FINITE BOOLEAN ALGEBRA. 1. Deconstructing Boolean algebras with atoms. Let B = <B,,,,,0,1> be a Boolean algebra and c B.

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1 FINITE BOOLEAN ALGEBRA 1. Decnstructing Blean algebras with atms. Let B = <B,,,,,0,1> be a Blean algebra and c B. The ideal generated by c, (c], is: (c] = {b B: b c} The filter generated by c, [c), is: [c) = {b B: c b} The ideal-relativizatin f B t c, (c], is the structure: (c] = <(c], (c], (c], (c], (c],0 (c],1 (c] >, where: 1. (c] = (c] 2. (c] = {<x, x c>: x (c]} 3. (c] = (c] 4. (c] = (c] 5. 0 (c] = (c] = c THEOREM 1: (c] is a Blean algebra. 1. <(c], (c], (c], (c],0 (c] > is a substructure f <B,,,,0>: namely, if a,b (c], then a,b c, hence (a b) c and (a b) c, hence (a b) (c] and (a b) (c]. Further 0 (c] = <(c], (c], (c], (c],0 (c],1 (c] > is bunded by 0 (c] and 1 (c]. We have seen already that 0 (c] is the minimum f (c]. 1 (c] = c, and c is, bviusly, the maximum f (c]. 3. We have prved under 1. that <(c], (c], (c], (c] > is a sublattice f <B,,, >. Since <B,,, > is distributive, it fllws that <(c], (c], (c], (c] > is distributive (since the class f distributive lattices is clsed under substructure). 4. Thus, we have prved that (c] is a bunded distributive lattice. S we nly need t prve that (c] is cmplemented, i.e. that (c] is cmplementatin n (c]. First: (c] is clsed under (c]. This is bvius, since bviusly x c c, fr any x B, hence als fr any x c. Secndly, (c] respects the laws f 0 (c] and 1 (c] : Let a (c]. a (c] (c] (a) = a ( a c) = (a a) (a c) = 0 (a c) = 0 a (c] (c] (a) = a ( a c) = (a a) (a c) = 1 (a c) = a c = c Thus, indeed (c] is cmplementatin n (c]. Nte: except fr the trivial case where c is 1, (c] is nt a sub-blean algebra f B, because 1 is nt preserved. It is a Blean algebra n a subset f B. 1

2 The filter-relativizatin f B t c, [c), is the structure: [c) = <B [c), [c), [c), [c), [c),0 [c),1 [c) >, where: 1. [c) = B [c) 2. [c) = {<x, x c>: x [c)} 3. [c) = [c) 4. [c) = [c) 5. 0 [c) = c 6. 1 [c) = 1 THEOREM 2: [c) is a Blean algebra. 1. <[c), [c), [c), [c),1 [c) > is a substructure f <B,,,,1>: namely, if a,b [c), then c a,b, hence c (a b) and c (a b), hence (a b) [c) and (a b) [c). Further 1 [c) = <[c), [c), [c), [c),0 [c),1 [c) > is bunded by 0 [c) and 1 [c). We have seen already that 1 [c) is the maximum f [c). 0 [c) = c, and c is, bviusly, the mimimum f [c). 3. We have prved under 1. that <[c), [c), [c), [c)> is a sublattice f <B,,, >. Since <B,,, > is distributive, it fllws that <[c), [c), [c), [c)> is distributive (since the class f distributive lattices is clsed under substructure). 4. Thus, we have prved that [c) is a bunded distributive lattice. S we nly need t prve that [c) is cmplemented, i.e. that [c) is cmplementatin n [c). First: [c) is clsed under [c). This is bvius, since bviusly c x c, fr any x B, hence als fr any x such that c x. Secndly, [c) respects the laws f 0 [c) and 1 [c) : Let a [c). a [c) [c) (a) = a ( a c) = (a a) (a c) = 0 (a c) = a c = c a [c) [c) (a) = a ( a c) = (a a) (a c) = 1 (a c) = 1 Thus, indeed [c) is cmplementatin n [c). LEMMA 3: If c 1 then (c] [ c) = Ø Let x (c] [ c). Then x c and c x. Then x c and and x c. Then x x c, hence c = 1. THEOREM 4: (c] and [ c) are ismrphic. If c =1, then (c] = [ c) = B. S clearly they are ismrphic. S let c 1. We define: h: (c] [ c) by: fr every x (c]: h(x) = x c. 1. Since fr every x B, c x c, als fr every x (c]: c x c. Hence fr every x (c]: h(x) [ c), hence h is indeed a functin frm (c] int [ c). 2

