Is forcing with open intervals on [0, 1] equivalent to Cohen forcing?

Question

When I first tried to understand random forcing (which according to Wikipedia is forcing with $(\operatorname{Bor}(I),\subseteq,I)$, where $I = [0,1]$ and $\operatorname{Bor}(I)$ is the collection of Borel subsets of $I$ having non-zero Lebesgue measure), I misunderstood Borel sets as finite unions of closed-open intervals.

But from my (limited) understanding, closed and open shouldn't matter that much, and finite unions should already been taken care of by using the poset formulation instead of the Boolean algebra formulation, so it seemed to me that I might as well just use open intervals on $[0, 1]$. Then, it seemed that I should be able to approximate any interval by intervals whose endpoints have a finite binary representation, and the latter type of intervals are essentially demanding the "new real number" $r$ to start with some specified bits (again, up to finite union and ignoring some interval endpoints), so it seemed that what I got was equivalent to Cohen forcing with the forcing poset $(\operatorname{Fin}(\omega,2),\supseteq,0)$.

Since then, reading on the internet have given me a better understanding of Borel sets, but I still wonder if my intuitions about my mistaken "random forcing notion" was correct.

Answer 1

In this context there is a crucial difference between open and closed sets. This is explained by the following observation: Any open set of positive measure contains a closed set of positive measure, but the converse is not true.

The forcing $\mathbb P$ of non-empty open subsets (or open intervals for that matter) of $[0, 1]$ is indeed equivalent to Cohen forcing. The reason is that $\mathbb P$ has a countable dense subset $\mathbb Q$, namely all nonempty open intervals with rational endpoints. So $\mathbb P$ is forcing equivalent to $\mathbb Q$. Now Cohen forcing densly embeds into any atomless countable partial order (it is a good exercise to prove this! As a warm up show that $(\mathrm{Fin}(\omega, \omega), \supseteq)$ densly embeds into $(\mathrm{Fin}(\omega, 2),\supseteq)$ and vice versa). So $\mathbb Q$ is equivalent to Cohen forcing.

If instead you consider $\mathbb P'$ consisting of closed subsets of $[0, 1]$ with positive measure then this gives you back random forcing, indeed $\mathbb P'$ is a dense subset of random forcing by inner regularity of the Lebesgue measure.