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A set of reals $X$ is $\textit{strong measure zero}$ if for any sequence of real numbers $ ( \epsilon_n ) _{n \in \omega }$ there is a sequence of open intervals $ ( a_n ) _{n \in \omega }$ which covers $X$ and such that each $ a_i $ has length less than $ \epsilon _i $.

The Borel Conjecture (BC) is the statement that a set of reals is strong measure zero iff it is countable. (It's easy to see that any countable set is strong measure zero, so BC just says that there are no uncountable strong measure zero sets.)

I heard somewhere that adding a Cohen real necessarily destroys BC, i.e., if $x$ is a Cohen real over $V$, then $V[x] \models \neg BC$. I can see why this is true for adding $\omega_1$ Cohen reals: for any $\epsilon$-sequence in the ground model, and any uncountable set of reals $X$ in the ground model (which is still uncountable after ccc forcing), we can use a single Cohen real to generically choose small intervals which will cover $X$ while staying within the constraints of the $\epsilon$-sequence. So when we add $\omega_1$ Cohen reals successively, every $\epsilon$-sequence shows up at some countable stage, and at the next stage we create a corresponding cover for $X$. Thus $X$ will be strong measure zero in the forcing extension. But here we are using the fact that every $\epsilon$-sequence shows up at some stage, which we then proceed to force over with a Cohen real; and that will not be true if we only add a single Cohen real.

So my question is: does adding a single Cohen real necessarily destroy BC?

And more generally: after adding a single Cohen real, what can we say about "the set of all reals from the ground model"? Is it meager? Measure zero? Does it contain no perfect sets?

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up vote 11 down vote accepted

(An attempt at an answer, and also my first posting here. Thanks to Andres Caicedo for the reformatting.)

I claim that a single Cohen real makes the set of old reals strong measure zero. Reals are functions from $\omega$ to 2.

Let ${\mathbb C}$ be Cohen forcing, and let $c$ be the name of the generic real.

Let $(n_k)$ be a sequence of ${\mathbb C}$-names for natural numbers. I will find a sequence $(s_k)$ of names for finite $01$-sequences ($s_k$ of length $n_k$) such that ${\mathbb C}$ forces: every old real is in some $[s_k]$.

Let $D_k$ be a dense open set deciding the value of $n_k$ and containing only conditions of length at least $k$. Say, each $q$ in $D_k$ decides that the value of $n_k$ is $f_k(q)$, where $f_k$ is a function in the ground model defined on $D_k$. Each $f_k$, and also the sequence $(f_k)$, is in $V$.

Now we work in the extension. (The point is that even though we now know the actual values of $n_k$, we will play stupid and use the names only, plus the minimal amount of information that we need from the generic real. This lets us gauge exactly how much information from the generic we need.)

In the extension I will define a sequence $(i_k)$ of natural numbers. Let $i_k$ be the minimal $i$ such that $c \mathord\upharpoonright i$ is in $D_k$, where $c\mathord\upharpoonright i = c$ restricted to $i$. (So $i_k$ is at least $k$.)

For each $k$ we now define a $01$-sequence $s_k$ of length $n_k$ as follows: Take $n_k$ successive bits from the Cohen real $c$, starting at position $i_k$. (Formally: $s_k(j) = c(i_k+j)$ for all $j\lt n_k$.)

I claim that "every old real is in some $[s_k]$" is forced. Assume not, so let $p$ force that $x$ is not covered. Let $k$ be larger than the length of $p$. So $p$ not in $D_k$. Extend $p$ to $q$ so that $q$ is in $D_k$, $q$ minimal. Let $l$ be the length of $q$. So $q$ forces that $i_k$ is exactly $l$. Also $q$ forces that $n_k = f_k(q)$. Now extend $q$ to $q'$, using the first $f_k(q)$ bits of $x$. So $q'$ is stronger than $q$, and $q'$ forces that $s_k$ is an initial segment of $x$.

