Cohen algebra (generalization)

Question

Let Bor($X$) = class of all borel subsets of $X$. Cohen algebra is defined as Bor(X) modulo the ideal of meager sets.

The Cohen algebra has a combinatorial : it is the unique atomless complete Boolean algebra with a countable dense subset, (i.e it is a completion of a countable dense Boolean subalgebra). This case is true when $X=2^\omega$.

(1) My first question is: does this characterization true when $X=2^\kappa$ for an infinite $\kappa$.

(2) Is there any characterization (unique like above one) for random algebra (= Bor($X$) modulo the ideal of Lebesgue measure zero set) even when when $X=2^\omega$.

Thanks in advance

Answer 1

Regarding question $1$, it seems that you want to know whether you've got the unique complete c.c.c. Boolean algebra with density $\kappa$. The answer is no.

On the one hand, the forcing notion $\text{Add}(\omega,\omega_1)$ to add $\omega_1$ many Cohen reals is c.c.c. and has density $\omega_1$. This is another way to describe your algebra in the case $\kappa=\omega_1$.

But meanwhile, forcing with a Suslin tree is also c.c.c. and has density $\omega_1$. But these two complete Boolean algebras are not isomorphic, since forcing with the Suslin tree adds no reals and in fact is $\leq\omega$-distributive, whereas clearly $\text{Add}(\omega,\omega_1)$ adds reals.

Of course, there may not be a Suslin tree, but consider the forcing to add a single Cohen real (which creates a Suslin tree), followed by the forcing to force with that Suslin tree. This is the iteration of two c.c.c. forcing notions, and is hence c.c.c., but under CH has density $\omega_1$. But this forcing is not isomorphic to $\text{Add}(\omega,\omega_1)$, since the latter forcing is absolutely c.c.c., but the former is not, as the branch through the Suslin tree creates an uncountable antichain in the extension.

One may omit the CH assumption by taking the case $\kappa=\mathfrak{c}$. That is, compare the forcing to add continuum many Cohen reals, versus the forcing to add this many Cohen reals, and afterwards also to force with the Suslin tree created by the first one. Both of these are c.c.c. and have density $\mathfrak{c}$, but they are not isomorphic since the former remains c.c.c. in the forcing extension and the latter does not.

Answer 2

There is a characterization of what you call a random algebra found in [1][Ch. 15. Sec 3]. This characterization involves the notion of a measure algebra.

We define a measure algebra to be a Boolean algebra $B$ along with a function $\mu:B\rightarrow[0,\infty)$ such that $\mu(a)=0$ if and only if $a=0$ and $\mu(\bigvee_{n}a_{n})=\sum_{n}\mu(a_{n})$ whenever $a_{i}\wedge a_{j}=0$ for $i\neq j$. Clearly, the random algebra is a measure algebra.

Every measure algebra $(B,\mu)$ can be endowed a metric $d$ where $d(a,b)=\mu((a\wedge b')\vee(a'\wedge b))$ for all $a,b\in B$. This metric is always a complete metric. Since $(B,\mu)$ becomes a metric space with the metric $d$, we may talk about topological properties of $(B,\mu)$.

Let $\mathcal{A}$ denote the $\sigma$-algebra of Lebesgue measurable sets on $[0,1]$(or $2^{\omega}$) and let $\mathcal{I}$ denote the ideal of sets of measure $0$. Let $m$ be the Lebesgue measure on $(\mathcal{A}/\mathcal{I},m)$. Then $(\mathcal{A}/\mathcal{I},m)$ is up to isomorphism of measure algebras the unique atomless separable (as a metric space) measure algebra with $m(1)=1$.

One can also give a purely combinatorial characterization of the random algebra that does not refer to measures. In the paper [2], Thomas Jech gives various characterizations of the Boolean algebras $B$ such that there is a function $\mu:B\rightarrow[0,\infty)$ where $(B,\mu)$ is a measure algebra. We shall therefore add conditions to Jech's characterization of measure algebras that guarantee that the measure algebra is isomorphic to the random algebra.

$\mathbf{Theorem}$ $\mathbf{1}$ A measure algebra $(B,\mu)$ is isomorphic to the random algebra if and only if $B$ is atomless and there is a countable subset $D\subseteq B$ such that for each $b\in B$ there are $R_{n}\subseteq D$ for natural numbers $n$ where $b=\bigwedge_{n}\bigvee R_{n}$.