3 2. Let y [ c). Then c y. Then y c, hence y (c]. Take the relative cmplement f y in (c]: y c. This is y c (c]. h(y c) = (y c) c = (y c) (c c) = (y c) 1 = y c = y. Hence h is nt. 3. Let h(x 1 ) = h(x 2 ). Then x 1 c = x 2 c. Then (x 1 c) = ( x 2 c). Then x 1 c = x 2 c. Since these are the relative cmplements f x 1 and x 2 in Blean algebra (c], it fllws that x 1 = x 2. Hence h is ne-ne. 4. h(0) = 0 c = c. h(c) = c c = 1 h(a b) = (a b) c = (a c) (b c) = h(a) h(b) h(a b) = (a b) c = (a c) (b c) = h(a) h(b) h( (c] (a)) = h( a c) = ( a c) c = (a c) c = [ c) (a c) = [ c) (h(a)). Thus, indeed, h is an ismrphism. LEMMA 5: If a is an atm in B, then fr every x B-{0}: a x r a x. Let a be an atm in B. Suppse that (a x). Then (a x) a. But a x a. Since a is is an atm, that means that a x = 0. But that means that a x. CORROLLARY 6: If a is an atm in B, then ( a] [a) = B. This fllws frm lemma 5: Let x B and x [a). Then (a x). Hence by lemma 5 a x, and that means that x a, hence x ( a]. All this has the fllwing cnsequence fr finite Blean algebras: THEOREM 7: Let B be a finite Blean algebra. Then B = 2 n, fr sme n 0. If B = 1 then B = 2 0. If B = 2 then B = 2 1. Let B > 2. We define fr B a decmpsitin tree DEC(B) in the fllwing way: tp(dec(b)) = <B,a 0,0>, with a 0 an atm in B. fr every nde <A,a,n> in DEC(B): if A >2 then daughters(<a,a,n>) = {<(a],a 1,n+1>,<[ a),a 2,n+1>}, with a 1 an atm in (a] and a 2 an atm in [ a). Let <A 1,a A1,k+1>, <A,a A,k> DEC(B) and let <A 1,a A1,k+1> be a daughter f <A,a A,k>. Then A = 2 A 1. Obviusly, this means that fr any nde <A,a A,k> DEC(B): B = 2 k A This means that if <A 1,a A1,k>, <A 2,a A2,k> DEC(B), then A 1 = A 2. 3

4 And that means that if <A 1,a A1,k>, <A 2,a A2,k> DEC(B), either bth A 1 and A 2 decmpse (if A 1 >2), r neither d (If a 1 2). Thus fr any k such that sme <A,a A,k> DEC(B): all ndes <A,a A,k> DEC(B) decmpse, r nne d. This means that fr sme k > 0: leave(dec(b)) = {<A,a A,k>: <A,a A,k> DEC(B)} (all leaves have the same level, and hence the same cardinality.) Let <A,a A,k> leave(dec(b). Then it fllws that B = 2 k A. Since <A,a A,k> leave(dec(b), A 2. Fr sme <C,a C,k-1> DEC(B), <A,a A,k> is the daughter f <C,a C,k-1>, hence C = 2 A, C > 2 and A 2. This means that A =2. Hence B = 2 k+1. Hence fr sme n>1: B = 2 n. We have nw prved that fr every finite Blean algebra B B = 2 n fr sme n 0. 4

