f p" by simp
+ }
+ thus "\<forall>p. f (p + Suc n) \<subseteq> f p" ..
+ qed
+ have subsetm: "\<forall>\<alpha> \<beta>. \<alpha> \<ge> \<beta> \<longrightarrow> (f \<alpha>) \<subseteq> (f \<beta>)"
+ proof ((rule allI)+, rule impI)
+ fix \<alpha>::nat and \<beta>::nat
+ assume "\<beta> \<le> \<alpha>"
+ hence "\<exists>k. \<alpha> = \<beta> + k" by (simp only: le_iff_add)
+ then obtain k where "\<alpha> = \<beta> + k" ..
+ moreover
+ from subsetd have "f (\<beta> + k) \<subseteq> f \<beta>" by simp
+ ultimately show "f \<alpha> \<subseteq> f \<beta>" by auto
+ qed
+
+ fix m
+ {
+ assume "m \<ge> n"
+ with subsetm have "f m \<subseteq> f n" by simp
+ with ain have "\<forall>x\<in>f m. x \<le> b" by auto
+ moreover
+ from gdef have "?g m \<in> f m \<and> (\<forall>x\<in>f m. ?g m \<le> x)" by simp
+ ultimately have "?g m \<le> b" by auto
+ }
+ moreover
+ {
+ assume "\<not>(m \<ge> n)"
+ hence "m < n" by simp
+ with subsetm have sub: "(f n) \<subseteq> (f m)" by simp
+ from closed obtain ma and mb where
+ "f m = closed_int ma mb \<and> ma \<le> mb" by blast
+ hence one: "ma \<le> mb" and fm: "f m = closed_int ma mb" by auto
+ from one alb sub fm int have "ma \<le> b" using closed_subset by blast
+ moreover have "(?g m) = ma"
+ proof -
+ from gdef have "?g m \<in> f m \<and> (\<forall>x\<in>f m. ?g m \<le> x)" ..
+ moreover from one have
+ "ma \<in> closed_int ma mb \<and> (\<forall>x\<in>closed_int ma mb. ma \<le> x)"
+ by (rule closed_int_least)
+ with fm have "ma\<in>f m \<and> (\<forall>x\<in>f m. ma \<le> x)" by simp
+ ultimately have "ma \<le> ?g m \<and> ?g m \<le> ma" by auto
+ thus "?g m = ma" by auto
+ qed
+ ultimately have "?g m \<le> b" by simp
+ }
+ ultimately have "?g m \<le> b" by (rule case_split)
+ }
+ with Adef have "\<forall>y\<in>A. y\<le>b" by auto
+ hence "A *<= b" by (unfold setle_def)
+ moreover have "b \<in> (UNIV::real set)" by simp
+ ultimately have "A *<= b \<and> b \<in> (UNIV::real set)" by simp
+ thus "isUb (UNIV::real set) A b" by (unfold isUb_def)
+ qed
+ with tdef show "t \<le> b" by (rule isLub_le_isUb)
+ qed
+ ultimately have "t \<in> closed_int a b" by (rule closed_mem)
+ with int show "t \<in> f n" by simp
+ qed
+ hence "t \<in> (\<Inter>n. f n)" by auto
+ thus ?thesis by auto
+qed
+
+subsection {* Generating the intervals *}
+
+subsubsection {* Existence of non-singleton closed intervals *}
+
+text {* This lemma asserts that given any non-singleton closed
+interval (a,b) and any element c, there exists a closed interval that
+is a subset of (a,b) and that does not contain c and is a
+non-singleton itself. *}
+
+lemma closed_subset_ex:
+ fixes c::real
+ assumes alb: "a < b"
+ shows
+ "\<exists>ka kb. ka < kb \<and> closed_int ka kb \<subseteq> closed_int a b \<and> c \<notin> (closed_int ka kb)"
+proof -
+ {
+ assume clb: "c < b"
+ {
+ assume cla: "c < a"
+ from alb cla clb have "c \<notin> closed_int a b" by (unfold closed_int_def, auto)
