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Let $\kappa$ be an infinite cardinal and let $B$ be the random forcing for adding $\kappa$-many random reals.

Question: What are the elements of $B$. More precisely given a condition $p \in B$, what are the domain and the range of $p$, if there are any?. What does it mean "a coordinate of $p$"?.

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    $\begingroup$ This question is more or less the same as this one: mathoverflow.net/questions/55832/what-is-random-real-forcing $\endgroup$
    – Eran
    Commented Oct 28, 2011 at 17:12
  • $\begingroup$ I think that now there is somewhat of a stronger set theorists base in math.stackexchange.com that this question can be asked there instead. $\endgroup$
    – Asaf Karagila ♦
    Commented Oct 28, 2011 at 18:34
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    $\begingroup$ @Eran: Not really, the question you linked to is about adding one random real. This one is more about the subtleties of the iteration. Note that there is at least two different ways of adding $\kappa$-many random reals. $\endgroup$
    – Ramiro de la Vega
    Commented Sep 9, 2014 at 13:20
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    $\begingroup$ I just realized this question already has an accepted answer. Who is voting to close it three years later? and why? $\endgroup$
    – Ramiro de la Vega
    Commented Sep 9, 2014 at 13:26

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There are different ways of representing $B$. One possibility is to let $B$ consist of all closed subsets of $2^\kappa$ of positive measure; another possibility is to allow all positive Borel sets. Two conditions are equivalent if their symmetric difference has measure zero.

Each basic clopen set has a finite set of coordinates. By ccc, each open set is almost the same as some open set that uses only countably many coordinates. So every closed/Borel set $p$ is equivalent to a set that uses only countably many coordinates. I guess that these are the "coordinates of $p$".

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    $\begingroup$ It might be worth adding that the complete Boolean algebra associated to this notion of forcing is the algebra of Borel subsets of $2^\kappa$ modulo sets of measure 0. In particular, thanks to the ccc, this algebra really is complete, not just countably complete as one might at first think. $\endgroup$
    – Andreas Blass
    Commented Oct 28, 2011 at 18:56

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