5 2. Cnstructing prduct Blean algebras. Let A and B be Blean algebras. The prduct f A and B, A B, is given by: A B = <B,,,,, 0, 1 > where 1. B = A B 2. x = {<<a 1,b 1 >:,<a 2,b 2 >>: a 1 A a 2 and b 1 B b 2 } 3. Fr every <a,b> A B: (<a,b>) = < A a, B b> 4. Fr every <a 1,b 1 >,<a 2,b 2 > A B: <a 1,b 1 > <a 2,b 2 > = <a 1 A a 2,b 1 B b 2 > 5. Fr every <a 1,b 1 >,<a 2,b 2 > A B: <a 1,b 1 > <a 2,b 2 > = <a 1 A a 2,b 1 B b 2 > 6. 0 = <0 A,0 B > 7. 1 x = <1 A,1 B > THEOREM 8: A B is a Blean algebra. 1. is a partial rder. reflexive: Since fr every a A: a A a and fr every b B: b B b, fr every <a,b> A B: <a,b> x <a,b>. antisymmetric: Let <a 1,b 1 > <a 2,b 2 > and <a 2,b 2 > <a 1,b 2 >. Then a 1 A a 2 and b 1 B b 2 and a 2 A a 1 and b 2 B b 1, hence a 1 = a 2 and b 1 = b 2, hence <a 1,b 2 > = <a 2,b 2 > transitive: Let <a 1,b 1 > <a 2,b 2 > and <a 2,b 2 > <a 3,b 3 >. Then a 1 A a 2 and b 1 B b 2 and a 2 A a 3 and b 2 B b 3, hence a 1 A a 3 and b 1 B b 3, hence <a 1,b 1 > <a 3,b 3 > 2. <a 1,b 1 > <a 2,b 2 > = <a 1 A a 2,b 1 B b 2 > a 1 A a 2 A a 1, a 1 A a 2 A a 2, b 1 B b 2 B b 1, b 1 B b 2 B b 2, hence <a 1,b 1 > <a 2,b 2 > <a 1,b 1 > and <a 1,b 1 > <a 2,b 2 > <a 2,b 2 >. Let <a,b> <a 1,b 1 > and <a,b> <a 2,b 2 >. Then a A a 1 and b B b 1 and a A a 2 and b B b 2, hence a A a 1 A a 2 and b B b 1 B b 2, hence <a,b> <a 1,b 1 > <a 2,b 2 >. Hence is meet in. 3. We shw that is jin in by a similar argument = <0 A,0 B >. Since fr every a A 0 A A a and fr every b B 0 B B b, fr every <a,b> A B <0 A,0 B > <a,b>. Hence 0 is the minumum under. Similarly 1 is the maximum under A 5

6 S A B is a bunded lattice. 5. <a 1,b 1 > (<a 2,b 2 > <a 3,b 3 >) = <a 1 A (a 2 A a 3 ),b 1 B (b 2 B b 3 )> = <(a 1 A a 2 ) A (a 1 A a 3 ),(b 1 B b 2 ) B (b 1 B b 3 )> = (<a 1 A a 2,b 1 B b 2 > <a 1 A a 3,b 1 B b 3 > = (<a 1,b 1 > <a 2,b 2 >) (<a 1,b 1 > x <a 2,b 2 >) S A B is distributive. 6. satisfies the laws f 0 and 1 : <a,b> (<a,b>) = <a,b> < A a, B b> = <a A A a,b B B b> = <0 A,0 B > = 0. <a,b> (<a,b>) = <a,b> < A a, B b> = <a A A a,b B B b> = <1 A,1 B > = 1. S A B is a Blean algebra. Let B 1 and B 2 be ismrphic Blean algebras such that B 1 B 2 = Ø, and let h be an ismrphism between B 1 and B 2. We definite the prduct f B 1 and B 2 under h, B h 1+2: B h 1+2 = < B h 1+2, 1+2, 1+2, 1+2, 1+2,0 1+2,1 1+2 > where: 1. B 1+2 = B 1 B = 1 2 {<b 1,b 2 >: h(b 1 ) 2 b 2 } is defined by: 2 (h(b)) if b B 1 Fr all b B 1+2: 1+2 (b) = 1 (h -1 (b)) if b B is defined by: a 1 b if a,b B 1 Fr all a,b B 1+2 : a 1+2 b = a 2 b if a,b B 2 a 1 h -1 (b) if a B 1 and b B is defined by: a 1 b if a,b B 1 Fr all a,b B 1+2 : a 1+2 b = a 2 b if a,b B 2 h(a) 1 b if a B 1 and b B = =

