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

Hi all,

Can someone please explain the idea and the main steps in a random real forcing? - what makes it (the new real) different from adding a Cohen real? - is there a good reference for it?

Thanks.

share|improve this question
add comment

1 Answer

up vote 10 down vote accepted

This is relatively easy to answer. One of main differences between Cohen forcing and random real forcing is that random real forcing does not add an unbounded real.
That is, every function $f:\omega\to\omega$ in the forcing extension is bounded by function $g:\omega\to\omega$ in the ground model, in the sense that for all $n\in\omega$, $f(n)\leq g(n)$.

Let me sketch you a proof of this fact. Let $p$ be a set of positive measure $\varepsilon$ and assume that $p$ forces $\dot f$ to be a name for a function from $\omega$ to $\omega$. Find a maximal antichain $p_n$, $n\in\omega$, below $p$ such that each $p_n$ decides $\dot f(0)$, i.e., for some $m\in\omega$ forces $\dot f(0)$ to be $m$. Now finitely many of the $p_n$ will union up to a subset of $p$ that has a measure close to the measure of $p$. Call this union $q_0$. Now $q_0$ does not decide $\dot f(0)$, but it gives you an upper bound, the maximum of the values forced on $\dot f(0)$ by the finitely many $p_n$ that union up to $q_0$.
Iterating this procedure we get a decreasing sequence $q_i$ of conditions such that $q_i$ forces upper bounds on all $\dot f(k)$, $k\leq i$. If each $q_{i+1}$ is a sufficiently large subset of $q_i$ in terms of measure, then the intersection $q$ of the $q_i$ will be of positive measure and gives us a ground model function that is forced by $q$ to be an upper bound for $\dot f$. Combining this with the usual density argument tells you that every new function is bounded by a ground model function.

This property is known as $\omega^\omega$-boundingness and can be formulated in terms of a distributivity property of the corresponding Boolean algebra (the completion of the forcing notion, in this case the measure algebra on the reals). Jech elaborates on that (i.e., proves the equivalence of these two properties) in his book.

Since Cohen forcing adds an unbounded real, the $\omega^\omega$-boundingness of random real forcing shows that it does not add Cohen reals, or algebraically speaking, that the measure algebra has no countable atomless regular subalgebra. Also, it can be shown the Cohen forcing does not add a random real.

A good source for these questions is Bartoszynski and Judah: Set Theory of the Real Line.

share|improve this answer
    
thank you for your answer. Before proving the boundedness property, can you describe what is the poset, and why does the generic results in a new real? –  Eran Feb 18 '11 at 21:41
    
Sorry. The poset is all Borel subsets of $\mathbb R$ of positive measure. Since every such set contains a compact set of positive measure, it is actually enough to consider compact (or just closed) sets of positive measure. If $G$ is a generic filter for this poset, then $\bigcap G$ has a single element, and this is the random real $r$. Now, it can be shown that $G$ consists precisely of those ground model Borel sets of positive measure that contain $r$. –  Stefan Geschke Feb 19 '11 at 5:30
    
Here I need to elaborate a little bit, however: When I say "Borel set" I really mean that there is a description of how the Borel set was constructed from, say, open intervals with rational endpoints. This is a Borel code of the set. Now, when I consider a Borel set in different models of set theory, what I mean is take the same Borel code an evaluate it (i.e., construct the Borel set according to the code) in the respective model of set theory. –  Stefan Geschke Feb 19 '11 at 5:33
    
Showing that $\bigcap G$ above is nonempty is easier when you consider the poset of compact sets of positive measure. Why is the random real new? Well, it codes the generic filter $G$, and $G$ is not in the ground model for the usual genericity reasons. Also, there is Solovays characterization of random reals: A real is random over the ground model iff it is not contained in a measure zero (Borel) set coded in the ground model. This in particular says that the random real is not in any ground model singleton. –  Stefan Geschke Feb 19 '11 at 8:57
    
OK this makes it more clear. Consulting Kanamori's book, it seems that coding Borel sets allows you to prove that the new real is in a Borel set (in the extension) only if its original Borel is in G. Since G is new => the real is new. –  Eran Feb 20 '11 at 8:05
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.