$\mathbf{Proof}$ $\leftarrow$. Let $\mathcal{B}$ be a countable basis for the topology on $[0,1]$. Then for each Borel set $B$ there is a sequence of open sets $U_{n}$ where $B\subseteq\bigcap_{n}U_{n}$ and $m((\bigcap_{n}U_{n})\setminus B)=0$. Since $\mathcal{B}$ is a basis, for all $n$ there is some $\mathcal{R}_{n}\subseteq\mathcal{B}$ with $U_{n}=\bigcup\mathcal{R}_{n}$. Therefore, the sets $B$ and $\bigcap_{n}\bigcup\mathcal{R}_{n}$ differ only by a set of measure zero. By taking the Borel sets modulo the measure zero sets, we conclude there is a countable subset $D$ of the random algebra $B$ such that for each $b\in B$ there is some $R_{n}\subseteq D$ for all $n$ such that $b=\bigwedge_{n}\bigvee R_{n}$.

$\rightarrow$. Assume that $(B,\mu)$ is an atomless measure algebra such that there is a countable dense subset $D\subseteq B$ where for each $b\in B$ there are $R_{n}\subseteq D$ for natural numbers $n$ where $b=\bigwedge_{n}\bigvee R_{n}$. To show that $(B,\mu)$ is isomorphic to the random algebra, it suffices to prove that $B$ is separable as a metric space. Let $E$ be the Boolean subalgebra of $D$ generated by $B$. Then $E$ is a countable subset of $B$. We claim that $E$ is a dense subset of $B$ as well.

Let $b\in B$. Then there are $R_{n}\subseteq D$ for each $n\geq 0$ such that $b=\bigwedge_{n=0}^{\infty}\bigvee R_{n}$. For all $n\geq 0$, let $S_{n}=\{a_{0}\wedge...\wedge a_{n}|a_{0}\in R_{0},...,a_{n}\in R_{n}\}$. Then $S_{n}\subseteq E$ for all $n$, and $\bigvee S_{n}=(\bigvee R_{0})\wedge...\wedge(\bigvee R_{n})$ for $n\geq 0$. Therefore the sequence $(\bigvee S_{n})_{n}$ is a decreasing sequence with $\bigwedge_{n}\bigvee S_{n}=\bigwedge_{n=0}^{\infty}\bigvee R_{n}=b$. Therefore, we have $^{Lim}_{n\rightarrow\infty}\bigvee S_{n}=b$ in the topology given by the metric defined on any measure algebra. Furthermore, for all $n$, we have $\lim_{T\subseteq S_{n},T\,\textrm{is finite}}\bigvee T=\bigvee S_{n}$, so there is a finite subset $T_{n}\subseteq S_{n}$ such that $d(\bigvee T_{n},\bigvee S_{n})<\frac{1}{n}$. Therefore, we have $^{\lim}_{n\rightarrow\infty}\bigvee T_{n}=b$. However, since $\bigvee T_{n}\in E$ for all $n$, we conclude that $E$ is a dense subset of $B$. We conclude that $(B,\mu)$ is a separable atomic measure algebra, so $(B,\mu)$ is isomorphic to the random algebra. QED

By combining the above characterization of the measure algebras which are isomorphic to the random algebra and Jech's characterizations of the measure algebras, we have the following algebraic characterizations of the random algebra that do not make any reference to measures, but we first need to give some definitions.

A Boolean algebra $B$ is said to be weakly distributive if for every sequence $(p_{n})_{n}$ of partitions of $B$ there is a partition $p$ of $B$ such that for each $a\in p$ the set $\{b\in p_{n}|a\wedge b>0\}$ is finite for each $n$.

A subset $r\subseteq B^{+}=B\setminus\{0\}$ is said to be a cellular family if $a\wedge b=0$ whenever $a,b\in r$ and $a\neq b$ (I should note that Jech calls cellular families antichains in [2]. ).