7 THEOREM 9: B h 1+2 is a Blean algebra is a partial rder. 1+2 is reflexive: If a B 1 : a 1 a, hence, a 1+2 a If a B 2 : a 2 a, hence, a 1+2 a 1+2 is antisymmetric. Let a 1+2 b and b 1+2 a. This is nly pssible if a,b B 1 r a,b B 2. In the first case a 1 b and b 1 a, hence a=b. In the secnd case a 2 b and b 1 a, hence a=b. 1+2 is transitive. Let a 1+2 b and b 1+2 c If a,b,c B 1, then a 1 b and b 1 c, hence a 1 c, and a 1+2 c If a,b,c B 2, then a 2 b and b 2 c, hence a 2 c, and a 1+2 c If a B 1 and b,c B 2, then h(a) 2 b and b 2 c. Then h(a) 2 c and a 1+2 c. If a,b B 1 and c B 2, then a 1 b and h(b) 2 c. Since h is an ismrphism, this means that h(a) 2 h(b), and hence h(a) 2 c. Hence a 1+2 c is meet under 1+2. If a,b B 1 : a 1+2 b = a 1 b, which is meet under 1, and 1 = 1+2 B 1. If a,b B 2 : a 1+2 b = a 2 b, which is meet under 2, and 2 = 1+2 B 2. If a B 1 and b B 2, then a 1+2 b = a 1 h -1 (b). a 1 h -1 (b) 1 a and a 1 h -1 (b) 1 h -1 (b). By definitin f 1+2, h -1 (b) 1+2 h(h -1 (b)). S h -1 (b) 1+2 b. Then a 1 h -1 (b) 1+2 b. This means that a 1+2 b 1+2 a and a 1+2 b 1+2 b. If x 1+2 a and x 1+2 b, then x 1 a and h(x) 2 b. Since h is an ismrphism, then h -1 (h(x)) 1 h -1 (b), i.e. x 1 h -1 (b). then x 1 a 1 h -1 (b) Hence x 1+2 a 1+2 b. S indeed 1+2 is meet under is jin under 1+2. If a,b B 1 : a 1+2 b = a 1 b, which is jin under 1, and 1 = 1+2 B 1. If a,b B 2 : a 1+2 b = a 2 b, which is jin under 2, and 2 = 1+2 B 2. If a B 1 and b B 2, then a 1+2 b = h(a) 2 b. b 2 h(a) 2 b and h(a) 2 h(a) 2 b. As we have seen a 1+2 h(a), hence a 1+2 h(a) 2 b. S a 1+2 a 1+2 b and b 1+2 a 1+2 b. If a 1+2 x and b 1+2 x, then h(a) 1+2 x, hence h(a) 2 b 2 x. 7

8 Hence a 1+2 b 1+2 x. S indeed 1+2 is jin under = 0 1. If a B 1, a. hence a. h is an ismrphism, s h(0 1 ) = 0 2. If a B 2, then h(0 1 2 a, hence a. S indeed is the minimum under 1+2. Similarly, is the maximum under 1+2. We have prved s far that B 1+2 h is a bunded lattice. 5. Distributivity: a 1+2 (b 1+2 c) = (a 1+2 b) 1+2 (a 1+2 c) a. Let a,b,c B 1 r a,b,c B 2, then distributivity fllws frm distributivity f 1 and 1 and f 2 and 2. b. Let a B 1 and b,c B 2 a 1+2 (b 1+2 c) = a 1 h -1 (b 2 c) = a 1 (h -1 (b) 1 h -1 (c)) = (a 1 h -1 (b)) 1 (a 1 h -1 (c)) = (a 1+2 b) 1+2 (a 1+2 c) c. Let b B 1 and a,c B 2 a 1+2 (b 1+2 c) = a 2 (h(b) 2 c) = (a 2 h(b)) 2 (a 2 c) If a B 2 and b B 1, then a 1+2 b = h -1 (a) 1 b. Then h(a 1+2 b) = h(h -1 (a) 1 b) = h(h -1 (a)) 2 h(b) = a 2 h(b) Hence (a 2 h(b)) 2 (a 2 c) = h(a 1+2 b) 2 (a 2 c) = (a 1+2 b) 1+2 (a 1+2 c) d. Let b,c B 1 and a B 2 a 1+2 (b 1+2 c) = a 1+2 (b 1 c) = h -1 (a) 1 (b 1 c) = (b 1 h -1 (a)) 1 (c 1 h -1 (a)) = (a 1+2 b) 1+2 (a 1+2 c) e. Let a,b B 1 and c B 2 a 1+2 (b 1+2 c) = a 1 ( h -1 (b 1+2 c)) If b B 1 and c B 2, then b 1+2 c = h(b) 2 c Hence h -1 (b 1+2 c) = h -1 (h(b) 2 c) = h -1 (h(b)) 1 h -1 (c) = b 1 h -1 (c) S a 1 ( h -1 (b 1+2 c)) = a 1 (b 1 h -1 (c)) = (a 1 b) 1 (a 1 h -1 (c)) = (a 1+2 b) 1+2 (a 1+2 c). These are all the relevant cases, s B 1+2 h is a distributive bunded lattice satisfies the laws f and If a B 1, a (a) = a 1 h -1 ( 1+2 (a)) = a 1 h -1 ( 2 (h(a))) = a 1 h -1 (h( 1 a)) = a 1 1 a = 0 1 =