mg*

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Welcome to MO, Martin. It might help some readers to point out that, when you say, after defining s_k, that it "is a 01-sequence of length n_k", you are implicitly shifting the domain of s_k from [i_k,i_k+n_k] to [0,n_k]. (Also, the length of this is n_k + 1, but who cares?) –  Andreas Blass May 6 '11 at 13:05
    
Thanks, Martin! –  Andres Caicedo May 6 '11 at 13:39
    
Thanks, Andreas, I reformulated it. –  Goldstern May 7 '11 at 20:11
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Andy:

A good reference for your questions is "Consequences of adding Cohen reals" by J. Steprans, in "Set Theory of the reals", Judah ed., Bar-Ilan University, 1993, pp. 583-617. Another reference is the excellent book "Set theory: on the structure of the real line" by Bartoszynski and Judah.

Adding a Cohen real makes the ground model reals have measure zero. I mentioned this in another answer, and it is similar to what you write in your post. Briefly: Any nontrivial countable forcing notion is equivalent to Cohen. For example, fix $\epsilon>0$ and let ${\mathbb P}_\epsilon$ be the collection of subsets $A$ of ${\mathbb R}$ that are the finite union of open intervals with rational endpoints such that the measure of $A$ is less than $\epsilon$. Ordered by (reverse) inclusion, this poset is equivalent to Cohen forcing. The generic is an open set of measure $\epsilon$ that covers the ground model reals. Since $\epsilon$ was arbitrary, the result follows. Steprans's paper contains a proof of a strengthening of this fact.

On the other hand, the ground model reals are not meager in the extension. In fact, any second category set of reals in the ground model is still second category after adding a Cohen real. (This is also proved in Steprans's paper.)

Finally, it is a theorem of Velickovic and Woodin that if $V\subseteq W$ are models of set theory, and for every countable $X\in W$ with $X\subseteq{\mathbb R}^V$ there is a $Y\in V$ countable in $V$ with $X\subseteq Y$, then, if there is a perfect set consisting of reals from $V$, then every real is in $V$. In particular, the ground model reals cannot contain a perfect set after adding a Cohen real. For a reference, see Velickovic-Woodin, "Complexity of reals in inner models of set theory", Annals of pure and applied logic, 92 (1998), 283-295.


[Edit: Martin Goldstern has now posted a nice argument by a more natural line of thought than my naive approach.]

Here is a sketch of an argument I think, if one manages to flesh out, should show that adding a Cohen real makes ${\mathbb R}^V$ be strong measure zero. By the way, I was unable to find an explicit mention of this result or a proof in any of the references I mentioned (or elsewhere). Laver's original paper ("On the consistency of Borel's conjecture", Acta Math., 137 (1976), no. 3-4, 151–169) only mentions in passing (page 155) that adding a Cohen real makes ${\mathbb R}^V$ strong measure zero. There may not be an explicit reference in print, actually.

We need the following basic property of Cohen forcing:

If $c$ is Cohen over $V$ and $g:\omega\to\omega$ in $V[c]$, then there is an $h:\omega\to\omega$ in $V$ such that $g(n)\lt h(n)$ infinitely often.

Let $c$ be Cohen over $V$ and let $\vec\varepsilon=(\varepsilon_0,\varepsilon_1,\dots)$ be a sequence in $V[c]$ of positive reals. As you observe, if $\vec\varepsilon\in V$, then $$ {\mathbb R}\cap V\subseteq \bigcup_n I^n_{c(n)} $$ where (for all $n$) the sequence $(I^n_m\mid m<\omega)$ lists (in $V$) all intervals with rational endpoints and length at most $\varepsilon_n$.

Suppose now that $\vec\varepsilon\in V[c]\setminus V$. Without loss, each $\varepsilon_m$ has the form $2^{-n_m}$ for some positive integer $n_m$, and the sequence is strictly decreasing. Fix $h\in V$ such that $A=\{m\mid n_m\le h(m)\}$ is infinite.