In a $\sigma$-complete Boolean algebra, we let $\overline{\lim}_{n\rightarrow\infty}a_{n}=\bigwedge_{n=0}^{\infty}\bigvee_{k=n}^{\infty}a_{k}$.`

A $\sigma$-complete Boolean algebra $B$ is said to be uniformly concentrated if there is a function $F$ where $f(c)\in c$ for each cellular family $c$ such that if $c_{n}$ is a finite cellular family with $|c_{n}|\geq 2^{n}$ for all $n$, then $^{\lim}_{n\rightarrow\infty}F(c_{n})=0$.

$\mathbf{Theorem}$ $\mathbf{2}$ Let $B$ be a $\sigma$-complete Boolean algebra. Then the following are equivalent.

I. $B$ is isomorphic to the random algebra.

II. $B$ satisfies the following properties.

$\phantom{spa}$ i. $B$ is atomless.

$\phantom{spa}$ ii. $B$ is weakly distributive

$\phantom{spa}$ iii. there is a countable subset $D\subseteq B$ such that if $b\in B$, then there is a sequence $(R_n)_n$ of subsets of $D$ such that $b=\bigwedge_{n}\bigvee R_{n}$.

$\phantom{spa}$ iv. there is a sequence $(C_{n})_{n}$ of subsets of $B$ with $B^{+}=\bigcup_{n}C_{n}$ and where

$\phantom{spaces}$ a. For all $n$ there is an integer $K(n)$ such that every cellular family in $C_{n}$ has at most $K(n)$ elements.

$\phantom{spaces}$ b. If $a_{n}\not\in C_{n}$ for all $n$, then $\overline{Lim}_{n\rightarrow\infty}a_{n}=0$.

III. The Boolean algebra $B$ satisfies the following.

$\phantom{spa}$ i. $B$ is atomless

$\phantom{spa}$ ii. there is a countable subset $D\subseteq B$ such that if $b\in B$, then there is a sequence $(R_n)_n$ of subsets of $D$ such that $b=\bigwedge_{n}\bigvee R_{n}$.

$\phantom{spa}$ iii. there is a sequence $(C_n)_n$ of subsets of $B$ with $B^{+}=\bigcup_{n}C_{n}$ such that

$\phantom{spaces}$ a. For all $n$ there is an integer $K(n)$ such that every cellular family in $C_{n}$ has at most $K(n)$ elements.

$\phantom{spaces}$ b. If $a_{n}\not\in C_{n}$ for all $n$, then $\overline{Lim}_{n\rightarrow\infty}a_{n}=0$.

$\phantom{spaces}$ c. For all $k$, if $\overline{Lim}_{n\rightarrow\infty}a_{n}=0$, then $a_{n}\in C_{k}$ for all but finitely many $n$.

IV. $B$ satisfies the following.

$\phantom{spa}$ i. $B$ is atomless.

$\phantom{spa}$ ii. there is a countable subset $D\subseteq B$ such that if $b\in B$, then there is a sequence $(R_{n})_{n}$ of subsets of $D$ such that $b=\bigwedge_{n}\bigvee R_{n}$.

$\phantom{spa}$ iii. $B$ is weakly distributive.

$\phantom{spa}$ iv. $B^{+}=\bigcup_{n}C_{n}$ for some sequence $C_{n}$ of sets where

$\phantom{spaces}$ a. For all $n$ there is an integer $K(n)$ such that every cellular family in $C_{n}$ has at most $K(n)$ elements.

$\phantom{spaces}$ b. If $a\vee b\in C_{n}$, then $a\in C_{n+1}$ or $b\in C_{n+1}$.

V. $B$ satisfies the following.

$\phantom{spa}$ i. $B$ is atomless.

$\phantom{spa}$ ii. there is a countable subset $D\subseteq B$ such that if $b\in B$, then there is a sequence $(R_{n})_{n}$ of subsets of $D$ such that $b=\bigwedge_{n}\bigvee R_{n}$.

$\phantom{spa}$ iii. $B$ is weakly distributive.

$\phantom{spa}$ iv. $B$ is uniformly concentrated.

$\mathbf{Proof}$ Simply combine the characterizations of measure algebras in [2] with Theorem 1. QED

We may also use [2] to get similar representations of the measure algebras produced by the standard measure on $2^{\kappa}$ for all cardinals $\kappa$ using Jech's results and Maharam's theorem. In order to state Maharam's theorem we will need a couple definitions.

