Copyright (c) 1995 Association of Mizar Users
environ vocabulary GROUP_1, ARYTM_3, PRALG_2, BOOLE, PRE_TOPC, FUNCT_1, FINSEQ_1, RELAT_1, ARYTM_1, CONNSP_2, TOPS_1, SUBSET_1, SETFAM_1, TARSKI, TOPMETR, COMPTS_1, URYSOHN1; notation TARSKI, XBOOLE_0, ZFMISC_1, SUBSET_1, NUMBERS, XREAL_0, DOMAIN_1, FUNCT_1, FUNCT_2, REAL_1, FINSEQ_1, NAT_1, TOPMETR, COMPTS_1, STRUCT_0, PRE_TOPC, CARD_4, TOPS_1, CONNSP_2; constructors DOMAIN_1, REAL_1, NAT_1, TOPMETR, COMPTS_1, CARD_4, TOPS_1, MEMBERED; clusters SUBSET_1, STRUCT_0, PRE_TOPC, RELSET_1, TOPS_1, ARYTM_3, XREAL_0, MEMBERED, ZFMISC_1; requirements REAL, NUMERALS, BOOLE, SUBSET, ARITHM; definitions TARSKI, CONNSP_2, XBOOLE_0; theorems TARSKI, AXIOMS, SUBSET_1, FUNCT_1, FUNCT_2, NAT_1, FINSEQ_1, ZFMISC_1, PRE_TOPC, TOPS_1, ENUMSET1, REAL_1, TOPMETR, SQUARE_1, COMPTS_1, NEWTON, REAL_2, HEINE, CONNSP_2, XBOOLE_0, XBOOLE_1, SCHEME1, XCMPLX_0, XCMPLX_1; schemes PRE_TOPC, REAL_1, NAT_1, FINSEQ_1, FUNCT_2; begin definition func R<0 ->Subset of REAL means for x being Real holds x in it iff x < 0; existence proof defpred P[Real] means $1 < 0; ex X being Subset of REAL st for x being Real holds x in X iff P[x] from SepReal; hence thesis; end; uniqueness proof let A1,A2 be Subset of REAL; assume that A1:(for x being Real holds x in A1 iff x < 0) and A2:(for x being Real holds x in A2 iff x < 0); for a being set holds a in A1 iff a in A2 proof let a be set; thus a in A1 implies a in A2 proof assume A3:a in A1; then reconsider a as Real; a < 0 by A1,A3; hence thesis by A2; end; thus a in A2 implies a in A1 proof assume A4:a in A2; then reconsider a as Real; a < 0 by A2,A4; hence thesis by A1; end; end; hence thesis by TARSKI:2; end; end; definition func R>1 ->Subset of REAL means for x being Real holds x in it iff 1 < x; existence proof defpred P[Real] means 1 < $1; ex X being Subset of REAL st for x being Real holds x in X iff P[x] from SepReal; hence thesis; end; uniqueness proof let A1,A2 be Subset of REAL; assume that A1:(for x being Real holds x in A1 iff 1 < x) and A2:(for x being Real holds x in A2 iff 1 < x); for a being set holds a in A1 iff a in A2 proof let a be set; thus a in A1 implies a in A2 proof assume A3:a in A1; then reconsider a as Real; 1 < a by A1,A3; hence thesis by A2; end; thus a in A2 implies a in A1 proof assume A4:a in A2; then reconsider a as Real; 1 < a by A2,A4; hence thesis by A1; end; end; hence thesis by TARSKI:2; end; end; definition let n be Nat; func dyadic(n) -> Subset of REAL means :Def3:for x being Real holds x in it iff ex i being Nat st (0 <= i & i <= (2|^n) & x = i/(2|^n)); existence proof defpred P[set] means ex i being Nat st (0 <= i & i <= (2|^n) & $1 = i/(2|^n)); ex X being Subset of REAL st for x being Real holds x in X iff P[x] from SepReal; hence thesis; end; uniqueness proof let A1,A2 be Subset of REAL; assume that A1:(for x being Real holds x in A1 iff ex i being Nat st (0 <= i & i <= (2|^n) & x = i/(2|^n))) and A2:(for x being Real holds x in A2 iff ex i being Nat st (0 <= i & i <= (2|^n) & x = i/(2|^n))); for a being set holds a in A1 iff a in A2 proof let a be set; thus a in A1 implies a in A2 proof assume A3:a in A1; then reconsider a as Real; ex i being Nat st (0 <= i & i <= (2|^n) & a = i/(2|^n)) by A1,A3; hence thesis by A2; end; thus a in A2 implies a in A1 proof assume A4:a in A2; then reconsider a as Real; ex i being Nat st (0 <= i & i <= (2|^n) & a = i/(2|^n)) by A2,A4; hence thesis by A1; end; end; hence thesis by TARSKI:2; end; end; definition func DYADIC -> Subset of REAL means :Def4:for a being Real holds a in it iff ex n being Nat st a in dyadic(n); existence proof defpred P[set] means ex i being Nat st $1 in dyadic(i); ex X being Subset of REAL st for x being Real holds x in X iff P[x] from SepReal; hence thesis; end; uniqueness proof let A1,A2 be Subset of REAL; assume that A1:(for x being Real holds x in A1 iff ex n being Nat st x in dyadic(n)) and A2:(for x being Real holds x in A2 iff ex n being Nat st x in dyadic(n)); for a being set holds a in A1 iff a in A2 proof let a be set; thus a in A1 implies a in A2 proof assume A3:a in A1; then reconsider a as Real; ex n being Nat st a in dyadic(n) by A1,A3; hence thesis by A2; end; thus a in A2 implies a in A1 proof assume A4:a in A2; then reconsider a as Real; ex n being Nat st a in dyadic(n) by A2,A4; hence thesis by A1; end; end; hence thesis by TARSKI:2; end; end; definition func DOM -> Subset of REAL equals :Def5: R<0 \/ DYADIC \/ R>1; coherence; end; definition let T be TopSpace; let A be non empty Subset of REAL; let F be Function of A,bool the carrier of T; let r be Element of A; redefine func F.r -> Subset of T; coherence proof F.r is Subset of T; hence thesis; end; end; canceled 4; theorem Th5: for n being Nat holds for x being Real st x in dyadic(n) holds 0 <= x & x <= 1 proof let n be Nat; let x be Real; assume x in dyadic(n); then consider i being Nat such that A1:0 <= i & i <= (2|^n) & x = i/(2|^n) by Def3; 0 < (2|^n) by HEINE:5; then A2:0/(2|^n) <= i/(2|^n) & i/(2|^n) <= (2|^n)/(2|^n) by A1,REAL_1:73; (2|^n) <> 0 by HEINE:5; hence thesis by A1,A2,XCMPLX_1:60; end; theorem Th6: dyadic(0) = {0,1} proof A1: 2|^0 = 1 by NEWTON:9; A2:for x being Real holds x in dyadic(0) iff ex i being Nat st (0 <= i & i <= 1 & x = i) proof let x be Real; A3:x in dyadic(0) implies ex i being Nat st (0 <= i & i <= 1 & x = i) proof assume x in dyadic(0); then consider i being