+ with alb have
+ "a < b \<and> closed_int a b \<subseteq> closed_int a b \<and> c \<notin> closed_int a b"
+ by auto
+ hence
+ "\<exists>ka kb. ka < kb \<and> closed_int ka kb \<subseteq> closed_int a b \<and> c \<notin> (closed_int ka kb)"
+ by auto
+ }
+ moreover
+ {
+ assume ncla: "\<not>(c < a)"
+ with clb have cdef: "a \<le> c \<and> c < b" by simp
+ obtain ka where kadef: "ka = (c + b)/2" by blast
+
+ from kadef clb have kalb: "ka < b" by auto
+ moreover from kadef cdef have kagc: "ka > c" by simp
+ ultimately have "c\<notin>(closed_int ka b)" by (unfold closed_int_def, auto)
+ moreover from cdef kagc have "ka \<ge> a" by simp
+ hence "closed_int ka b \<subseteq> closed_int a b" by (unfold closed_int_def, auto)
+ ultimately have
+ "ka < b \<and> closed_int ka b \<subseteq> closed_int a b \<and> c \<notin> closed_int ka b"
+ using kalb by auto
+ hence
+ "\<exists>ka kb. ka < kb \<and> closed_int ka kb \<subseteq> closed_int a b \<and> c \<notin> (closed_int ka kb)"
+ by auto
+
+ }
+ ultimately have
+ "\<exists>ka kb. ka < kb \<and> closed_int ka kb \<subseteq> closed_int a b \<and> c \<notin> (closed_int ka kb)"
+ by (rule case_split)
+ }
+ moreover
+ {
+ assume "\<not> (c < b)"
+ hence cgeb: "c \<ge> b" by simp
+
+ obtain kb where kbdef: "kb = (a + b)/2" by blast
+ with alb have kblb: "kb < b" by auto
+ with kbdef cgeb have "a < kb \<and> kb < c" by auto
+ moreover hence "c \<notin> (closed_int a kb)" by (unfold closed_int_def, auto)
+ moreover from kblb have
+ "closed_int a kb \<subseteq> closed_int a b" by (unfold closed_int_def, auto)
+ ultimately have
+ "a < kb \<and> closed_int a kb \<subseteq> closed_int a b \<and> c\<notin>closed_int a kb"
+ by simp
+ hence
+ "\<exists>ka kb. ka < kb \<and> closed_int ka kb \<subseteq> closed_int a b \<and> c \<notin> (closed_int ka kb)"
+ by auto
+ }
+ ultimately show ?thesis by (rule case_split)
+qed
+
+subsection {* newInt: Interval generation *}
+
+text {* Given a function f:@{text "\<nat>\<Rightarrow>\<real>"}, newInt (Suc n) f returns a
+closed interval such that @{text "newInt (Suc n) f \<subseteq> newInt n f"} and
+does not contain @{text "f (Suc n)"}. With the base case defined such
+that @{text "(f 0)\<notin>newInt 0 f"}. *}
+
+subsubsection {* Definition *}
+
+consts newInt :: "nat \<Rightarrow> (nat \<Rightarrow> real) \<Rightarrow> (real set)"
+primrec
+"newInt 0 f = closed_int (f 0 + 1) (f 0 + 2)"
+"newInt (Suc n) f =
+ (SOME e. (\<exists>e1 e2.
+ e1 < e2 \<and>
+ e = closed_int e1 e2 \<and>
+ e \<subseteq> (newInt n f) \<and>
+ (f (Suc n)) \<notin> e)
+ )"
+
+subsubsection {* Properties *}
+
+text {* We now show that every application of newInt returns an
+appropriate interval. *}
+
+lemma newInt_ex:
+ "\<exists>a b. a < b \<and>
+ newInt (Suc n) f = closed_int a b \<and>
+ newInt (Suc n) f \<subseteq> newInt n f \<and>
+ f (Suc n) \<notin> newInt (Suc n) f"
+proof (induct n)
+ case 0
+
+ let ?e = "SOME e. \<exists>e1 e2.
