This is a question about forcing. I have seen the following fact mentioned in multiple places, but have not been able to find a proof: if a random real is added to a transitive model of ZFC, then in the generic extension the set of reals in the ground model becomes meager.

My guess is that one should be able to, in some natural way, directly construct from a random real a countable sequence of nowhere dense sets covering the ground model reals, but I am not sure.

  • $\begingroup$ You mean arbitrary, not random, right? $\endgroup$ Commented Dec 9, 2009 at 0:09
  • $\begingroup$ Can you be more precise? I think I like this question... (click edit if you want to add details directly to your question). $\endgroup$ Commented Dec 9, 2009 at 0:34
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    $\begingroup$ No, he means a random real. This is a technical term in set-theoretic forcing, meaning that the real is V-generic for the forcing notion known as the measure algebra. $\endgroup$ Commented Dec 9, 2009 at 0:47
  • $\begingroup$ Ah, just making sure. $\endgroup$ Commented Dec 9, 2009 at 1:46

2 Answers 2


The proof is based on the fact that there is a decomposition ${\bf R}=A\cup B$ of the reals such that $A$, $B$ are (very simple) Borel sets, $A$ is meager, $B$ is of measure zero, and ${\bf R}=A\cup B$ even holds if after forcing we reinterpret the sets. Nos let $s$ be a random real. If $r\in {\bf R}$ is an old real, then $s\notin r+B$, so $s\in r+A$, that is, the meager $s-A$ contains all old reals.

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    $\begingroup$ ALthough there are, as Peter says, very simple ways to partition the reals into a meager set and a measure-zero set, a slightly less simple example may be better appreciated by non-set-theorists: Working with $2^\omega$ as our "reals", take $A$ to be the set of those sequences that obey the strong law of large numbers (the limit, as $n\to\infty$ of the proportion of 1's among the first $n$ terms is $1/2$). This $A$ contains almost all reals in the sense of measure but almost none in the sense of Baire category. (A generic real has, for infinitely many $n$, nothing but 0's from $n$ to $n!!$.) $\endgroup$ Commented Aug 17, 2010 at 16:43

Joerg Brendle mentions this fact and cites:

K. Kunen, Random and Cohen reals, Handbook of set-theoretic topology (K. Kunen and J. Vaughan, eds.), North{Holland, Amsterdam, 1984, pp. 887{911. MR 86d:03049

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    $\begingroup$ It is Theorem 3.20 in Kunen´s article, which also proves (at once) that after adding a Cohen real, the ground model reals become null. $\endgroup$ Commented Mar 6, 2013 at 19:39

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