*This is a crossposting of https://math.stackexchange.com/questions/1446602/anti-random-reals, which has not gotten any answers; after thinking about the problem, I've become more convinced that it belongs here instead.*

For a function $f: \mathbb{N}\rightarrow\mathbb{R}_{>0}$ and a set $X\subseteq \mathbb{R}$, an *$f$-cover of $X$* is a sequence $(I_n)_{n\in\mathbb{N}}$ of open intervals with rational endpoints such that

$X\subseteq\bigcup I_n$, and

$\mu(I_n)<f(n)$.

Say that a set $X\subseteq\mathbb{R}$ is *$f$-small* if $X$ has an $f$-cover.

Talking about $f$-covers provides us with many different refinements of the notion of "measure zero": e.g., a set is *strong measure zero* if it has an $f$-cover for *every* function $f$.

Say that a set of reals $X$ is *computably strong measure zero* (csmz) if there is an $e$ such that $\Phi_e^f$ is an $f$-cover of $X$ whenever $f$ is a function from $\mathbb{N}$ to $\mathbb{R}_{>0}$. (This is sadly *not* the same as *effective strong measure zero*, a notion introduced by Kihara in his thesis; see also http://www.sciencedirect.com/science/article/pii/S016800721400044X.)

In computability theory, a real is said (informally) to be "random" if it is not in any "simple" measure-zero set; there are of course many ways to formalize this, but this is the basic theme. Csmz sets provide a very strong notion of non-randomness: say that an *individual* real $r$ is **antirandom** if $\{r\}$ is csmz - equivalently, if $r$ is contained in some csmz set. My question is:

What are the antirandom reals?

I strongly suspect that every antirandom real is computable, but I can't prove it. **EDIT: As Joe Miller's answer below shows, my guess was really wrong!**

**Comment 1**. The set of antirandom reals is countable, but the proof is surprisingly nontrivial. There is a countable set $\{A_i: i\in\omega\}$ of csmz sets such that every csmz set is contained in one of the $A_i$s; specifically, take $A_i$ to be the largest csmz set witnessed to be csmz by $\Phi_i$. Now, it can be forced that every strong measure zero set is countable (this is *Borel's conjecture*) - in such a forcing extension, the antirandoms are countable. Meanwhile, antirandomness is a $\Pi^1_1$ property. This means the statement "there are countably many antirandom reals" is $\Sigma^1_3$, and so absolute assuming large cardinals.

This argument is probably overkill, but (1) we do seem to need Borel's conjecture for coanalytic sets, which is independent of ZFC, and (2) we do seem to need $\Sigma^1_3$-absoluteness for proper forcing, which has nontrivial large cardinal strength (see http://www.logic.univie.ac.at/~sdf/papers/bagfried.pdf).

**Comment 2**. Thomas Andrews, in the comments section to the original question, proposed looking at "c-csmz" sets - these are sets which are csmz, but where we only allow *computable* sequences of epsilons. Say that a real $r$ is *weakly antirandom* if $\{r\}$ is a c-csmz set. I believe that there are noncomputable weakly antirandom reals, but I haven't been able to prove that yet; I would be interested in these reals as well. Note that the countability proof breaks down: a c-csmz set is *not* strong measure zero.

1more comments