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?



Cohen reals and strong measure zero sets
(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\*