9 a (a) = h(a) (a) = h(a) 2 2 (h(a) = 1 2 = If a B 2, a (a) = h -1 (a) (a) = h -1 (a) 1 1 (h -1 (a) = 0 1 = a (a) = a 2 h( 1+2 (a)) = a 2 h( 1 (h -1 (a)) = a 2 2 (h(h -1 (a)) = a 2 2 a = 1 2 = Thus B 1+2 h is a Blean algebra. THEOREM 10:Let B 1 and B 2 be ismrphic Blean algebras such that B 1 B 2 = Ø, and let h be an ismrphism between B 1 and B 2.. Let {0,1} be a Blean algebra f cardinality 2. B 1+2 h is ismrphic t B 1 {0,1} We define functin k frm B 1 B 2 int B 1 {0,1): Fr all x B 1 : k(x) = <x,0> Fr all x B 2 : k(x) = <h -1 (x),1> 1. Since h is an ismrphism between B 1 and B 2, and B 1 B 2 = Ø, k is bviusly a bijectin between B 1 B 2 and B 1 {0,1). 2. If x B 1, k( 1+2 (x)) = <h -1 ( 1+2 (x)),1> = <h -1 ( 2 (h(x)), {0,1} 0> = <h -1 (h( 1 (x))), {0,1} 0> = < 1 (x), {0,1} 0> = <x,0> 3. k(0 1+2 ) = k(0 1 ) = <0 1,0> = 0. k(1 1+2 ) = k(1 2 ) = <h -1 (1 2 ),1> = <1 1,1> = k preserves meet: If a,b B 1 then k(a 1+2 b) = k(a 1 b) = <a 1 b,0> = <a,0> <b,0> = k(a) k(b). If a,b B 2 then k(a 1+2 b) = k(a 2 b) = <h -1 (a 2 b),1> = <h -1 (a) 1 h -1 (b),1> = <h -1 (a),1> <h -1 (b),1> = k(a) k(b). If a B 1 and b B 2 then k(a 1+2 b) = k(a 1 h -1 (b)) = <a 1 h -1 (b),0> = <a 1 h -1 (b),0 {0,1} 1> <a,0> <h -1 (b),1> = k(a) k(b). 5. k preserves jin: If a,b B 1 then k(a 1+2 b) = k(a 1 b) = <a 1 b,0> = <a,0> <b,0> = k(a) k(b). If a,b B 2 then k(a 1+2 b) = k(a 2 b) = <h -1 (a 2 b),1> = <h -1 (a) 1 h -1 (b),1> = <h -1 (a),1> <h -1 (b),1> = k(a) k(b). 9

10 If a B 1 and b B 2 then k(a 1+2 b) = k(h(a) 2 b) = <h -1 (h(a) 2 b),1> = <a 1 h -1 (b),1> = <a 1 h -1 (b),0 {0,1} 1> = <a,0> <h -1 (b),1> = k(a) k(b). Thus indeed k is an ismrphism. THEOREM 11: Let B be a Blean algebra f cardinality larger than 2 and let a be an atm in B. Let h: (a] [ a) be the ismrphism defined by: fr every x (a]: h(x) = x a. Then B (a]+[ a) h = B. 1. B (a]+[ a) = (a]+[ a) = B. 2. (a]+[ a) = B. a. Let <x,y> (a]+[ a). Either x,y (a], then x B y, r x,y [ a), then x B y, r x (a] and y [ a) and h(x) [ a) y. Since h(x) = x B a, then bviusly x B y. S in all cases <x,y> B. b. Let <x,y> B, i.e. x B y. Either x,y (a], then <x,y> (a]+[ a), r x,y [ a), then <x,y> (a]+[ a). It can't be the case that y (a] and x [ a), because then y [ a), but then the intersectin f (a] and [ a) wuld nt be empty, and it is. This leaves nly the case that x (a] and y [ a). Nw x B y and a B y, hence x B a B y. But h(x) = x B a. Hence <x,y> (a]+[ a). Thus, indeed (a]+[ a) = B. This means that B (a]+[ a) h and B are the same partial rder. That means, f curse, that they have identical jins and meets, and this means that they are the same bunded distributive lattice. Since in a bunded distributive lattice, each element has a unique cmplement and since (a]+[ a) and B map every element nt its cmplement, (a]+[ a) = B. Hence, indeed, B (a]+[ a) h and B are the same Blean algebra. 10

