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(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 $k$.) Now let $s_k$ be the Cohen real restricted to the interval $[i_k , i_k+n_k]$. This is a $01$-sequence of length $n_k$.

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*

(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*

(An attempt at an answer, and also my first posting here. My browser has problems with rendering, so I write in pseudo-TeX.)

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 $c\upharpoonright i$ is in $D_k$, where $c\upharpoonright i = c$ restricted to $i$. (So $i_k$ is at least $k$.) Now let $s_k$ be the Cohen real restricted to the interval $[i_k , i_k+n_k]$. This is a $01$-sequence of length $n_k$.

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*

(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 $k$.) Now let $s_k$ be the Cohen real restricted to the interval $[i_k , i_k+n_k]$. This is a $01$-sequence of length $n_k$.

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*

(An attempt at an answer, and also my first posting here. My browser has problems with rendering, so I write in pseudo-TeX.)

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

Let C be Cohen forcing, and let c be the name of the generic real.

Let (n_k) be a sequence of 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 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|i is in D_k, where c|i = c restricted to i. (So i_k is at least k.) Now let s_k be the Cohen real restricted to the interval [i_k , i_k+n_k]. This is a 01-sequence of length 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*

(An attempt at an answer, and also my first posting here. My browser has problems with rendering, so I write in pseudo-TeX.)

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 $c\upharpoonright i$ is in $D_k$, where $c\upharpoonright i = c$ restricted to $i$. (So $i_k$ is at least $k$.) Now let $s_k$ be the Cohen real restricted to the interval $[i_k , i_k+n_k]$. This is a $01$-sequence of length $n_k$.

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*