If $B$ is a Boolean algebra, then recall that $B\upharpoonright a$ is the relative Boolean algebra $\{b\in B|b\leq a\}$.

If $B$ is a complete Boolean algebra, then we say that $X\subseteq B$ completely generates $B$ if whenever $X\subseteq C\subseteq B$ and $C$ is a complete subalgebra of $B$, then $C=B$. Let $\tau(B)=\min\{|X|:X\subseteq B,X\,\textrm{completely generates}\,B\}$.

We say that a complete Boolean algebra $B$ is $\tau$-homogeneous if $\tau(B)=\tau(B\upharpoonright a)$ for each $a\in B^{+}$.

Let $\mu_{\kappa}$ be the standard measure on $2^{\kappa}$. Let $Bor_{\kappa}$ denote the set of Borel subsets of $2^{\kappa}$ and let $I_{\kappa}$ be the Borel subsets of $2^{\kappa}$ of measure zero. Let $P_{\kappa}=Bor_{\kappa}/I_{\kappa}$, and let $\tilde{\mu}_{\kappa}$ denote the measure on the Boolean algebra $P_{\kappa}$ induced by $\mu_{\kappa}$.

$\mathbf{Theorem}$ $\mathbf{3}$(Maharam)(see [3][Ch. 22 Sec. 3]) If $(B,\mu)$ is a $\tau$-homogeneous measure algebra with $\tau(B)=\kappa$ and $\mu(1)=1$. Then $(B,\mu)$ is isomorphic as a measure algebra to $(P_{\kappa},\tilde{\mu}_{\kappa})$.

Therefore, if we combine Maharam's theorem with the characterizations of the measure algebras given by Jech in [2], then we obtain purely algebraic characterizations of the algebras $P_{\kappa}$.

References.

1. H.L. Royden, Real analysis, Third edition (1988).

2. Jech, Thomas. Algebraic characterizations of measure algebras. Proc. Amer. Math. Soc. 136 (2008), no. 4, 1285–1294.

3. Monk, J. Donald, Robert Bonnet, and Sabine Koppelberg. Handbook of Boolean Algebras. Vol. 1-3. Amsterdam: North-Holland, 1989.

Answer 3

I was skimming through the handbook of set theory, and I stumbled upon a combinatorial characterization of the Cohen algebras $Ro(2^{\kappa})$ where $\kappa$ is any infinite cardinal and $2^{\kappa}$ is given the product topology. Recall that a Boolean subalgebra $A$ of a Boolean algebra $B$ is a regular subalgebra if $\bigvee^{A}R=1$ implies that $\bigvee^{B}R=1$ as well.

Suppose that $\kappa$ is a regular cardinal and $X$ is a set with $\kappa\leq|X|$. Let $[X]^{<\kappa}=\{R\subseteq X:|R|<\kappa\}$. A set $L\subseteq[X]^{<\kappa}$ is said to be unbounded if whenever $R\in[X]^{<\kappa}$ there is some $S\in[X]^{<\kappa}$ with $R\subseteq S$. A set $L\subseteq[X]^{<\kappa}$ is said to be closed if whenever $\lambda<\kappa$, $R_{\alpha}\in L$ for $\alpha<\lambda$, and the sequence $(R_{\alpha})_{\alpha<\kappa}$ is increasing, then $\bigcup_{\alpha<\lambda}R_{\alpha}\in L$ as well. A set $C\subseteq[X]^{<\kappa}$ is said to be a closed set if it is closed and unbounded.

A Boolean algebra $B$ is isomorphic to $Ro(2^{\kappa})$ if and only if there is a club set $C\subseteq[B]^{<\aleph_{1}}$ such that each $R\in C$ is a regular Boolean subalgebra of $B$ and whenever $R,S\in C$, we have $\langle R\cup S\rangle\in C$ as well (here $\langle R\cup S\rangle$ is the subagebra generated by the union $R\cup S$.)

1 Semi-Cohen Boolean algebras, Bohuslav Balcar, Thomas Jech, Jindřich Zapletal (1997)

[2] Handbook of set theory M. Foreman - Akihiro Kanamori - Springer - 2010

[3] KOPPELBERG, S. (1993), Characterizations of Cohen Algebras. Annals of the New York Academy of Sciences, 704: 222–237. doi: 10.1111/j.1749-6632.1993.tb52525.x