+ e1 < e2 \<and>
+ e = closed_int e1 e2 \<and>
+ e \<subseteq> closed_int (f 0 + 1) (f 0 + 2) \<and>
+ f (Suc 0) \<notin> e"
+
+ have "newInt (Suc 0) f = ?e" by auto
+ moreover
+ have "f 0 + 1 < f 0 + 2" by simp
+ with closed_subset_ex have
+ "\<exists>ka kb. ka < kb \<and> closed_int ka kb \<subseteq> closed_int (f 0 + 1) (f 0 + 2) \<and>
+ f (Suc 0) \<notin> (closed_int ka kb)" .
+ hence
+ "\<exists>e. \<exists>ka kb. ka < kb \<and> e = closed_int ka kb \<and>
+ e \<subseteq> closed_int (f 0 + 1) (f 0 + 2) \<and> f (Suc 0) \<notin> e" by simp
+ hence
+ "\<exists>ka kb. ka < kb \<and> ?e = closed_int ka kb \<and>
+ ?e \<subseteq> closed_int (f 0 + 1) (f 0 + 2) \<and> f (Suc 0) \<notin> ?e"
+ by (rule someI_ex)
+ ultimately have "\<exists>e1 e2. e1 < e2 \<and>
+ newInt (Suc 0) f = closed_int e1 e2 \<and>
+ newInt (Suc 0) f \<subseteq> closed_int (f 0 + 1) (f 0 + 2) \<and>
+ f (Suc 0) \<notin> newInt (Suc 0) f" by simp
+ thus
+ "\<exists>a b. a < b \<and> newInt (Suc 0) f = closed_int a b \<and>
+ newInt (Suc 0) f \<subseteq> newInt 0 f \<and> f (Suc 0) \<notin> newInt (Suc 0) f"
+ by simp
+next
+ case (Suc n)
+ hence "\<exists>a b.
+ a < b \<and>
+ newInt (Suc n) f = closed_int a b \<and>
+ newInt (Suc n) f \<subseteq> newInt n f \<and>
+ f (Suc n) \<notin> newInt (Suc n) f" by simp
+ then obtain a and b where ab: "a < b \<and>
+ newInt (Suc n) f = closed_int a b \<and>
+ newInt (Suc n) f \<subseteq> newInt n f \<and>
+ f (Suc n) \<notin> newInt (Suc n) f" by auto
+ hence cab: "closed_int a b = newInt (Suc n) f" by simp
+
+ let ?e = "SOME e. \<exists>e1 e2.
+ e1 < e2 \<and>
+ e = closed_int e1 e2 \<and>
+ e \<subseteq> closed_int a b \<and>
+ f (Suc (Suc n)) \<notin> e"
+ from cab have ni: "newInt (Suc (Suc n)) f = ?e" by auto
+
+ from ab have "a < b" by simp
+ with closed_subset_ex have
+ "\<exists>ka kb. ka < kb \<and> closed_int ka kb \<subseteq> closed_int a b \<and>
+ f (Suc (Suc n)) \<notin> closed_int ka kb" .
+ hence
+ "\<exists>e. \<exists>ka kb. ka < kb \<and> e = closed_int ka kb \<and>
+ closed_int ka kb \<subseteq> closed_int a b \<and> f (Suc (Suc n)) \<notin> closed_int ka kb"
+ by simp
+ hence
+ "\<exists>e. \<exists>ka kb. ka < kb \<and> e = closed_int ka kb \<and>
+ e \<subseteq> closed_int a b \<and> f (Suc (Suc n)) \<notin> e" by simp
+ hence
+ "\<exists>ka kb. ka < kb \<and> ?e = closed_int ka kb \<and>
+ ?e \<subseteq> closed_int a b \<and> f (Suc (Suc n)) \<notin> ?e" by (rule someI_ex)
+ with ab ni show
+ "\<exists>ka kb. ka < kb \<and>
+ newInt (Suc (Suc n)) f = closed_int ka kb \<and>
+ newInt (Suc (Suc n)) f \<subseteq> newInt (Suc n) f \<and>
+ f (Suc (Suc n)) \<notin> newInt (Suc (Suc n)) f" by auto
+qed
+
+lemma newInt_subset:
+ "newInt (Suc n) f \<subseteq> newInt n f"
+ using newInt_ex by auto
+
+
+text {* Another fundamental property is that no element in the range
+of f is in the intersection of all closed intervals generated by
+newInt. *}
+
+lemma newInt_inter:
+ "\<forall>n. f n \<notin> (\<Inter>n. newInt n f)"
+proof
+ fix n::nat
+ {
+ assume n0: "n = 0"
+ moreover have "newInt 0 f = closed_int (f 0 + 1) (f 0 + 2)" by simp
+ ultimately have "f n \<notin> newInt n f" by (unfold closed_int_def, simp)
+ }
+ moreover
+ {
+ assume "\<not> n = 0"
+ hence "n > 0" by simp
+ then obtain m where ndef: "n = Suc m" by (auto simp add: gr0_conv_Suc)
+
+ from newInt_ex have
+ "\<exists>a b. a < b \<and> (newInt (Suc m) f) = closed_int a b \<and>
+ newInt (Suc m) f \<subseteq> newInt m f \<and> f (Suc m) \<notin> newInt (Suc m) f" .