Nat such that A4:0 <= i & i <= 1 & x = i/1 by A1,Def3; thus thesis by A4; end; (ex i being Nat st (0 <= i & i <= 1 & x = i)) implies x in dyadic(0) proof given i being Nat such that A5:0 <= i & i <= 1 & x = i; x = i/1 by A5; hence thesis by A1,A5,Def3; end; hence thesis by A3; end; dyadic(0) = {0,1} proof for x being set holds x in dyadic(0) iff x in {0,1} proof let x be set; A6:x in dyadic(0) implies x in {0,1} proof assume A7:x in dyadic(0); then reconsider x as Real; consider i being Nat such that A8:0 <= i & i <= 1 & x = i by A2,A7; 0 <= i & i <= 0+1 by A8; then x = 0 or x = 1 by A8,NAT_1:27; hence thesis by TARSKI:def 2; end; x in {0,1} implies x in dyadic(0) proof assume x in {0,1}; then x = 0 or x = 1 by TARSKI:def 2; hence thesis by A2; end; hence thesis by A6; end; hence thesis by TARSKI:2; end; hence thesis; end; theorem dyadic(1) = {0,1/2,1} proof A1: 2|^1 = 2 by NEWTON:10; for x being set holds x in dyadic(1) iff x in {0,1/2,1} proof let x be set; A2:x in dyadic(1) implies x in {0,1/2,1} proof assume A3:x in dyadic(1); then reconsider x as Real; consider i being Nat such that A4:0 <= i & i <= 2 & x = i/2 by A1,A3,Def3; 0 <= i & (i <= 1 or i = 1+1) by A4,NAT_1:26; then i = 0 or i = 0+1 or i = 2 by NAT_1:27; hence thesis by A4,ENUMSET1:14; end; x in {0,1/2,1} implies x in dyadic(1) proof assume A5: x in {0,1/2,1}; per cases by A5,ENUMSET1:13; suppose x = 0; then x = 0/2; hence thesis by A1,Def3; suppose x = 1/2; hence thesis by A1,Def3; suppose A6:x = 1; then reconsider x as Real; x = 2/2 by A6; hence thesis by A1,Def3; end; hence thesis by A2; end; hence thesis by TARSKI:2; end; definition let n be Nat; cluster dyadic(n) -> non empty; coherence proof consider x being Real such that A1:x = 0/(2|^n); ex i being Nat st 0 <= i & i <= (2|^n) & x = i/(2|^n) proof take 0; thus thesis by A1,HEINE:5; end; hence thesis by Def3; end; end; theorem for x, n being Nat holds x|^n is Nat; theorem Th9: for n being Nat holds ex FS being FinSequence st (dom FS = Seg ((2|^n)+1) & for i being Nat st i in dom FS holds FS.i = (i-1)/(2|^n)) proof let n be Nat; deffunc F(Nat)=($1-1)/(2|^n); consider FS being FinSequence such that A1:len FS = (2|^n)+1 & for i being Nat st i in Seg((2|^n)+1) holds FS.i = F(i) from SeqLambda; A2:dom FS = Seg((2|^n)+1) by A1,FINSEQ_1:def 3; take FS; thus thesis by A1,A2; end; definition let n be Nat; func dyad(n) -> FinSequence means :Def6:dom it = Seg((2|^n)+1) & for i being Nat st i in dom it holds it.i = (i-1)/(2|^n); existence by Th9; uniqueness proof let F1,F2 be FinSequence; assume that A1:dom F1 = Seg((2|^n)+1) & for i being Nat st i in dom F1 holds F1.i = (i-1)/(2|^n) and A2:dom F2 = Seg((2|^n)+1) & for i being Nat st i in dom F2 holds F2.i = (i-1)/(2|^n); consider X being set such that A3:X = dom F1; for k being Nat st k in X holds F1.k = F2.k proof let k be Nat; assume A4: k in X; hence F1.k = (k-1)/(2|^n) by A1,A3 .= F2.k by A1,A2,A3,A4; end; hence thesis by A1,A2,A3,FINSEQ_1:17; end; end; theorem for n being Nat holds dom dyad(n) = Seg((2|^n)+1) & rng dyad(n) = dyadic(n) proof let n be Nat; A1:dom (dyad(n)) = Seg((2|^n)+1) & for i being Nat st i in dom (dyad(n)) holds (dyad(n)).i = (i-1)/(2|^n) by Def6; for x being set holds x in rng (dyad(n)) iff x in dyadic(n) proof let x be set; A2:x in rng (dyad(n)) implies x in dyadic(n) proof assume x in rng (dyad(n)); then consider y being set such that A3:y in dom (dyad(n)) & x = (dyad(n)).y by FUNCT_1:def 5; A4:y in Seg((2|^n)+1) by A3,Def6; reconsider y as Nat by A3; A5:1 <= y & y <= (2|^n) + 1 by A4,FINSEQ_1:3; then 1 - 1 <= y - 1 & y - 1 <= (2|^n) + 1 - 1 by REAL_1:92; then A6: 1 + (-1) <= y - 1 & y - 1 <= (2|^n) + (1 - 1) by XCMPLX_1:29; consider i being Nat such that A7:1 + i = y by A5,NAT_1:28; A8:i = y - 1 by A7,XCMPLX_1:26; x = (y-1)/(2|^n) by A3,Def6; hence thesis by A6,A8,Def3; end; x in dyadic(n) implies x in rng (dyad(n)) proof assume A9:x in dyadic(n); then reconsider x as Real; consider i being Nat such that A10:0 <= i & i <= (2|^n) & x = i/(2|^n) by A9,Def3; A11: 0 + 1 <= i + 1 & i + 1 <= (2|^n) + 1 by A10,AXIOMS:24; consider y being Nat such that A12:y = i + 1; A13:y in Seg((2|^n)+1) by A11,A12,FINSEQ_1:3; then A14:y in dom (dyad(n)) by Def6; x = (y-1)/(2|^n) by A10,A12,XCMPLX_1:26; then x = (dyad(n)).y by A1,A13; hence thesis by A14,FUNCT_1:def 5; end; hence thesis by A2; end; hence thesis by Def6,TARSKI:2; end; definition cluster DYADIC -> non empty; coherence proof consider x being set such that A1:x in dyadic(0) by XBOOLE_0:def 1; thus thesis by A1,Def4; end; end; definition cluster DOM -> non empty; coherence proof consider x being set such that A1:x in DYADIC by XBOOLE_0:def 1; x in R<0 \/ DYADIC by A1,XBOOLE_0:def 2; hence thesis by Def5,XBOOLE_0:def 2; end; end; theorem Th11: for n being Nat holds dyadic(n) c= dyadic(n+1) proof let n be Nat; let x be set; assume A1:x in dyadic(n); ex i being Nat st (0 <= i & i <= (2|^(n+1)) & x = i/(2|^(n+1))) proof reconsider x as Real by A1; consider i being Nat such that A2:0 <= i & i <= (2|^n) & x = i/(2|^n) by A1,Def3; A3: 0*2 <= i*2 by A2,AXIOMS:25; A4:2|^(n+1) = (2|^n)*2 by NEWTON:11; take i*2; thus thesis by A2,A3,A4,AXIOMS:25,XCMPLX_1:92; end; hence thesis by Def3; end; theorem Th12: for n being Nat holds 0 in dyadic(n) & 1 in dyadic(n) proof defpred P[Nat] means 0 in dyadic($1) & 1 in dyadic($1); A1:P[0] by Th6,TARSKI:def 2; A2:for k being Nat st P[k] holds P[(k+1)] proof let