Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question
    
Also in math.SE: math.stackexchange.com/questions/374492/… –  Asaf Karagila Apr 27 '13 at 19:34
    
Although there is an accepted answer, it only answers question (1). Question (2) - whether there is a measure-free algebraic/combinatorial characterisation of the measure algebra - seems to me an interesting question, so I am highlighting here that it is still unanswered. –  Alex Simpson Apr 30 '13 at 9:36
    
Using the paper "Algebraic characterizations of measure algebras" by Thomas Jech, I developed a few purely algebraic combinatorial characterizations of the random algebra when $X=2^{\omega}$. –  Joseph Van Name Apr 30 '13 at 14:38
    
@Alex ... Yes, 2nd one still unanswered. Let's wait for the answer (there maybe partial answer to it in measure theory point of view. Let's wait for some other mathematicians and their ideas. @Joseph, I saw that paper .. thanks for info. –  Rina Shora Apr 30 '13 at 16:30
    
@Joseph. Thanks for now extending your answer with the combinatorial characterisations answering (2). –  Alex Simpson May 1 '13 at 9:36
add comment

2 Answers

up vote 6 down vote accepted

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.

share|improve this answer
    
I see now that you didn't actually insist on c.c.c.; but if you drop this, of course there are even more counterexamples. –  Joel David Hamkins Apr 27 '13 at 21:27
    
Meanwhile, there is a characterization of $\Add(\kappa,1)$, which is like $2^\kappa$ but using $\lt\kappa$ support instead of finite support. If that is what you had meant, then I can post something about this. –  Joel David Hamkins Apr 27 '13 at 21:29
    
@Joel , thanks for your reply, very nice examples. Yes both algebras satisfy ccc. Is it possible to find an approach ( to prove it under strong assumption) to Question 1 without forcing methods. I will be pleased to share with me any result you know, or even refer me to the references –  Rina Shora Apr 28 '13 at 0:22
    
If you don't want to think about forcing, then I suggest the example of the regular open algebra arising from a Suslin line (the existence of such a space is consistent with ZFC, but not provable). This will be c.c.c. and $\omega_1$-dense, but the Boolean algebra is $(\omega,\infty)$-distributive, and so it is not isomorphic to the regular open algebra you suggest for the case of $\kappa=\omega_1$. –  Joel David Hamkins Apr 28 '13 at 0:54
    
Exellent example (consistency in ZFC is enough). What about the second question (random algebra) even when $\kappa$ is the cardinality of continuum ($X=\mathbb{R}$). Is there any characterization. –  Rina Shora Apr 28 '13 at 11:02
show 2 more comments

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.

share|improve this answer
    
@Joseph Thank you for your reply. I am aware of measure algebra, but do not you think that random algebra is little bit stronger than measure algebra, because random = Bor(X)/ideal of null sets, measure algebra=measurable (Lebesgue) set of X/ideal of null set, and every Borel is measurable but not vice versa. Yes that result of true, but I thought there maybe some other results stronger that these. I got the book you recommended, it is very good one. I will be pleased to hear anything else from you. –  Rina Shora Apr 29 '13 at 11:18
1  
For every measurable set $X$ there exist Borel sets $A,B$ of the same measure as $X$ with $A \subseteq X \subseteq B$. So your "random algebra" is isomorphic to the measure algebra. –  Alex Simpson Apr 29 '13 at 11:48
    
Dear Alex ... Thanks for this clarification. –  Rina Shora Apr 29 '13 at 23:47
    
@Joseph Thanks again, this is really a nice characterization. I appreciate you help and the time you spent following these results. –  Rina Shora May 2 '13 at 1:06
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.