+ then have "f (Suc m) \<notin> newInt (Suc m) f" by auto
+ with ndef have "f n \<notin> newInt n f" by simp
+ }
+ ultimately have "f n \<notin> newInt n f" by (rule case_split)
+ thus "f n \<notin> (\<Inter>n. newInt n f)" by auto
+qed
+
+
+lemma newInt_notempty:
+ "(\<Inter>n. newInt n f) \<noteq> {}"
+proof -
+ let ?g = "\<lambda>n. newInt n f"
+ have "\<forall>n. ?g (Suc n) \<subseteq> ?g n"
+ proof
+ fix n
+ show "?g (Suc n) \<subseteq> ?g n" by (rule newInt_subset)
+ qed
+ moreover have "\<forall>n. \<exists>a b. ?g n = closed_int a b \<and> a \<le> b"
+ proof
+ fix n::nat
+ {
+ assume "n = 0"
+ then have
+ "?g n = closed_int (f 0 + 1) (f 0 + 2) \<and> (f 0 + 1 \<le> f 0 + 2)"
+ by simp
+ hence "\<exists>a b. ?g n = closed_int a b \<and> a \<le> b" by blast
+ }
+ moreover
+ {
+ assume "\<not> n = 0"
+ then have "n > 0" by simp
+ then obtain m where nd: "n = Suc m" by (auto simp add: gr0_conv_Suc)
+
+ have
+ "\<exists>a b. a < b \<and> (newInt (Suc m) f) = closed_int a b \<and>
+ (newInt (Suc m) f) \<subseteq> (newInt m f) \<and> (f (Suc m)) \<notin> (newInt (Suc m) f)"
+ by (rule newInt_ex)
+ then obtain a and b where
+ "a < b \<and> (newInt (Suc m) f) = closed_int a b" by auto
+ with nd have "?g n = closed_int a b \<and> a \<le> b" by auto
+ hence "\<exists>a b. ?g n = closed_int a b \<and> a \<le> b" by blast
+ }
+ ultimately show "\<exists>a b. ?g n = closed_int a b \<and> a \<le> b" by (rule case_split)
+ qed
+ ultimately show ?thesis by (rule NIP)
+qed
+
+
+subsection {* Final Theorem *}
+
+theorem real_non_denum:
+ shows "\<not> (\<exists>f::nat\<Rightarrow>real. surj f)"
+proof -- "by contradiction"
+ assume "\<exists>f::nat\<Rightarrow>real. surj f"
+ then obtain f::"nat\<Rightarrow>real" where "surj f" by auto
+ hence rangeF: "range f = UNIV" by (rule surj_range)
+ -- "We now produce a real number x that is not in the range of f, using the properties of newInt. "
+ have "\<exists>x. x \<in> (\<Inter>n. newInt n f)" using newInt_notempty by blast
+ moreover have "\<forall>n. f n \<notin> (\<Inter>n. newInt n f)" by (rule newInt_inter)
+ ultimately obtain x where "x \<in> (\<Inter>n. newInt n f)" and "\<forall>n. f n \<noteq> x" by blast
+ moreover from rangeF have "x \<in> range f" by simp
+ ultimately show False by blast
+qed
+
+end