k be Nat; assume A3:0 in dyadic(k) & 1 in dyadic(k); dyadic(k) c= dyadic(k+1) by Th11; hence thesis by A3; end; for k be Nat holds P[k] from Ind(A1,A2); hence thesis; end; theorem Th13: for n,i being Nat st 0 < i & i <= 2|^n holds (i*2-1)/(2|^(n+1)) in dyadic(n+1) \ dyadic(n) proof let n,i be Nat; assume A1:0 < i & i <= 2|^n; then A2: 0*2 < i*2 by REAL_1:70; then consider k being Nat such that A3:i*2 = k + 1 by NAT_1:22; A4:k = i*2 - 1 by A3,XCMPLX_1:26; i*2 <= (2|^n)*2 by A1,AXIOMS:25; then i*2 <= 2|^(n+1) by NEWTON:11; then 0 <= k & k <= 2|^(n+1) by A2,A3,NAT_1:38; then A5:(i*2-1)/(2|^(n+1)) in dyadic(n+1) by A4,Def3; not (i*2-1)/(2|^(n+1)) in dyadic(n) proof assume (i*2-1)/(2|^(n+1)) in dyadic(n); then consider s being Nat such that A6:0 <= s & s <= 2|^n & (i*2-1)/(2|^(n+1)) = s/(2|^n) by Def3; A7:2|^n <> 0 & 2|^(n+1) <> 0 by HEINE:5; then (i*2-1)*(2|^n) = s*(2|^(n+1)) by A6,XCMPLX_1:96; then (i*2-1)*(2|^n) = s*((2|^n)*2) by NEWTON:11; then (i*2-1)*(2|^n) = s*2*(2|^n) by XCMPLX_1:4; then (i*2-1)/(2|^n) = s*2/(2|^n) by A7,XCMPLX_1:95; then i*2-1 = s*2 by A7,XCMPLX_1:53; then i*2+(-1) = s*2 by XCMPLX_0:def 8; then i*2 = s*2 - (-1) by XCMPLX_1:26; then i*2 = s*2 + (-(-1)) by XCMPLX_0:def 8; then 2*i + 0 = 2*s + 1; hence thesis by NAT_1:43; end; hence thesis by A5,XBOOLE_0:def 4; end; theorem Th14: for n,i being Nat st 0 <= i & i < 2|^n holds (i*2+1)/(2|^(n+1)) in dyadic(n+1) \ dyadic(n) proof let n,i be Nat; assume A1:0 <= i & i < 2|^n; consider s being Nat such that A2:s = i + 1; 0 + 1 <= i + 1 by A1,AXIOMS:24; then A3:0 < s by A2,AXIOMS:22; A4: s <= 2|^n by A1,A2,NAT_1:38; s*2 - 1 = i*2 + 1*2 - 1 by A2,XCMPLX_1:8 .= i*2 + 2 + (-1) by XCMPLX_0:def 8 .= i*2 + (2 + (-1)) by XCMPLX_1:1 .= i*2 + (1 + 0); hence thesis by A3,A4,Th13; end; theorem Th15: for n being Nat holds 1/(2|^(n+1)) in dyadic(n+1) \ dyadic(n) proof let n be Nat; A1: 2*0 + 1 = 1; 0 < 2|^n by HEINE:5; hence thesis by A1,Th14; end; definition let n be Nat; let x be Element of dyadic(n); func axis(x,n) -> Nat means :Def7:x = it/(2|^n); existence proof consider i being Nat such that A1:0 <= i & i <= (2|^n) & x = i/(2|^n) by Def3; take i; thus thesis by A1; end; uniqueness proof let i1,i2 be Nat; assume that A2:x = i1/(2|^n) and A3:x = i2/(2|^n); A4:(2|^n) <> 0 by HEINE:5; then i1 = i1/(2|^n)*(2|^n) by XCMPLX_1:88 .= i2 by A2,A3,A4,XCMPLX_1:88; hence thesis; end; end; theorem Th16: for n being Nat holds for x being Element of dyadic(n) holds x = axis(x,n)/(2|^n) & 0 <= axis(x,n) & axis(x,n) <= (2|^n) proof let n be Nat; let x be Element of dyadic(n); ex i being Nat st 0 <= i & i <= 2|^n & x = i/(2|^n) by Def3; hence thesis by Def7; end; theorem for n being Nat holds for x being Element of dyadic(n) holds (axis(x,n)-1)/(2|^n) < x & x < (axis(x,n)+1)/(2|^n) proof let n be Nat; let x be Element of dyadic(n); A1:x = axis(x,n)/(2|^n) by Def7; -1 + axis(x,n) < 0 + axis(x,n) & 0 + axis(x,n) < 1 + axis(x,n) by REAL_1:67; then A2:axis(x,n) - 1 < axis(x,n) & axis(x,n) < axis(x,n) + 1 by XCMPLX_0:def 8 ; 0 < 2|^n by HEINE:5; hence thesis by A1,A2,REAL_1:73; end; theorem Th18: for n being Nat holds for x being Element of dyadic(n) holds ((axis(x,n)-1)/(2|^n) < (axis(x,n)+1)/(2|^n)) proof let n be Nat; let x be Element of dyadic(n); -1 + axis(x,n) < 1 + axis(x,n) by REAL_1:67; then A1:axis(x,n) -1 < axis(x,n) + 1 by XCMPLX_0:def 8; 0 < 2|^n by HEINE:5; hence thesis by A1,REAL_1:73; end; canceled; theorem Th20: for n being Nat holds for x being Element of dyadic(n+1) holds (not x in dyadic(n) implies ((axis(x,n+1)-1)/(2|^(n+1)) in dyadic(n) & (axis(x,n+1)+1)/(2|^(n+1)) in dyadic(n))) proof let n be Nat; let x be Element of dyadic(n+1); assume A1:not x in dyadic(n); thus (axis(x,n+1)-1)/(2|^(n+1)) in dyadic(n) proof consider a being Real such that A2:a = (axis(x,n+1)-1)/(2|^(n+1)); ex i being Nat st (0 <= i & i <= (2|^n) & a = i/(2|^n)) proof consider s being Nat such that A3:0 <= s & s <= (2|^(n+1)) & x = s/(2|^(n+1)) by Def3; consider k being Nat such that A4:s = 2 * k or s = 2 * k + 1 by SCHEME1:1; now per cases by A4; case s = k * 2; then 0 <= k * 2 & k * 2 <= (2|^n)* 2 & x = (k * 2)/((2|^n)* 2) by A3,NEWTON:11; then 0 <= k & k <= ((2|^n)* 2)/2 & x = k/(2|^n) by REAL_2:177,SQUARE_1:25,XCMPLX_1:92; then 0 <= k & k <= (2|^n)*(2/2) & x = k/(2|^n) by XCMPLX_1:75; hence thesis by A1,Def3; case A5:s = k * 2 + 1; then A6:a = (k * 2 + 1-1)/(2|^(n+1)) by A2,A3,Def7 .= (k * 2 + 1 + (-1))/(2|^(n+1)) by XCMPLX_0:def 8 .= (k * 2 + (1 + (-1)))/(2|^(n+1)) by XCMPLX_1:1 .= k * 2 /((2|^n)* 2) by NEWTON:11 .= (k/(2|^n))*(2/2) by XCMPLX_1:77 .= k/(2|^n); take k; 0 <= k & k <= (2|^n) proof thus 0 <= k by NAT_1:18; A7: k * 2 <= (2|^(n+1)) - 1 by A3,A5,REAL_1:84; (2|^(n+1)) - 1 <= (2|^(n+1)) by REAL_2:174; then k * 2 <= (2|^(n+1)) by A7,AXIOMS:22; then k * 2 <= (2|^n)* 2 by NEWTON:11; then k <= ((2|^n)* 2)/2 & 2 <> 0 by REAL_2:177; then k <= (2|^n)*(2/2) by XCMPLX_1:75; hence thesis; end; hence thesis by A6; end; hence thesis; end; hence thesis by A2,Def3; end; thus (axis(x,n+1)+1)/(2|^(n+1)) in dyadic(n) proof consider a being Real such that A8:a = (axis(x,n+1)+1)/(2|^(n+1)); ex i being Nat st (0 <= i & i <= (2|^n) & a = i/(2|^n)) proof consider s being Nat such that A9:0 <= s & s <= (2|^(n+1)) & x = s/(2|^(n+1)) by Def3; consider k being Nat such that A10:s = 2 * k or s = 2 * k + 1 by SCHEME1:1; now per cases by A10; case s = k * 2; then 0 <= k * 2 & k * 2 <= (2|^n)* 2 & x = (k * 2)/((2|^n)* 2) by A9,NEWTON:11; then 0 <= k & k <= ((2|^n)* 2)/2 & x = k/(2|^n) by REAL_2:177,SQUARE_1:25,XCMPLX_1:92; then 0 <= k & k <= (2|^n)*(2/2) & x = k/(2|^n) by XCMPLX_1:75; hence thesis by A1,Def3; case A11:s = k * 2 + 1; consider l being Nat such that A12:l = k + 1; A13: s <> (2|^(n+1)) proof assume A14:s = (2|^(n+1)); 2|^(n+1) <> 0 by HEINE:5; then x = 1 by A9,A14,XCMPLX_1:60; hence thesis by A1,Th12; end; A15:a = (k * 2 + 1 + 1)/(2|^(n+1)) by A8,A9,A11,Def7 .