11 THEOREM 12: Let B 1 and B 2 be Blean algebras, a 1 an atm in B 1 and a 2 an atm in B 2, and let (a 1 ] be ismrphic t (a 2 ]. Then B 1 and B 2 are ismrphic. Let h 1 be the ismrphism between (a 1 ] and [ a 1 ) defined by: fr all x (a 1 ]: h 1 (x) = x B1 B1 (a 1 ) Let h 2 be the ismrphism between (a 2 ] and [ a 2 ) defined by: fr all x (a 2 ]: h 2 (x) = x B2 B2 (a 2 ) Let k the ismrphism between (a 1 ] and (a 2 ]. Let {0,1} be a tw element Blean algebra. B 1 = B (a1] + [ a1) h1 and B 2 = B (a2] + [ a2) h2, by therem11. B 1 is ismrphic t (a 1 ] {0,1} and B 2 is ismrphic t (a 2 ] {0,1}, by therem 10. Define g: (a 1 ] {0,1} (a 2 ] {0,1} by: fr every <a,b> (a 1 ] {0,1}: g(<a,b>) = <k(a),b>. It is straightfrward t prve that g is an ismrphism between B 1 and B 2. All this has the fllwing cnsequences fr finite Blean algebras: THEOREM 13: Any tw finite Blean algebras f the same cardinality are ismrphic. 1. Obviusly, up t ismrphism, there is nly ne Blean algebra f cardinality 1 r cardinality 2. Up t ismrphism, there is nly ne partial rder f cardinality 1, hence als nly ne Blean algebra. Up t ismrphism there are tw partial rders f cardinality 2: <{0,1},{<0,0>,<1,1>}> and <{0,1},{<0,0>,<0,1>,<1,1>}>. Only the secnd is a lattice and a Blean algebra. 2. If all Blean algebras f cardinality 2 n, n>0 are ismrphic, then all Blean algebras f cardinality 2 n+1 are ismrphic. This fllws frm therem 12. Let B 1 and B 2 be Blean algebras f cardinality 2 n+1, and assume that all Blean algebras f cardinality 2 n are ismrphic. Let a 1 be an atm in B 1 and a 2 be an atm in B 2. (a 1 ] and (a 2 ] are Blean algebras f cardinality 2 n, hence, by assumptin they are ismrphic. Then, by therem12, B 1 and B 2 are ismrphic. 1 and 2 tgether prve that any tw finite Blean algebras f the same cardinality are ismrphic. CORROLLARY 14: Up t ismrphism the finite Blean algebras are exactly the finite pwerset Blean algebras. Fr every n 1, if X =n then <pw(x),-,,,x,ø> is a pwerset Blean algebra f cardinality 2 n. By therem13, every Blean algebra f cardinality 2 n is ismrphic t it. 11

12 S, we can cnstruct every finite Blean algebra as a pwerset Blean algebra. We can als use the prduct cnstructin under an ismrphism t cnstruct all finite Blean algebras. Cardinality 1: 1 a pint in 0-dimensinal space. Cardinality 2: Take tw nn-verlapping Blean algebras f cardinality 1 and an ismrphism, and cnstruct the prduct: {<1,2>} 1 a pint mved alng a new dimensin: a line in 1-dimensinal space. 2 Cardinality 4: Take tw nn-verlapping Blean algebras f cardinality 2 and an ismrphism, and cnstruct the prduct: {<1,3>,<2,4>} A line mved alng a new dimensin. A square in 2-dimensinal space Cardinality 8: Take tw nn-verlapping Blean algebras f cardinality 4 and an ismrphism, and cnstruct the prduct: {<1,5>,<2,6>,<3,7>,<4,8>}

13 8 A square mved alng a new dimensin. A cube in 3-dimensinal space Cardinality 16: Take tw nn-verlapping Blean algebras f cardinality 8 and an ismrphism, and cnstruct the prduct: {<1,9>,<2,10>,<3,11>,<4,12>,<5,13>,<6,14>,<7,15>,<8,16>} A cube mved alng a new dimensin

14 A 4-dimensinal bject: