ATOMLESS BOOLEAN ALGEBRAS AND ALGEBRAS OF SOLIDS
An addition to Structures for Semantics
Fred Landman
TelAvivUniversity
2000
1. ATOMLESS BOOLEAN ALGEBRAS
1.1. Intervals.
I collect together here all the relevant notions in the theory of intervals that we will rely on in this addition.
Let B be a non-empty set and R a binary relation on B (R BB), and A B.
The restriction of R to A, R⌠A = {<a,b> R: a,b A}
R is reflexive iff for every a B: R(a,a).
R is antisymmetric iff for every a,b B: if R(a,b) and R(b,a) then a=b.
R is transitive iff for every a, b, c B: if R(a,b) and R(b,c) then R(a,c).
R is a partial order on B iff R is reflexive, antisymmetric, and transitive.
We write <B,≤> for a partial order, and define: a < b iff a ≤ b and a b. <B,> is the strict partial order corresponding to <B,≤>. We choose ≤ or < depending on which is more appropriate in the context.
≤ is a linear orderon B iff ≤ is a partial order on B and
for every a,b B: a ≤ b or b ≤ a.
Let <B,≤> be a partial order, and let C B.
C is a chain in B iff <C,≤⌠C> is a linear order.
C is a maximal chain in B iff C is a chain in B and for every chain D in B
such that C D: C=D.
Let <B,> be a linear order.
< is dense iff for every a,b B such that a < b,
there is a c B such that a <c < b.
Let <B,≤> be a partial order, a,b B:
a covers b iff a < b and there is no c B: a < c < b.
Let <B,≤> be a partial order, X B, and a,b B.
a is the minimum of X (if there is one) iff a X and for every x X: a ≤ x.
a is the maximum of X (if there is one) iff a X and for every x X: x ≤ a.
The minimum of B, if there is one, is called 0.
The maximum of B, if there is one, is called 1.
The set of upper bounds of X, UB(X) = {a B: for every x X: x ≤ a}.
The set of lower bounds of X, LB(X) = {a B: for every x X: a ≤ x}.
V(X) is the join of X iff V(X) is the minumum of UB(X).
Λ(X) is the meet of X iff Λ(X) is the maximum of LB(X).
(a b) = V ({a,b})
(a b) = Λ ({a,b})
A set is lower bounded if it has a meet, upperbounded if it has a join, bounded if it has both.
N is the set of natural numbers, Q the set of rational numbets, R the set of real numbers. These are regarded with their natural linear order <.
Let <B, > be a linear order.
An interval in B is a set i B such that:
for every n,m i, for every k B: if n ≤ k ≤ m then k i.
I will write i and i for the meet and join of interval i under <.
To avoid confusion, let me note the following: I will use one-place small and only to indicate the bounds of intervals, or more generally, sets of points. Otherwise, I will use two-place small and , or one-place largeV and Λ.
So:
Let i be an interval in B.
i is lowerbounded in B iff i B.
i is upper bounded in B iff i B.
i is bounded in B iff i,i B.
Let i be a bounded interval in B.
i is closed iff i,i i
i is open iff i,i i
The empty set Ø counts as an open interval.
If i is a non-empty open interval, we write: i = (i,i)
If i is a non-empty closed interval, we write: i = [i,i]
Similarly, [i,i) and (i,i] stand for half-open intervals.
We can restrict the notions of closed and open to bounded intervals here, because in all our applications here we will only be concerned with bounded intervals.
Note too that the notion of interval itself is tailored to linear orders. In a partial order, the corresponding notion of a set closed under intermediate elements is a convex set (see Structures for Semantics).
Note too that in a partial order, a covers b iff a ≤ b and (a,b) = Ø, or alternatively,
a ≤ b and [a,b] = {a,b}.
Let <A,<A> and <B,<B> be partial orders.
A and B are isomorphic iff there is a function h:A B such that:
1. h is a bijection, a one-one function from A onto B.
2. For every a,b A: a <A b iff h(a) <B h(b)
THEOREM 1: CANTOR'S THEOREM
Consider the following four substructures of the rational numbers <Q,> (with
the order the obvious restriction):
[0,1],>, <[0,1),>, <(0,1],>, <(0,1),>.
Every countable dense linear order is isomorphic to one of these four
structures.
i.e.: The countable dense linear orders with end points are isomorphic to [0,1].
The countable dense linear orders without end points are isomorphic to (0,1).
The countable dense linear orders with a minimum but no maximum are
isomorphic to [0,1).
The countable dense linear orders with a maximum but no minimum are
isomorphic to (0,1].
PROOF: Cantor's zig-zag construction, the proof for countable dense linear orders without end points is given in the appendix of chapter 1, in Structures for Semantics.
The others are straightforward variants of this. ◄
1.2. Boolean Algebras.
Let <B,≤> be a partial order.
<B,≤> is a lattice iff for every a,b B: (a b), (a b) B
Let <B,≤> be a lattice.
<B,≤> is bounded iff 0,1 B
Let <B,≤> be a bounded lattice.
a is the complement of a in B (if there is one) iff a is the unique element of
B such that (a a) = 0 and (a a) = 1.
<B,≤> is a complemented bounded lattice iff for every a B: a B.
Let <B,≤> be a lattice.
<B,≤> is distributive iff for every a,b,c B:
(a (b c)) = ((a b) (a c)) or
(a (b c)) = ((a b) (a c))
<B,≤> is a Boolean Latticeiff <B,≤> is a distributive complemented bounded
lattice.
<B,≤,,,,0,1> is a Boolean Algebra iff
1. <B,≤> is a Boolean Lattice with minimum 0 and maximum 1.
2. is the one-place operation on B such that for every a B:
(a) = a.
3. is the two-place operation on B such that for every a,b B:
(a,b) = (a b).
3. is the two-place operation on B such that for every a,b B:
(a,b) = (a b).
Let B be a Boolean algebra, and let a B.
B is complete iff for every X B: V(X), Λ(X) B.
(In fact, for completeness of Boolean algebras you only need to require closure under one of V and Λ, see Structures for Semantics.)
a is an atom in B iff 0 covers a.
a is a dual atom in B iff a covers 1.
ATB = {a B: a is an atom in B}.
B is atomic iff for every b B-{0}: there is an a ATB: a ≤ b.
B is atomlessiff ATB = Ø.
Let B be a Boolean algebra and a,b B
The relative complement of a with respect tob, b(a), is:
b(a) = a b
The symmetric difference of a and b, a + b, is:
a + b = b(a) a(b)
LEMMA 2:If a ≤ b then a + b = b(a)
PROOF: a + b = b(a) a(b) = (a b) (b a)
If a ≤ b then b a = 0. Thus, a + b = b(a) 0 = b(a) ◄
Let B = <B,≤,,,,0,1> be a Boolean 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 relativization of B to c, Bc, is the structure:
Bc = <Bc,≤c,c,c,c,0c,1c>, where:
1. Bc = (c]
2. ≤c = ≤⌠Bc
3. c = c⌠Bc where cis relative complement on B.
4. c = ⌠Bc
5. c = ⌠Bc
6. 0c = 0
7. 1c = c
LEMMA 3: Bc is a Boolean algebra.
PROOF:
1. <Bc,≤c,c,c,0c> is a substructure of <B,≤,,,0>: namely, if a,b Bc, then
a,b ≤ c, hence (a b) ≤ c and (a b) ≤ c, hence (a b) Bc and (a b) Bc. Further 0c = 0.
2. <Bc,≤c,c,c,0c,1c> is bounded by 0c and 1c. We have seen already that 0c is the minimum of Bc. 1c = c, and c is, obviously, the maximum of (c].
3. We have proved under 1. that <Bc,≤c,c,c> is a sublattice of <B,≤,,>. Since <B,≤,,> is distributive, it follows that <Bc,≤c,c,c> is distributive (since the class of distributive lattices is closed under substructure).
4. Thus, we have proved that Bc is a bounded distributive lattice. So we only need to prove that Bc is complemented, i.e. that c is complementation on Bc.
First: Bc is closed under c.
Let a Bc. c(a) = a c. Since a c ≤ c, c(a) Bc.
Secondly, c respects the laws of 0 and 1c:
Let a Bc. a cc(a) = a (a c) = (a a) (a c) = 0 (a c) = 0
a cc(a) = a (a c) = (a a) (a c) = 1 (a c) = a c = c
Thus, indeed c is complementation on Bc. ◄
Note: except for the trivial case where c is 1, Bc is not a sub-Boolean algebra of B, because 1 is not preserved. It is a Boolean algebra on a subset of B.
Of course, this theorem dualizes to [c): this too forms a Boolean algebra with maximum 1 and minumum c, and the operation of complementation appropriately dualized.
Let B be a Boolean algebra, let a,b B and a ≤ b.
LEMMA 4: a covers b iff a + b ATB
PROOF:
1. Let a cover b. This means that a is a dual atom in Bb. But that means that b(a) is an atom in Bb. Since ATBb = ATB Bb, it follows that b(a) ATB.
Since a ≤ b, a + b = b(a), hence a + b ATB.
2. Let a + b ATB. Since a ≤ b, a + b = b(a). Hence b(a) ATB. Since b(a) = a b, b(a) Bb, hence b(a) ATBb.
Then b(b(a)) is a dual atom in Bb, which means that b(b(a)) covers b.
But b(b(a)) = b(a b) = (a b) b = (a b) b = (a b) b =
(a b) (b b) = (a b) 0 = a b. Since a ≤ b, a b = a.
Hence b(b(a)) = a.
Hence a covers b. ◄
CORROLLARY 5:If B is an atomless Boolean algebra, then every maximal chain in B is a dense linear order with endpoints.
PROOF:
If B is atomless, then for no a,b B: a covers b (else a + b ATB). Hence indeed
every maximal chain is dense. The endpoints of any maximal chain are, of course, 0
and 1. ◄
Let A and B be Boolean algebras.
A and B are isomorphiciff there is a function h;A B such that:
1. h is a bijection, a one-one function from A onto B.
2. for every a,b A: a ≤A b iff h(a) ≤B h(b).
3. for every a A: h(A(a)) = B(h(a)).
4. for every a,b A: h(a A b) = h(a) B h(b).
5. for every a,b A: h(a A b) = h(a) Bh(b).
6. h(0A) = 0B.
7. h(1A) = 1B.
1.3. Boolean algebras +-generated by chains.
Let B be a Boolean algebra and let C B be a chain in B with 0 C.
C +-generates B iff for every b B there are some c1,...,cn C such that:
b = c1+ ... + cn.
So chain C +-generates B iff every element of B can be generated as the symmetric difference of (finitely many) elements of C.
Let R be a relation on B.
The transitive closure of R, [R]TR, is given by:
[R]TR = R {<a,c> B B: for some bB: <a,b>R and <b,c>R}.
Let B1 and B2 be finite Boolean algebras of cardinality 2n, such that B1 B2 = Ø, and let h be an isomorphism between B1 and B2.
We definite the +-product of B1 and B2 under h, Bh1+2:
Bh1+2 = <Bh1+2,≤1+2,1+2,1+2,1+2,01+2,11+2> where:
1. B1+2 = B1 B2.
2. ≤1+2 = [≤1 ≤2 {<b,h(b)>: b B1}]TR.
3. 1+2 is meet under ≤1+2.
4. 1+2 is join under ≤1+2.
5. 01+2 = 01.
6. 11+2 = 11.
7. 1+2 is defined by:
2(h(b))if b B1
For all b B1+2+: 1+2(b) =
1(h-1(b))if b B2
LEMMA 6: Bh1+2is a Boolean algebra of cardinality 2n+1.
PROOF: This is a special instance of the construction of the product Boolean algebra. ◄
Let B be a finite Boolean algebra of cardinality 2n+1, n>0.
Let C be a maximal chain in B. Since n>0, C-{1} is a dual atom in B, C-{1} is an atom in B (in particular, neither is identical to 0 or to 1).
Look at B1 = (BC-{1}] and B2 = [BC-{1}). Both are Boolean algebras. That B1 is a Boolean algebra we have proved above. That B2 is a Boolean algebra follows, as expressed, from that proof with duality.
Let h be the function from B2 to B1 given by:
for every b B1: h(b) = B1(Bb)
LEMMA 7: B = Bh-11+2
PROOF:
1. B1 B2 = Ø.
Namely, suppose b B1 B2. Then b ≤ C-{1} and C-{1} ≤ b, which means
b ≤ C-{1}, which means that b b ≤ C-{1}, hence C-{1} = 1. But, of course, it isn't, C-{1} is a dual atom in B, not the maximum.
2. B1 B2 = B
Since B1 is a prime ideal in B, if b B1, b B1. But if b B1, b ≤ C-{1}, hence C-{1} ≤ b, hence b B2.
3. h (and hence h-1) is an isomorphism between B1 and B2.
-h is one-one:
As we know, if b B2, b B1. We map b onto the relative complement in B1 of its complement in B. This is, of course an element of B1. Now, assume that for a,b B2 h(a)=h(b). That means that (a ) C-{1} = (b) C-{1}, which means that
a C-{1} = b C-{1}. Since a,b ≤ C-{1}, it follows that a=b.
-h is onto:
Let a B1. There is obviously some b B2 such that h(b)=a, namely B(B1(a)).
This means that h is a bijection between B1 and B2. Thus B1 and B2 have the same cardinality, and hence they are isomorphic, because all finite Boolean algebras of the same cardinality are isomorphic.
Since B1 and B2 partition B and have the same cardinality, it obviously follows that B1 and B2 have cardinality 2n.
3. Given these facts, we can form Bh-11+2. Obviously Bh-11+2 and B are isomorphic. Moreover B1+2 = B, which means that Bh-11+2 and B are automorphic. It is not hard to see that in fact ≤1+2 = ≤B, thus, indeed Bh-11+2 = B ◄
We will use this lemma to prove a theorem:
THEOREM 8: Every finite Boolean algebra B is +-generated by every maximal
chain in B.
PROOF: With induction.
1. {0,1} is trivially +-generated by maximal chain {0,1}, which is the only maximal chain in {0,1}.
2. Assume that every Boolean algebra of cardinality 2n, n N is +-generated by every maximal chain in it. Let B be a Boolean algebra of cardinality 2n+1, n N. Let C be any maximal chain in B. Let B1 and B2 be as in Lemma 7 the ideal generated by
C-{1} and the filter generated by C-{1}. Then, by induction hypothesis, B1 and B2 are +-generated by every maximal chain in them. By lemma 7, B = Bh-11+2.
Claim: C +-generates Bh-11+2.
Namely: C-{1} is a maximal chain in B1, hence, by induction hypothesis, C-{1}
+-generates B1. But if C-{1} generates B1, of course, so does C.
For every b B2: b = 1+2(b) + 11+2,, where 1+2(b) B1. Since C-{1} +-generates 1+2(b), C +-generates b. Hence C +-generates B.
Since C was an arbitrary maximal chain in B, it follows that B is indeed +-generated by every maximal chain in B. ◄
Two more facts without proof:
FACT 9: If B is a Boolean algebra and every maximal chain in B +-generates
B then B is finite.
FACT 10: If B is a complete Boolean algebra +-generated by a chain then B is finite.
Thus, besides finite Boolean algebras, Boolean algebras +-generated by a chain can only be incomplete. Furthermore, if there are such Boolean algebras (and we will see below that there are), they are +-generated by some maximal chain, but not by every maximal chain.
ILLUSTRATIONS
Let B be the following Boolean algebra:
1
o
ca
oo bo
oo b o
a c
o
0
Let C be {0,a,c,1}
B1 = {0,a,b,c}.
B1(c) = c c = 0
B1(0) = 0 c = c
B1(a) = a c = b
B1(b) = b c = a
B2 = {c,b,a,1}
B2(1) = 1 c = c
B2(b) = b c = a
B2(a) = a c = b
B2(c) = c c = 1
1
o
ca
oo bo
oo b o
a c
o
0
h: B2 B1 is given as follows:
h(1) = B1(B 1) = B1(0) = c
h(b) = B1(Bb) = B1(b) = a
h(a) = B1(Ba) = B1(a) = b
h(c) = B1(Bc) = B1(c) = 0
Hence h-1 = {<0,c>,<a,b>,<b,a>,<c,1>}:
1
o
ca
oo bo
oo b o
a c
o
0
≤B1 = {<0,0>,<a,a>,<b,b>,<c,c>,<0,a>,<0,b>,<0,c>,<a,c>,<b,c>}
≤B2 = {<c,c>,<b,b>,<a,a>,<1,1>,
<c, b>,<c,a>,<c,1>,<b,1>,<a,1>}
We add to that: {<0,c>,<a,b>,<b,a>,<c,1>}
The transitive closure adds to this:
{<a,1>,<b,1>,<0,1>,<0,b>,<0,a>}
The result is, of course, just ≤B.
Now look at C-{1}, which is {0,a,c}. This is, of course, a maximal chain in B1.
It generates B1.
First all the elements in the chain are +-generated by the chain.
This follows from the following fact:
FACT: Let B be a Boolean algebra and b B.
b+0 = a
Namely: b + 0 = (b 0) (0 b) = 0 (1 b) = b
Since 0 is in the chain by definition, every element c in the chain is +-generated by the chain as c + 0.
Next, b is +-generated by the chain, because b = a + c
a + c = (a c) (c a) = b 0 = b.
So we get, so far:
1
o
c+0a
oo bo
oo a+c o
a+0 c
o
0+0
When we now look at C itself, then 1 is of course +-generated by C as 1 + 0.
For every b B2: b = b + 1
1 = 0 + 1, we already saw that.
Take as an example a.
a = a 1 = (a 1) 0 = (a 1) (1 a) =
(a 1) (1 a) = a + 1.
Thus we can write these elements as follows:
1 = 0 + 1
b = b + 1 = b + 1 = a + c + 1
a = a + 1 = a + 1
c = c + 1 = c + 1
Now we have written all elements of B as generated from elements in chain C:
1+0
o
c+0a+1
oo a + c + 1o
oo a+c o
a+0 c+1
o
0+0
THEOREM 11: Every countable Boolean algebra is +-generated by a maximal
chain.
PROOF: Let B be a countable Boolean algebra, enumerated as:
B = {b0,b1,b2,....,bn,....}, where b0 = 0
Let Bn be the subalgebra of B generated by {b0,...,bn} under the Boolean operations.
Since {b0,...,bn} is finite, Bn is finite. Hence, Bn is +-generated by every one of its maximal chains.
We make the following construction:
1. C0 = {0,1}, the maximal chain +-generation B0.
2. For every n N, choose a maximal chain Cn in Bn +-generating Bn and a
maximal chain Cn+1 in Bn+1 +-generating Bn+1 such that Cn Cn+1.
3. C = n N Cn
Claim 1: C is a maximal chain in B.
C is a chain in B. Let a,b C. Then for some n N: a,b Cn, hence a ≤ b or b ≤ a holds in Bn, and hence it holds in B.
C is maximal. Suppose C C' and C' is a chain. Then, for some element c C',
c C. c occurs somewhere in the enumeration, say, c = bk, k N. Look at Bk and Ck. Since for every b C: b ≤ c or c ≤ b, it follows that for every b Ck: b ≤ c or
c ≤ b. But then Ck {c} is a chain in Bk. But since c C, c Ck. Hence Ck is not a maximal chain in Bk. But it is! Hence C is maximal.
Claim 2: C +-generates B.
Let b B. Again, b occurs somewhere in the enumeration, say b = bm, m N.
This means that b Bm. That means that Cm +-generates b. But, since Cm C, this means that C +-generates b. Hence, indeed C +-generates B. ◄
Let B be a Boolean algebra and C a chain with 0 in B.
B is freely +-generated by C iff
1. B is +-generated by C.
2. For every Boolean algebra B' which is +-generated by C, there is a
homomorphism h: B B' such that:
a. h is onto.
b. for every c C: h(c)=c
FACT 12: For every linear order <C,≤,0> (i.e. every chain with 0) there is a Boolean
algebra which is freely +-generated by C.
The proof, which is omitted, is a straightforward adaptation of the proof for the existence of free lattices generated by a partial order in an equational class of lattices (and that proof was omitted in Structures for Semantics because it is too complicated.)
THEOREM 13: Any two Boolean algebras +-generated by chain C with 0 are
isomorphic.
PROOF: Let B be a Boolean algebra freely +-generated by C, and let B1 be a Boolean algebra +-generated by C. And let h: B B1 be a homomorphism which is onto B1 and is identity for C.
Claim: h is an isomorphism.
Since h is a homomorphism and h is onto, we only need to prove that h is one-one.
Let a,b B and assume that h(a)=h(b).
Since B is +-generated by C, we know that:
a = c1 + ... + cn, for some c1,...,cn C
b = d1 + ... + dm, for some d1,...,dm C
Hence:
h(a) = h(b)
h(c1 + ... + cn) = h(d1 + ... + dm)[a,b are +-generated]
h(c1) + ... + h(cn) = h(d1) + ... + h(dm) [h is a homomorphism]
c1 + ... + cn = d1 + ... + dm [h is identity on C]
a = b ◄
This means that, up to isomorphism, for any chain C with 0, there is exactly one Boolean algebra +-generated by C, namely the Boolean algebra freely +-generated by C. Standard considerations about free algebras give the following generalization:
CORROLLARY 14: If chain C1 +-generates Boolean algebra B1 and chain C2
+-generates Boolean algebra B2, and C1 and C2 are isomorphic,
then B1 and B2 are isomorphic.
THEOREM 15: Every two countable atomless Boolean algebras are isomorphic.
PROOF: Let B1 and B2 be countable atomless Boolean algebras. By theorem 11, we know that, since B1 is countable, it is +-generated by a maximal chain, say, C1, and similarly, B2 is +-generated by a maximal chain, say, C2.
Corrollary 5, tells us that C1 and C2 are dense linear orders with endpoints.
Theorem 1 tells us that, hence, C1 and C2 are isomorphic.
Corrollary 14, then, tells us that B1 and B2 are isomorphic. ◄
Let B be a Boolean algebra.
B is homogenous iff for every b B-{0}: B is isomorphic to Bb.
CORROLLARY 16: The countable atomless Boolean algebra is homogenous.
PROOF: Let B be the countable atomless Boolean algebra, and let b B. Look at Bb. We have proved before that Bb is a Boolean algebra. Since B is atomless, Bb is atomless as well. This is, because any atom in Bb would be an atom in B as well, and B doesn't have any atoms. Since every finite Boolean algebra is atomic, it follows that Bb is not finite. Since B is countable, and Bb B, it follows that Bb is countable. Hence Bb is also a countable atomless Boolean algebra. Hence, by theorem 15, B and Bb are isomorphic, hence B is homogenous. ◄
1.4. Completions of Boolean algebras.
Let A and B be Boolean algebras.
B is a completion of A iff
1. B is complete.
2. A is a subalgebra of B.
3. For every X A: if VAX A, then VAX = VBX
if Λ AX A, then Λ AX = Λ BX
B is a free completion of A iff
1. B is a completion of A.
2. For every completion B1 of A, there is a homomorphism h: B B1
such that:
a. h is onto.
b. for every a A: h(a)=a
FACT 17: For every Boolean algebra A there is a Boolean algebra B which is a free
completion of A.
The proof is, once again, a straightforward adaptation of the proof of the existence of lattices freely generated by a partial order in an equational class.
THEOREM 18: any two completions of a Boolean algebra are isomorphic.
PROOF: Let B be a free completion of A. Let B1 be a completion of A and let
h: B B1 be a homomorphism which is onto and identity on A.
Claim: h is an isomorphism.
This means we need to shown that h is one-one.
Let a B.
Since B is a completion of A, we know that:
either a A or a = VBX, for some X A or a = ΛBX, for some X A.
h is a homomorphism which is onto and identity on A. This means that for some substructure B' of B, h⌠B' is an isomorphism between B' and B1 which is identity on A. This means that complete joins and meets in B1 are preserved in B. h⌠B' -1 is an isomorphism g between B1 and B' which is identity on A and for every Z B1:
g(VB1Z) = VB{g(z): z Z}. and g(ΛB1Z) = ΛB{g(z): z Z}.
Now we argue as follows.
Since B is a completion of A, we know that:
either a A or a = VBX, for some X A or a = ΛBX, for some X A.
Look at h:
-if a A, then h(a)=a.
-if a = VBX, for some X A, then h(a) = h(VBX) = VB1{h(x): x X} = VB1X .
(because h is identity on A).
-if a = ΛBX, for some X A, h(a) = ΛB1X.
Similarly,
-if b A, then h(b)=b.
-if b = VBY, for some Y A, then h(b) = VB1Y.
-if b = ΛBY, for some Y A, then h(b) = ΛB1Y.
Assume a,b B and h(a)=h(b).
This means that we have one of the following cases (ignoring equivalent cases):
1. a = b
2. VB1X = b, for some X A
3. ΛB1X = b, for some X A
4. VB1X = VB1Y, for some X,Y A
5. VB1X = ΛB1Y, for some X,Y A
In cases, 2 and 3, it follows that VB1X (ΛB1X) is in A, but that means, by definition of completion that it is identical to VAX (ΛAX), and so is VBX (ΛBX). Since the latter is a, a = b follows.
For cases 4 and 5, we use the fact that g = h⌠B'-1 is an isomorphism between B1 and subalgebra B' of B which is identity on A. From the statement in 4 it follows that
VB{g(x): x X} = VB{g(x): x Y}. Since g is identity on A and X and Y are subsets of A, it follows that VBX = VBY, hence a = b. Case 5 goes just the same.
Thus we have shows that if h(a) = h(b), in all possible cases a = b.
Hence h is an isomorphism. ◄
CORROLLARY 19: If A is a homogenous Boolean algebra, and B is the
completion of A, then B is homogenous.
PROOF: We note that theorem 18 generalizes along standard lines to: if two Boolean algebras are isomorphic then their completions are isomorphic.
Let A be a homogenous Boolean algebra, and B its completion. Let B' be Bb for some b B. Obviously, if b A, then Bb is the completion of Ab, and since A and Ab are isomorphic, so are B and Bb. The case left is the case where b is not in A.
In that case, b = VBX for some X A or b = ΛBX , for some X A.
To prove this case, you need to generalize some of the theorems discussed so far.
If A is a Boolean algebra, then A-{1} is a Boolean meet semilattice. If A is a homogenous Boolean algebra, and (X] an ideal in A which does not have a maximum, then (X] is isomorphic to A-{1}. Like Boolean algebras, Boolean meet semilattices have unique completions, the completion of A-{1} is just the completion of A (this is, because Boolean meet semilattices are just Boolean algebras with 1 removed, but all the remaining structure in tact, and by completing the Boolean meet semilattice, you get a complete Boolean algebra).
Now let b = VBX (the case for ΛBX is basically similar). Look at (VBX], and form (VBX] A. (VBX] A is an ideal in A which does not have a maximum (since, by assumption the maximum is in B, but not in A). Thus, (VBX] A is isomorphic to
A-{1}. This means that the completion of (VBX] A is isomorphic to the completion of A-1. But the completion of (VBX] A is (VBX], and the completion of A-{1} is B. Hence it follows that (VBX] is isomorphic to B. ◄