Random forcing does not add Cohen real.

Question

I know that random forcing does not add unbounded reals. Is a Cohen real always unbounded?

I can proof that the product of two random forcings adds a Cohen real. How can I show that the two step iteration of random forcings does not add cohen real?

Answer 1

Cohen reals are unbounded, this is fairly straightforward to show when presenting the Cohen forcing with $\omega^{<\omega}$ rather than $2^{<\omega}$.

Suppose that $f\colon\omega\to\omega$ is in the ground model, and let $\dot c$ be the canonical name for the Cohen real, so for any $p\in\omega^{<\omega}$ and $n<\omega$ we can find $m>n$ and $q\leq p$ such that $q(m)>f(m)$, simply by letting $m=\max\operatorname{dom} p+1$ and setting $q=p\cup\{(m,f(m)+1)\}$, for example.

This shows that $\bf 1$ already forces that $\check f$ does not dominate $\dot c$. Therefore this holds for any $f$ in the ground model, so $\dot c$ is unbounded. $\square$

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For the iteration of random reals, show that it is isomorphic to adding a single random real by showing that the measure algebra is the same measure algebra.