= (k * 2 + (1 + 1))/(2|^(n+1)) by XCMPLX_1:1 .= (k * 2 + 2 * 1)/(2|^(n+1)) .= (k + 1 )* 2 /(2|^(n+1)) by XCMPLX_1:8 .= (k + 1) * 2 /((2|^n)* 2) by NEWTON:11 .= ((k + 1)/(2|^n))*(2/2) by XCMPLX_1:77 .= l/(2|^n) by A12; take l; 0 <= l & l <= 2|^n proof k * 2 + 1 + (1 + (-1)) < (2|^(n+1)) by A9,A11,A13,REAL_1:def 5; then (k * 2 + 1 + 1) + (-1) < (2|^(n+1)) by XCMPLX_1:1; then k * 2 + (1 + 1) + (-1) < (2|^(n+1)) by XCMPLX_1:1; then k * 2 + 1*2 + (-1) < (2|^(n+1)); then l * 2 + (-1) < (2|^(n+1)) by A12,XCMPLX_1:8; then l * 2 + (-1) + 1 < (2|^(n+1)) + 1 by REAL_1:53; then l * 2 + ((-1) + 1) < (2|^(n+1)) + 1 by XCMPLX_1:1; then l * 2 <= (2|^(n+1)) by NAT_1:38; then l * 2 <= (2|^n)* 2 by NEWTON:11; then l <= ((2|^n)* 2)/2 & 2 <> 0 by REAL_2:177; then l <= (2|^n)*(2/2) by XCMPLX_1:75; hence thesis by NAT_1:18; end; hence thesis by A15; end; hence thesis; end; hence thesis by A8,Def3; end; end; theorem Th21: for n being Nat holds for x1,x2 being Element of dyadic(n) st x1 < x2 holds axis(x1,n) < axis(x2,n) proof let n be Nat; let x1,x2 be Element of dyadic(n); A1: x1 = axis(x1,n)/(2|^n) & x2 = axis(x2,n)/(2|^n) by Th16; 0 < (2|^n) by HEINE:5; hence thesis by A1,REAL_1:73; end; theorem Th22: for n being Nat holds for x1,x2 being Element of dyadic(n) st x1 < x2 holds x1 <= (axis(x2,n)-1)/(2|^n) & (axis(x1,n)+1)/(2|^n) <= x2 proof let n be Nat; let x1,x2 be Element of dyadic(n); assume A1:x1 < x2; then axis(x1,n) < axis(x2,n) by Th21; then axis(x1,n) + 1 <= axis(x2,n) by NAT_1:38; then A2:axis(x1,n) <= axis(x2,n) - 1 by REAL_1:84; 0 < (2|^n) by HEINE:5; then A3: axis(x1,n)/(2|^n) <= (axis(x2,n) - 1)/(2|^n) by A2,REAL_1:73; axis(x1,n) < axis(x2,n) by A1,Th21; then A4:axis(x1,n) + 1 <= axis(x2,n) by NAT_1:38; 0 < (2|^n) by HEINE:5; then (axis(x1,n)+ 1)/(2|^n) <= axis(x2,n)/(2|^n) by A4,REAL_1:73; hence thesis by A3,Th16; end; theorem Th23: for n being Nat holds for x1,x2 being Element of dyadic(n+1) st x1 < x2 & not x1 in dyadic(n) & not x2 in dyadic(n) holds (axis(x1,n+1)+1)/(2|^(n+1)) <= (axis(x2,n+1)-1)/(2|^(n+1)) proof let n be Nat; let x1,x2 be Element of dyadic(n+1); assume A1:x1 < x2 & not x1 in dyadic(n) & not x2 in dyadic(n); consider k1 being Nat such that A2:axis(x1,n+1) = 2 * k1 or axis(x1,n+1) = 2 * k1 + 1 by SCHEME1:1; consider k2 being Nat such that A3:axis(x2,n+1) = 2 * k2 or axis(x2,n+1) = 2 * k2 + 1 by SCHEME1:1; A4:not axis(x1,n+1) = k1 * 2 proof assume axis(x1,n+1) = k1 * 2; then A5:x1 = (k1 * 2)/(2|^(n+1)) & 0 <= k1 * 2 & k1 * 2 <= (2|^(n+1)) by Th16; then A6:x1 = (k1 * 2)/((2|^n) * 2) by NEWTON:11 .= (k1/(2|^n))*(2/2) by XCMPLX_1:77 .= k1/(2|^n); k1 * 2 <= (2|^n) * 2 by A5,NEWTON:11; then k1 <= ((2|^n)* 2)/2 by REAL_2:177; then k1 <= (2|^n)*(2/2) by XCMPLX_1:75; then 0 <= k1 & k1 <= (2|^n) by NAT_1:18; hence thesis by A1,A6,Def3; end; A7:axis(x2,n+1) <> k2 * 2 proof assume axis(x2,n+1) = k2 * 2; then A8:x2 = (k2 * 2)/(2|^(n+1)) & 0 <= k2 * 2 & k2 * 2 <= (2|^(n+1)) by Th16; then A9:x2 = (k2 * 2)/((2|^n) * 2) by NEWTON:11 .= (k2/(2|^n))*(2/2) by XCMPLX_1:77 .= k2/(2|^n); k2 * 2 <= (2|^n) * 2 by A8,NEWTON:11; then k2 <= ((2|^n)* 2)/2 by REAL_2:177; then k2 <= (2|^n)*(2/2) by XCMPLX_1:75; then 0 <= k2 & k2 <= (2|^n) by NAT_1:18; hence thesis by A1,A9,Def3; end; A10: (2|^n) <> 0 & 0 < (2|^(n+1)) by HEINE:5; k1 * 2 + 1 < k2 * 2 + 1 by A1,A2,A3,A4,A7,Th21; then k1 * 2 + 1 + (-1) < k2 * 2 + 1 + (-1) & k1 * 2 + 1 + (-1) = k1 * 2 + (1 + (-1)) & k2 * 2 + 1 + (-1) = k2 * 2 + (1 + (-1)) by REAL_1:53,XCMPLX_1:1; then (k1 * 2)/2 < (k2 * 2)/2 & (k1 * 2)/2 = k1 * (2/2) & (k2 * 2)/2 = k2 * (2/2) by REAL_1:73,XCMPLX_1:75; then k1 + 1 <= k2 by NAT_1:38; then (k1 + 1) * 2 <= k2 * 2 by AXIOMS:25; then k1 * 2 + 1 * 2 <= k2 * 2 + 0 by XCMPLX_1:8; then k1 * 2 + 2 <= k2 * 2 + (1 + (-1)); then k1 * 2 + (1 + 1) <= k2 * 2 + 1 + (-1) by XCMPLX_1:1; then k1 * 2 + 1 + 1 <= k2 * 2 + 1 + (-1) by XCMPLX_1:1; then axis(x1,n+1) + 1 <= axis(x2,n+1) - 1 by A2,A3,A4,A7,XCMPLX_0:def 8; hence thesis by A10,REAL_1:73; end; begin :: :: Normal TopSpaces :: definition let T be non empty TopSpace; let x be Point of T; redefine mode a_neighborhood of x means ex A being Subset of T st A is open & x in A & A c= it; compatibility proof let S be Subset of T; hereby assume A1: S is a_neighborhood of x; take N = Int S; thus N is open; thus x in N by A1,CONNSP_2:def 1; thus N c= S by TOPS_1:44; end; assume ex A being Subset of T st A is open & x in A & A c= S; hence x in Int S by TOPS_1:54; end; synonym Nbhd of x,T; end; theorem Th24: for T being non empty TopSpace for A being Subset of T holds A is open iff (for x being Point of T holds x in A implies ex B being Nbhd of x,T st B c= A) proof let T be non empty TopSpace; let A be Subset of T; thus A is open implies for x being Point of T st x in A ex B being Nbhd of x,T st B c= A proof assume A1: A is open; let x be Point of T; assume x in A; then ex B being Subset of T st B is a_neighborhood of x & B c= A by A1,CONNSP_2:9; hence thesis; end; assume