# Cantor-Bernstein theorem for $\sigma$-complete Boolean algebras.

I am working on problem 7.28 from Jech's Set Theory:

Let A and B be σ-complete Boolean algebras. Let a and b be elements of A and B respectively. If A is isomorphic to B$\upharpoonright$b and B is isomorphic to A$\upharpoonright$a, then A and B are isomorphic.

The relevant version of the CSB theorem is:

Theorem 3.2 (Cantor-Bernstein). If |A|≤|B| and |B|≤|A|, then |A| = |B|.

Proof. If f:A → B and g:B → A are one-to-one, then if we let B = g(B) and $A_{1}$ = g(f(A)), we have $A_{1}$ ⊂ B ⊂ A and |$A_{1}$| = |A|. Thus we may assume that $A_{1}$ ⊂ B ⊂ A and that f is a one-to-one function of A onto $A_{1}$; we will show that |A| = |B|. We deﬁne (by induction) for all n ∈ N: $A_{0}$ = A, $A_{n+1}$ = f($A_{n}$), $B_{0}$ = B, $B_{n+1}$ = f($B_{n}$). Let h be the function on A deﬁned as follows: h(x) = f(x) if x ∈ $A_{n}$ −$B_{n}$ for some n, x otherwise. Then h is a one-to-one mapping of A onto B, as the reader will easily verify. Thus |A| = |B|.

In my attempt to prove 7.28, I did not consider B as a subalgebra of A, so I used $B_{0}=ran(g)$ and I defined h by

h(x)=f(x), if x ∈ $A_{n}$ −$B_{n}$ for some n

h(x)=$g^{-1}$(x), otherwise.

I copied this construction, letting f and g be isomorphic embedding from A to B and from B to A respectively, and with ran(f)=B$\upharpoonright$b and ran(g)=A$\upharpoonright$a. I showed that the constructed function is 1-1 and surjective. However, showing that + (or *) is preserved appears to break down into several cases, some of which are trivial and the rest of which have completely stumped me.

I would appreciate any suggestions, especially clever hints that do not explicitly spell out a solution.

• What are a and b? Commented Nov 17, 2013 at 11:07
• a and b are non-0 elements of A and B respectively. Commented Nov 17, 2013 at 18:24
• A↾a denotes the set of members of A that are less than or equal to a. Commented Nov 17, 2013 at 18:25

Hint:It suffices to show that if $a,b\in A$ are such that there is an isomorphism $f:A\rightarrow A\upharpoonright a$ and $a\leq b$, then $A\simeq A\upharpoonright b$.

By recursion, define $a_0=1$, $b_0=b$, and $a_{n+1}=f(a_n)$ and $b_{n+1}=f(b_n)$. Put $c=\sum_{n<\omega} (a_n-b_n)$ and $c_1=\sum_{n=1}^\infty (a_n-b_n)$, notice that $c_1\leq a$, $c_1\leq c$ and $f(c)=c_1$.

Now define $g:A\longrightarrow A\upharpoonright b$ by $g(x)=f(x)\cdot c+x\cdot(-c)$, and notice that for all $x\leq b$ we have that $x\cdot c=x\cdot c_1$. Prove that $g$ is an isomorphism from $A$ onto $A\upharpoonright b$ (nontrivial). This is the simplest hint I could came up with, although it is not that trivial to continue from here.

However, if you cannot continue with the hint above, here is how it's done:

To see $g$ is an homomorphism notice that it is clear that $g(x+y)=g(x)+g(y)$, and $g(x\cdot y)=f(x\cdot y)\cdot c+(x\cdot y)\cdot(-c)=[f(x)\cdot c+x\cdot (-c)][f(y)\cdot c+y\cdot (-c)]$ using distributivity and $c\cdot(-c)=0$. $g$ is one-to-one, as $f$ is one-to-one and for all $x,y,w,z\in A$, $x\cdot c+y\cdot(-c)=w\cdot c+z\cdot(-c)$ iff $x=w$ and $y=z$. We have that $d\cdot c=d\cdot c_1$ for $d\leq b$ follows from $\sigma$-distributivity and $d\cdot(a_0-b_0)=0$; $\sigma$-complete Boolean algebras are always $\sigma$-distributive. Now let us check $g$ is onto $A\upharpoonright b$. Let $d\leq b$. We have $d\cdot c=d\cdot c_1$, but as $c_1\leq a$ and $f$ is onto $A\upharpoonright a$, there is $d'\in A$ with $f(d')=d\cdot c_1$, but $c_1=f(c)$, hence $d'\leq c$, in consequence $g(d')=f(d')\cdot c+d'\cdot(-c)=f(d')\cdot c=d\cdot c_1=d\cdot c$. Also $d\cdot(-c)\leq-c$, in consequence $f(d\cdot(-c))\leq f(-c)=-c_1$, thus $f(d\cdot(-c))\cdot c_1=0$, and we get $g(d\cdot(-c))=f(d\cdot(-c))\cdot c+d\cdot(-c)=f(d\cdot(-c))\cdot c_1+d\cdot(-c)=d\cdot(-c)$, thus $f(d'+d\cdot(-c))=d\cdot c+d\cdot(-c)=d$. Therefore $g$ is onto $A\upharpoonright b$.

• Thanks so much! This was very helpful. I noticed that things are a little simpler if you just define c to be negative of the inf of b, f(b), f(f(b)),... . Commented Nov 25, 2013 at 4:27
• @AriHerman glad you could come up with a nicer way of doing it. You're welcome. Commented Nov 25, 2013 at 4:30
• @CamiloArosemena-Serrato I have a question about your proof, more precisely, with the part where you prove that $g$ is one-to-one. When you say that $x\cdot c+y\cdot(-c)=w\cdot c+z\cdot(-c)$ iff $x=w$ and $y=z$, Is this holds only in this particular case? Becasuse I think that it's not true in general.
– YCB
Commented Jun 13, 2020 at 23:25

Let $$f: A \to B \restriction b$$ and $$g: B \to A \restriction a$$ be isomorphisms, and assume $$0 < a < 1, 0 < b < 1$$. It suffices to show that $$A \simeq A\restriction a \simeq B$$.

Define sequences of elements in $$A$$, $$a_n, b_n, n \in N$$: $$a_0 = 1, b_0 = g(1)$$, $$a_{n+1} = g(f(a_n))$$, $$b_{n+1} = g(f(b_n))$$, $$\forall n \in N$$. In particular, $$a_0 = 1 > b_0 = a > a_1 = g(b) > b_1 = g(f(a))$$. Notice that $$g \circ f$$ is an isomorphism, $$A \restriction a_i \simeq_{g \circ f} A\restriction a_{i+1}$$, $$\forall i$$.

Moreover, $$\forall n, a_{n+1} < b_n < a_n$$, by injectivity of $$f$$ and $$g$$.

Now let $$a^* = \Pi_{n \in N}a_n = \Pi_{n\in N}b_n$$, which exists because of $$\sigma$$-completeness of A and B. $$\forall x \leq a^*$$, $$(g \circ f)(x) = x$$ follows from the definition of $$\Pi$$.

We will define an isomorphism $$h: A \to A \restriction a$$, which for any $$x \in A$$ sends the part of $$x$$ that is $$\leq a^*$$ to itself, the part of $$x$$ that is $$\leq a_i - b_i$$ for some $$i \in N$$ to $$g(f(x))$$ and the part of $$x$$ that is $$\leq b_i - a_{i+1}$$ again to itself.

Let $$a_A = \Sigma_{n} (a_n - b_n)$$, $$a_B = \Sigma_{n} (b_n - a_{n+1})$$, both of which exist by $$\sigma$$-completeness and $$1 = a^* + a_A + a_B$$ is a partition. Let $$x \in A, x = x\cdot a^* + x \cdot a_A + x \cdot a_B = x^* + x_A + x_B$$, with $$x^* \leq a^*, x_A \leq a_A, x_B \leq a_B$$ and $$x^* \cdot x_A = x_A \cdot x_B = x^* \cdot x_B = 0$$. Define $$h: A \to A \restriction a$$ by $$h(x) = x^* + x_B + g(f(x_A))$$.

Notice that $$h(1) = a^* + a_B + g(f(a_A)) = a^* + \Sigma_{n} (b_n - a_{n+1}) + \Sigma_{n \geq 1} (a_n - b_n) = a$$, and $$h(0) = 0$$.

It follows that $$h$$ is a Boolean algebra homomorphism: $$h(x) + h(y)$$ = $$x^* + x_B + g(f(x_A))$$ $$+$$ $$y^* + y_B$$ $$+$$ $$g(f(y_A))$$ $$=$$ $$(x^* + y^*) + (x_B + y_B)$$ $$+$$ $$(g(f(x_A))$$ $$+$$ $$g(f(y_A)))$$ $$=$$ $$(x^* + y^*)$$ $$+$$ $$(x_B + y_B)$$ $$+$$ $$g(f(x_A +y_A))$$ $$=$$ $$h(x + y)$$. Similarly for $$h(x \cdot y)$$ by using the fact that $$(x^* + x_A + x_B)$$ $$\cdot$$ $$(y^* + y_A + y_B)$$ $$=$$ $$x^* \cdot y^* + x_A \cdot y_A + x_B \cdot y_B$$ because all other terms are $$=0$$. Finally $$h(\lnot x)$$ $$=$$ $$h(1-(x^* + x_B + x_A))$$ $$=$$ $$h((a^* - x^*) + (a_B - x_B) + (a_A - x_A))$$ $$=$$ $$(a^* - x^*)$$ $$+$$ $$(a_B - x_B)$$ $$+$$ $$(g(f(a_A)) - g(f(x_A)))$$ $$=$$ $$h(1) - h(x) = \lnot h(x)$$.

It is immediate that $$h$$ is injective: if $$x^* + x_B + g(f(x_A))$$ $$=$$ $$y^* + y_B + g(f(y_A))$$ then $$x^* = y^*$$, $$x_B$$ $$=$$ $$y_B$$, and $$g(f(x_A)) = g(f(y_A)) \implies x_A = y_A$$.

Finally, $$h$$ is onto: let $$x \in A \restriction a$$ i.e., $$x \leq a$$. Let $$x = x^* + x_A + x_B$$. Because $$x \leq a$$, it follows that $$x_A \leq g(b)$$: anything $$\leq 1-a$$ cannot be part of $$x$$. Therefore, $$(g \cdot f)^{-1}$$ is defined for $$x_A$$, and we can find $$z = x^* + x_B + (g \cdot f)^{-1}(x_A)$$ with $$h(z) = x$$.