For each $n$, let $(I_{n,m}\mid m\in\omega)$ list (in $V$) all intervals with rational endpoints and length at most $2^{-h(m)}$. The argument "reduces" then to proving the following key fact:

A genericity argument should show that, in fact, $$ {\mathbb R}\cap V\subseteq \bigcup_{n\in A}I^n_{c(n)}. $$

If we manage to prove this, since the sequence of numbers $n_m$ is increasing, we are done.

Note that another basic property of Cohen forcing gives us that there is a real $t$ in the ground model coding a Borel function $f$ such that $A=f(c)$. To make the argument work, something stronger (like $f$ recursive with the real $t$ as oracle) seems needed. I don't see how to make the result go through for arbitrary Borel $f$.

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@Andres: I know that one of the key differences between Random and Cohen reals is that Random reals are dominated by some function in the ground model, while Cohen reals are not (as you mentioned in your answer). Is this the key property for the ground model reals becoming of measure zero? (i.e. if we add one Random real, does it "nullify" (in a measure theoretic sense) the real numbers of the ground model?) –  Asaf Karagila May 1 '11 at 15:23
    
@Asaf: Adding a random real does not make the ground model reals null. In fact, the outer measure of a set is preserved by forcing with a Random algebra. That said, I am not sure that "non-dominance" is the key in the argument above: Note that the sketch also exploits that conditions are finite, and so they can be extended very easily, with much more freedom than with random forcing. –  Andres Caicedo May 1 '11 at 15:58
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I tried an approach very similar to Andres's, but I couldn't make it work. I fear there may be a similar problem in Andres's argument, where he says that a genericity argument shows that $\mathbb{R}\cap V$ is covered by the intervals $I^n_{c(n)}$ for $n\in A$. The trouble is that $A$ can depend on the Cohen real $c$, and that (Murphy's law?) the dependence will w.l.o.g. conspire against what we want to happen. So I don't yet see how to do this genericity argument. –  Andreas Blass May 1 '11 at 20:20
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Here is an argument showing that adding a single Cohen reals yields an uncountable strong measure zero set.

In the spirit of the previous comments, let $r_n$ for $n\in\omega$ enumerate the rationals. Let Cohen forcing be represented by the set of finite partial functions from $\omega$ to $\omega$ under inclusion and let $G$ be a name for the generic function $G:\omega \to \omega$. For any pair of functions $f:\omega \to \mathbb N$ and $F:\omega \to \omega$ define $B(f,F)=\bigcup_{n\in\omega}(r_{G(F(n))} - 1/f(n),r_{G(F(n))} + 1/f(n))$. It suffice to show that for any Cohen name $\dot{f}$ there is $F:\omega \to \omega$ such that it is forced that the ground model reals are contained in $B(\dot{f},F)$.

To this end, for each Cohen condition $p$ define $F_p(m)$ to be some integer $k$ such that there is $q:k\to \omega$ such that $q\supset p$ and $q$ decides a value for $\dot{f}(m)$. Let $F:\omega \to \omega$ be such that $F\geq^* F_p$ for each $p$. To see that it is forced that $B(\dot{f},F)$ contains the ground model reals let $p$ be an arbitrary Cohen condition and $x$ a ground model real. Let $m$ be such that $F(m)>F_p(m)$. Choose $q:k \to \omega$ witnessing that $F_p(m) = k$ and suppose that $q$ forces $\dot{f}(m)=j$. Let $i\in\omega$ be such that $x\in (r_i- 1/j,r_i+1/j)$. Let $q'= q\cup\{(F(m),i)\}$. Then $q'$ forces that $G(F(m))=i$ and hence that $x\in (r_{G(F(m))}- 1/\dot{f}(m),r_{G(F(m))}+1/\dot{f}(m))\subseteq B(\dot{f},F)$.

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