A2: for x being Point of T holds x in A implies ex B being Nbhd of x,T st B c= A; for x being Point of T st x in A ex B being Subset of T st B is a_neighborhood of x & B c= A proof let x be Point of T; assume x in A; then ex B being Nbhd of x,T st B c= A by A2; hence thesis; end; hence A is open by CONNSP_2:9; end; canceled; theorem for T being non empty TopSpace holds for A being Subset of T holds (for x being Point of T st x in A holds A is Nbhd of x,T) implies A is open proof let T be non empty TopSpace; let A be Subset of T; assume A1:for x being Point of T st x in A holds A is Nbhd of x,T; for x being Point of T holds x in A implies ex B being Nbhd of x,T st B c= A proof let x be Point of T; assume A2:x in A; ex B being Nbhd of x,T st B c= A proof A is Nbhd of x,T by A1,A2; then consider B being Nbhd of x,T such that A3:B = A; take B; thus thesis by A3; end; hence thesis; end; hence thesis by Th24; end; definition let T be TopStruct; attr T is being_T1 means :Def9:for p,q being Point of T st not p = q ex W,V being Subset of T st W is open & V is open & p in W & not q in W & q in V & not p in V; synonym T is_T1; end; theorem Th27: for T being non empty TopSpace holds T is_T1 iff for p being Point of T holds {p} is closed proof let T be non empty TopSpace; thus T is_T1 implies for p being Point of T holds {p} is closed proof assume A1:T is_T1; for p being Point of T holds {p} is closed proof let p be Point of T; A2:{p}` = [#](T) \ {p} by PRE_TOPC:17; consider B being Subset of T such that A3:B = {p}`; A4:for q being Point of T st q in B holds ex V being Subset of T st V is open & q in V & not p in V proof let q be Point of T; assume q in B; then q in [#](T) & not q in {p} by A2,A3,XBOOLE_0:def 4; then q in [#](T) & not q = p by TARSKI:def 1; then ex V,W being Subset of T st V is open & W is open & q in V & not p in V & p in W & not q in W by A1,Def9; then consider V being Subset of T such that A5:V is open & q in V & not p in V; take V; thus thesis by A5; end; defpred Q[Subset of T] means ex q being Point of T st (q in B & for V being Subset of T st $1 = V holds (V is open & q in V & not p in V )); consider F being Subset-Family of T such that A6:for C being Subset of T holds C in F iff Q[C] from SubFamExS; A7:for C being Subset of T holds (C in F iff (ex q being Point of T st (q in B & C is open & q in C & not p in C))) proof let C be Subset of T; A8:C in F implies (ex q being Point of T st (q in B & C is open & q in C & not p in C)) proof assume C in F; then consider q being Point of T such that A9:q in B & for V being Subset of T st C = V holds (V is open & q in V & not p in V ) by A6; take q; thus thesis by A9; end; (ex q being Point of T st (q in B & C is open & q in C & not p in C)) implies C in F proof assume A10:ex q being Point of T st (q in B & C is open & q in C & not p in C); ex q being Point of T st (q in B & for V being Subset of T st C = V holds (V is open & q in V & not p in V )) proof consider q being Point of T such that A11:q in B & C is open & q in C & not p in C by A10; take q; thus thesis by A11; end; hence thesis by A6; end; hence thesis by A8; end; A12:B = union F proof A13:B c= union F proof for x being set holds x in B implies x in union F proof let x be set; assume A14:x in B; then reconsider x as Point of T; consider C being Subset of T such that A15:C is open & x in C & not p in C by A4,A14; ex C being set st x in C & C in F proof take C; thus thesis by A7,A14,A15; end; hence thesis by TARSKI:def 4; end; hence thesis by TARSKI:def 3; end; union F c= B proof for x being set holds x in union F implies x in B proof let x be set; assume x in union F; then consider C being set such that A16:x in C & C in F by TARSKI:def 4; reconsider C as Subset of T by A16; consider q being Point of T such that A17:q in B & C is open & q in C & not p in C by A7,A16; C c= [#](T) by PRE_TOPC:14; then C c= [#](T) \ {p} by A17,ZFMISC_1:40; hence thesis by A2,A3,A16; end; hence thesis by TARSKI:def 3; end; hence thesis by A13,XBOOLE_0:def 10; end; F c= the topology of T proof for x being set holds x in F implies x in the topology of T proof let x be set; assume A18:x in F; then reconsider x as Subset of T; consider q being Point of T such that A19:q in B & x is open & q in x & not p in x by A7,A18; thus thesis by A19,PRE_TOPC:def 5; end; hence thesis by TARSKI:def 3; end; then B in the topology of T by A12,PRE_TOPC:def 1; then B is open by PRE_TOPC:def 5; hence thesis by A2,A3,PRE_TOPC:def 6; end; hence thesis; end; assume A20:for p being Point of T holds {p} is closed; for p,q being Point of T st not p = q ex W,V being Subset of T st W is open & V is open & p in W & not q in W & q in V & not p in V proof let p,q be Point of T; assume A21:not p = q; A22:{p}` = [#](T) \ {p} & {q}` = [#](T) \ {q} by PRE_TOPC:17; consider V,W being Subset of T such that A23:V = {p}`& W = {q}`; {p} is closed & {q} is closed by A20; then A24:V is open & W is open by A22,A23,PRE_TOPC:def 6; p in W & not q in W & q in V & not p in V proof A25:p in [#](T) & q in [#](T) by PRE_TOPC:13; A26: not p in {q} & not q in {p} by A21,TARSKI:def 1; p in {p} & q in {q} by TARSKI:def 1; hence thesis by A22,A23,A25,A26,XBOOLE_0:def 4; end; hence thesis by A24; end; hence thesis by Def9; end; definition let T be non empty TopSpace; let F be map of T,R^1; let x be Point of T; redefine func F.x -> Real; coherence by FUNCT_2:7,TOPMETR:24; end; theorem Th28: for T being non empty TopSpace st T is_T4 holds for A,B being open Subset of T st A <> {} & Cl(A) c=B holds ex C being Subset of T st C <> {} & C is open & Cl(A) c= C & Cl(C) c= B proof let T be non empty TopSpace; assume A1:T is_T4; let A,B be open Subset of T; assume A2: A <> {} & Cl(A) c=B; now per cases; case A3:B <> [#](T); reconsider W = Cl(A), V = [#](T) \ B as Subset of T; B = [#](T) \ V by PRE_TOPC:22; then A4:V is closed by PRE_TOPC:def 6; A5:W <> {} & V <> {} proof consider x being set such that A6:x in A by A2,XBOOLE_0:def 1; A7: A c= Cl(A) by PRE_TOPC:48; [#](T) \ B <> {} proof assume [#](T) \ B = {}; then B = [#](T) \ {} by PRE_TOPC:22; hence thesis by A3; end; hence thesis by A6,A7; end; W misses V proof assume W meets V; then consider x being set such that A8:x in W /\ V by XBOOLE_0:4; x in Cl(A) & x in [#](T) \ B by A8,XBOOLE_0:def 3; hence thesis by A2,XBOOLE_0:def 4; end; then consider C,Q being Subset of T such that A9:C is open & Q is open & W c= C & V c= Q & C misses Q by A1,A4,A5,COMPTS_1:def 6; take C; C <> {} & Cl(A) c= C & Cl(C) c= B proof consider x being set such that A10:x in W by A5,XBOOLE_0:def 1; A11:Q` is closed by A9,TOPS_1:30; Q c= [#](T) & C c= [#](T) by PRE_TOPC:14; then consider Q0,C0 being Subset of [#](T) such that A12:Q0 = Q & C0 = C; A13:C0 c= Q0` by A9,A12,SUBSET_1:43; C c= Q` proof for x being set holds x in C implies x in Q` proof let x be set; assume x in C; then x in Q0` by A12,A13; then x in [#](T) \ Q0 by SUBSET_1:def 5; hence thesis by A12,PRE_TOPC:17; end; hence thesis by TARSKI:def 3; end; then Cl(C) c= Q` by A11,TOPS_1:31; then Cl(C) misses Q by PRE_TOPC:21; then V misses Cl(C) by A9,XBOOLE_1:63; then A14: (B`) misses Cl(C) by PRE_TOPC:17; B`` = [#](T) \ B` by PRE_TOPC:17 .= [#](T) \ ([#](T) \ B) by PRE_TOPC:17 .= B by PRE_TOPC:22; hence thesis by A9,A10,A14,PRE_TOPC:21; end; hence thesis by A9; case A15:B = [#](T); consider C being Subset of T such that A16:C = [#](T); take C; C <> {} & C is open & Cl(A) c= C & Cl(C) c= B by A15,A16,PRE_TOPC:14 ; hence thesis; end; hence thesis; end; theorem for T being non empty TopSpace holds T is_T3 iff for A being open Subset of T for p being Point of T st p in A holds ex B being open Subset of T st p in B & Cl(B) c= A proof let T be non empty TopSpace; thus T is_T3 implies (for A being open Subset of T for p being Point of T st p in A holds ex B being open Subset of T st p in B & Cl(B) c= A) proof assume A1:T is_T3; thus for A being open Subset of T for p being Point of T st p in A holds ex B being open Subset of T st p in B & Cl(B) c= A proof let A be open Subset of T; let p be Point of T; assume A2:p in A; thus ex B being open Subset of T st p in B & Cl(B) c= A proof reconsider P = A` as Subset of T; now per cases; case A3:P <> {}; A = [#](T) \ ([#](T) \ A) by PRE_TOPC:22 .= [#](T) \ P by PRE_TOPC:17; then not p in P & P is closed by A2,XBOOLE_0:def 4; then consider W,V being Subset of T such that A4:W is open & V is open & p in W & P c= V & W misses V by A1,A3,COMPTS_1:def 5; take W; Cl(W) c= A proof W /\ V = {} by A4,XBOOLE_0:def 7; then V /\ Cl(W) c= Cl({} T) by A4,TOPS_1:40; then V /\ Cl(W) c= {} by PRE_TOPC:52; then V /\ Cl(W) = {} by XBOOLE_1:3; then V misses Cl(W) by XBOOLE_0:def 7; then A5: P misses Cl(W) by A4,XBOOLE_1:63; A`` = [#](T) \ A` by PRE_TOPC:17 .= [#](T) \ ([#](T) \ A) by PRE_TOPC:17 .= A by PRE_TOPC:22; hence thesis by A5,PRE_TOPC:21; end; hence thesis by A4; case P = {}; then [#](T) \ A = {} by PRE_TOPC:17; then A6:A = [#](T) by PRE_TOPC:23; take A; Cl(A) c= A by A6,TOPS_1:27; hence thesis by A2; end; hence thesis; end; end; end; assume A7:for A being open Subset of T for p being Point of T st p in A holds ex B being open Subset of T st p in B & Cl(B) c= A; thus T is_T3 proof for p being Point of T for P being Subset of T st P <> {} & P is closed & not p in P ex W,V being Subset of T st W is open & V is open & p in W & P c= V & W misses V proof let p be Point of T; let P be Subset of T; assume A8:P <> {} & P is closed & not p in P; thus ex W,V being Subset of T st W is open & V is open & p in W & P c= V & W misses V proof consider A being Subset of T such that A9:A = P`; A10:A is open by A8,A9,TOPS_1:29; A11:A =[#](T) \ P by A9,PRE_TOPC:17; p in the carrier of T; then p in [#](T) & not p in P by A8,PRE_TOPC:12; then p in A by A11,XBOOLE_0:def 4; then consider B being open Subset of T such that A12:p in B & Cl(B) c= A by A7,A10; consider C being Subset of T such that A13:C = [#](T) \ Cl(B); reconsider B,C as Subset of T; A14: (Cl(B))` is open; take B; take C; p in B & P c= C & B misses C proof thus p in B by A12; P`` = [#](T) \ P` by PRE_TOPC:17 .= [#](T) \ ([#](T) \ P) by PRE_TOPC:17 .= P by PRE_TOPC:22; then A15: P = [#](T) \ A by A9,PRE_TOPC:17; B c= Cl(B) & Cl(B) misses C by A13,PRE_TOPC:48,XBOOLE_1:79; hence thesis by A12,A13,A15,XBOOLE_1:34,63; end; hence thesis by A13,A14,PRE_TOPC:17; end; end; hence thesis by COMPTS_1:def 5; end; end; theorem for T being non empty TopSpace st T is_T4 & T is_T1 holds for A being open Subset of T st A <> {} holds ex B being Subset of T st B <> {} & Cl(B) c= A proof let T be non empty TopSpace; assume A1:T is_T4 & T is_T1; let A be open Subset of T; assume A2:A <> {}; now per cases; case A3:A <> [#](T); consider x being set such that A4:x in A by A2,XBOOLE_0:def 1; reconsider x as Point of T by A4; consider W being set such that A5:W = {x}; reconsider W as Subset of T by A5; A6:W is closed by A1,A5,Th27; reconsider V = [#](T) \ A as Subset of T; A = [#](T) \ V by PRE_TOPC:22; then A7:V is closed by PRE_TOPC:def 6; A8: [#](T) \ A <> {} proof assume [#](T) \ A = {}; then A = [#](T) \ {} by PRE_TOPC:22; hence thesis by A3; end; W misses V proof assume W meets V; then consider z being set such that A9:z in W /\ V by XBOOLE_0:4; z in W & z in V by A9,XBOOLE_0:def 3; then z in A & not z in A by A4,A5,TARSKI:def 1,XBOOLE_0:def 4; hence thesis; end; then consider B,Q being Subset of T such that A10:B is open & Q is open & W c= B & V c= Q & B misses Q by A1,A5,A6,A7,A8,COMPTS_1:def 6; take B; B <> {} & Cl(B) c= A proof consider x being set such that A11:x in W by A5,XBOOLE_0:def 1; A12:Q` is closed by A10,TOPS_1:30; B c= Q` by A10,PRE_TOPC:21; then Cl(B) c= Q` by A12,TOPS_1:31; then Cl(B) misses Q by PRE_TOPC:21; then V misses Cl(B) by A10,XBOOLE_1:63; then A13: (A`) misses Cl(B) by PRE_TOPC:17; A`` = [#](T) \ A` by PRE_TOPC:17 .= [#](T) \ ([#](T) \ A) by PRE_TOPC:17 .= A by PRE_TOPC:22; hence thesis by A10,A11,A13,PRE_TOPC:21; end; hence thesis; case A14:A = [#](T); consider B being Subset of T such that A15:B = [#](T); take B; B <> {} & B is open & Cl(B) c= A by A14,A15,PRE_TOPC:14; hence thesis; end; hence thesis; end; theorem for T being non empty TopSpace st T is_T4 for A,B being Subset of T st A is open & B is closed & B <> {} & B c= A ex C being Subset of T st C is open & B c= C & Cl(C) c= A proof let T be non empty TopSpace; assume A1:T is_T4; let A,B be Subset of T such that A2:A is open and A3:B is closed & B <> {} & B c= A; per cases; suppose A4:A <> [#](T); reconsider V = [#](T) \ A as Subset of T; A5: A = [#](T) \ V by PRE_TOPC:22; then A6:V is closed by A2,PRE_TOPC:def 6; A7: [#](T) \ A <> {} by A4,A5; B misses V proof assume B /\ V <> {}; then consider z being set such that A8:z in B /\ V by XBOOLE_0:def 1; z in B & z in V by A8,XBOOLE_0:def 3; hence thesis by A3,XBOOLE_0:def 4; end; then consider C,Q being Subset of T such that A9:C is open & Q is open & B c= C & V c= Q & C misses Q by A1,A3,A6,A7,COMPTS_1:def 6; take C; thus C is open & B c= C by A9; A10:Q` is closed by A9,TOPS_1:30; C c= Q` by A9,PRE_TOPC:21; then Cl(C) c= Q` by A10,TOPS_1:31; then V c= Q & Q misses Cl(C) by A9,PRE_TOPC:21; then V misses Cl(C) by XBOOLE_1:63; then A11: (A`) misses Cl(C) by PRE_TOPC:17; A`` = A; hence Cl(C) c= A by A11,PRE_TOPC:21; suppose A12:A = [#](T); take [#](T); thus thesis by A12,PRE_TOPC:14; end; begin :: :: Some increasing family of sets in normal space :: definition let T be non empty TopSpace; let A,B be Subset of T; assume A1:T is_T4 & A <> {} & A is open & B is open & Cl(A) c= B; mode Between of A,B -> Subset of T means :Def10: it <> {} & it is open & Cl(A) c= it & Cl(it) c= B; existence by A1,Th28; end; theorem for T being non empty TopSpace st T is_T4 for A,B being closed Subset of T st A <> {} & A misses B for n being Nat for G being Function of dyadic(n),bool the carrier of T st A c= G.0 & B = [#](T) \ G.1 & for r1,r2 being Element of dyadic(n) st r1 < r2 holds G.r1 is open & G.r2 is open & Cl(G.r1) c= G.r2 ex F being Function of dyadic(n+1),bool the carrier of T st A c= F.0 & B = [#](T) \ F.1 & for r1,r2,r being Element of dyadic(n+1) st r1 < r2 holds F.r1 is open & F.r2 is open & Cl(F.r1) c= F.r2 & (r in dyadic(n) implies F.r = G.r) proof let T be non empty TopSpace such that A1:T is_T4; let A,B be closed Subset of T such that A2: A <> {} & A /\ B = {}; let n be Nat; let G be Function of dyadic(n),bool the carrier of T such that A3: A c= G.0 & B = [#](T) \ G.1 and A4: for r1,r2 being Element of dyadic(n) st r1 < r2 holds G.r1 is open & G.r2 is open & Cl(G.r1) c= G.r2; A5: ex x being set st x in A by A2,XBOOLE_0:def 1; A6: 0 in dyadic(n) & 1 in dyadic(n) by Th12; A7:for r being Element of dyadic(n) holds G.r <> {} proof let r be Element of dyadic(n); per cases by Th5; suppose 0 = r; hence thesis by A3,A5; suppose A8:0 < r; reconsider q1 = 0 as Element of dyadic(n) by Th12; A c= G.q1 & G.q1 c= Cl(G.q1) & Cl(G.q1) c= G.r by A3,A4,A8,PRE_TOPC:48; then A c= Cl(G.q1) & Cl(G.q1) c= G.r by XBOOLE_1:1; then A c= G.r by XBOOLE_1:1; hence thesis by A5; end; reconsider S = dyadic(n+1) \ dyadic(n) as non empty set by Th15; defpred P[Element of S,Element of bool the carrier of T] means for r being Element of dyadic(n+1) st r in S & $1 = r holds for r1,r2 being Element of dyadic(n) st r1 = (axis(r,n+1)-1)/(2|^(n+1)) & r2 = (axis(r,n+1)+1)/(2|^(n+1)) holds $2 is Between of G.r1, G.r2; A9:for x being Element of S ex y being Element of bool the carrier of T st P[x,y] proof let x be Element of S; A10:x in dyadic(n+1) & not x in dyadic(n) by XBOOLE_0:def 4; reconsider x as Element of dyadic(n+1) by XBOOLE_0:def 4; (axis(x,n+1)-1)/(2|^(n+1)) in dyadic(n) & (axis(x,n+1)+1)/(2|^(n+1)) in dyadic(n) by A10,Th20; then consider q1,q2 being Element of dyadic(n) such that A11:q1 = (axis(x,n+1)-1)/(2|^(n+1)) and A12:q2 = (axis(x,n+1)+1)/(2|^(n+1)); consider Q0 being Between of G.q1, G.q2; take Q0; thus thesis by A11,A12; end; consider D being Function of S,bool the carrier of T such that A13:for x being Element of S holds P[x,D.x] from FuncExD(A9); defpred W[Element of dyadic(n+1),Element of bool the carrier of T] means for r being Element of dyadic(n+1) st $1 = r holds ((r in dyadic(n) implies $2 = G.r) & (not r in dyadic(n) implies $2 = D.r)); A14:for x being Element of dyadic(n+1) ex y being Element of bool the carrier of T st W[x,y] proof let x be Element of dyadic(n+1); per cases; suppose x in dyadic(n); then reconsider x as Element of dyadic(n); consider y being Element of bool the carrier of T such that A15:y = G.x; take y; thus thesis by A15; suppose A16:not x in dyadic(n); then reconsider x as Element of S by XBOOLE_0:def 4; consider y being Element of bool the carrier of T such that A17:y = D.x; take y; thus thesis by A16,A17; end; consider F being Function of dyadic(n+1),bool the carrier of T such that A18:for x being Element of dyadic(n+1) holds W[x,F.x] from FuncExD(A14); take F; 0 in dyadic(n+1) & 1 in dyadic(n+1) by Th12; hence A c= F.0 & B = [#](T) \ F.1 by A3,A6,A18; let r1,r2,r be Element of dyadic(n+1) such that A19: r1 < r2; thus F.r1 is open & F.r2 is open & Cl(F.r1) c= F.r2 proof now per cases; suppose A20:r1 in dyadic(n) & r2 in dyadic(n); then A21:F.r1 = G.r1 & F.r2 = G.r2 by A18; reconsider r1,r2 as Element of dyadic(n) by A20; r1 < r2 by A19; hence thesis by A4,A21; suppose A22:r1 in dyadic(n) & not r2 in dyadic(n); then A23:r2 in S by XBOOLE_0:def 4; A24:F.r2 = D.r2 by A18,A22; (axis(r2,n+1)-1)/(2|^(n+1)) in dyadic(n) & (axis(r2,n+1)+1)/(2|^(n+1)) in dyadic(n) by A22,Th20; then consider q1,q2 being Element of dyadic(n) such that A25:q1 = (axis(r2,n+1)-1)/(2|^(n+1)) & q2 = (axis(r2,n+1)+1)/(2|^(n+1)); D.r2 is Between of (G.q1),(G.q2) by A13,A23,A25; then consider D0 being Between of (G.q1),(G.q2) such that A26:D0 = D.r2; q1 < q2 by A25,Th18; then A27: G.q1 <> {} & G.q1 is open & G.q2 is open & Cl(G.q1) c= G.q2 by A4,A7; then A28: F.r2 <> {} & F.r2 is open & Cl(G.q1) c= F.r2 & Cl(F.r2) c= G.q2 by A1,A24,A26,Def10; A29: r1 <= q1 by A19,A25,Th22; now per cases by A29,AXIOMS:21; suppose r1 = q1; hence thesis by A18,A27,A28; suppose A30:r1 < q1; consider q0 being Element of dyadic(n) such that A31:q0 = r1 by A22; G.q0 is open & G.q1 is open & Cl(G.q0) c= G.q1 & A c= G.0 & B = [#](T) \ G.1 by A3,A4,A30,A31; then F.r1 is open & F.r2 is open & Cl(F.r1) c= G.q1 & G.q1 c= Cl(G.q1) & Cl(G.q1) c= F.r2 by A1,A18,A24,A26,A27,A31,Def10,PRE_TOPC:48; then F.r1 is open & F.r2 is open & Cl(F.r1) c= Cl(G.q1) & Cl(G.q1) c= F.r2 by XBOOLE_1:1; hence thesis by XBOOLE_1:1; end; hence thesis; suppose A32:not r1 in dyadic(n) & r2 in dyadic(n); then A33:r1 in S by XBOOLE_0:def 4; A34:F.r1 = D.r1 by A18,A32; (axis(r1,n+1)-1)/(2|^(n+1)) in dyadic(n) & (axis(r1,n+1)+1)/(2|^(n+1)) in dyadic(n) by A32,Th20; then consider q1,q2 being Element of dyadic(n) such that A35:q1 = (axis(r1,n+1)-1)/(2|^(n+1)) & q2 = (axis(r1,n+1)+1)/(2|^(n+1)); D.r1 is Between of (G.q1),(G.q2) by A13,A33,A35; then consider D0 being Between of (G.q1),(G.q2) such that A36:D0 = D.r1; q1 < q2 by A35,Th18; then A37:G.q1 <> {} & G.q1 is open & G.q2 is open & Cl(G.q1) c= G.q2 by A4,A7; then A38:F.r1 <> {} & F.r1 is open & Cl(G.q1) c= F.r1 & Cl(F.r1) c= G.q2 by A1,A34,A36,Def10; A39: q2 <= r2 by A19,A35,Th22; now per cases by A39,AXIOMS:21; suppose q2 = r2; hence thesis by A18,A37,A38; suppose A40:q2 < r2; consider q3 being Element of dyadic(n) such that A41:q3 = r2 by A32; G.q2 is open & G.q3 is open & Cl(G.q2) c= G.q3 & A c= G.0 & B = [#](T) \ G.1 by A3,A4,A40,A41; then F.r1 is open & F.r2 is open & Cl(F.r1) c= G.q2 & G.q2 c= Cl(G.q2) & Cl(G.q2) c= F.r2 by A1,A18,A34,A36,A37,A41,Def10,PRE_TOPC:48; then F.r1 is open & F.r2 is open & Cl(F.r1) c= Cl(G.q2) & Cl(G.q2) c= F.r2 by XBOOLE_1:1; hence thesis by XBOOLE_1:1; end; hence thesis; suppose A42:not r1 in dyadic(n) & not r2 in dyadic(n); then A43:r1 in S by XBOOLE_0:def 4; A44:F.r1 = D.r1 by A18,A42; (axis(r1,n+1)-1)/(2|^(n+1)) in dyadic(n) & (axis(r1,n+1)+1)/(2|^(n+1)) in dyadic(n) by A42,Th20; then consider q11,q21 being Element of dyadic(n) such that A45:q11 = (axis(r1,n+1)-1)/(2|^(n+1)) & q21 = (axis(r1,n+1)+1)/(2|^(n+1)); D.r1 is Between of (G.q11),(G.q21) by A13,A43,A45; then consider D01 being Between of (G.q11),(G.q21) such that A46:D01 = D.r1; q11 < q21 by A45,Th18; then A47: G.q11 <> {} & G.q11 is open & G.q21 is open & Cl(G.q11) c= G.q21 by A4,A7; A48:r2 in S by A42,XBOOLE_0:def 4; A49:F.r2 = D.r2 by A18,A42; (axis(r2,n+1)-1)/(2|^(n+1)) in dyadic(n) & (axis(r2,n+1)+1)/(2|^(n+1)) in dyadic(n) by A42,Th20; then consider q12,q22 being Element of dyadic(n) such that A50:q12 = (axis(r2,n+1)-1)/(2|^(n+1)) & q22 = (axis(r2,n+1)+1)/(2|^(n+1)); D.r2 is Between of (G.q12),(G.q22) by A13,A48,A50; then consider D02 being Between of (G.q12),(G.q22) such that A51:D02 = D.r2; q12 < q22 by A50,Th18; then A52: G.q12 <> {} & G.q12 is open & G.q22 is open & Cl(G.q12) c= G.q22 by A4,A7; hence F.r1 is open & F.r2 is open by A1,A44,A46,A47,A49,A51,Def10; A53: q21 <= q12 by A19,A42,A45,A50,Th23; now per cases by A53,AXIOMS:21; suppose q21 = q12; then Cl(F.r1) c= G.q21 & G.q21 c= Cl(G.q21) & Cl(G.q21) c= F.r2 by A1,A44,A46,A47,A49,A51,A52,Def10,PRE_TOPC:48; then Cl(F.r1) c= Cl(G.q21) & Cl(G.q21) c= F.r2 by XBOOLE_1:1 ; hence Cl(F.r1) c= F.r2 by XBOOLE_1:1; suppose q21 < q12; then Cl(F.r1) c= G.q21 & G.q21 c= Cl(G.q21) & G.q12 c= Cl(G.q12) & Cl(G.q21) c= G.q12 & Cl(G.q12) c= F.r2 by A1,A4,A44,A46,A47,A49,A51,A52,Def10,PRE_TOPC:48; then Cl(F.r1) c= Cl(G.q21) & Cl(G.q21) c= G.q12 & G.q12 c= F.r2 by XBOOLE_1:1; then Cl(F.r1) c= G.q12 & G.q12 c= F.r2 by XBOOLE_1:1; hence Cl(F.r1) c= F.r2 by XBOOLE_1:1; end; hence Cl(F.r1) c= F.r2; end; hence thesis; end; thus thesis by A18; end;