TITLE: Is every non-recursive set in $\Sigma_1$ complete in $\Sigma_1$ (relatively to many-to-one reductions)?
QUESTION [6 upvotes]: Most well known sets in $\Sigma_1 \setminus\Delta_0$, such as the
  Halting problem, are complete in $\Sigma_1$, relatively to the
  many-to-one reduction. In fact I don't know any example of a (non recursive) set in $\Sigma_1$ that
  is not complete.
Is every set in $\Sigma_1\setminus\Delta$ complete in $\Sigma_1$?

If so, why the trouble of proving completeness for such problems?
If not (is there some Ladner-type Theorem?), can anyone give
natural examples of $\Sigma_1\setminus\Delta_0$ sets that are not complete?
Of course the question generalizes for other levels of the arithmetic hierarchy.
I apologize if this question was already answered here, or if the answer is well known.

REPLY [4 votes]: Let $\Omega$ be the halting probability, and let $X$ be the set of rational numbers less than $\Omega$.  I think I can reasonably claim that $X$ is a natural set.
Since $\Omega$ is left-c.e., $X$ is $\Sigma^0_1$.  Also, $X$ is non-computable since it's Turing equivalent to $\Omega$.  However, $X$ cannot be many-one complete for $\Sigma^0_1$ sets.  Indeed, no convex set of rationals (or finite union of convex sets of rationals) can be many-one complete for $\Sigma^0_1$ sets.
Suppose $X$ were complete.  Note that $\emptyset'$ is uniformly many-one complete: from a $\Sigma^0_1$ index for a set, we can compute an index for its reduction to $\emptyset'$.  Since $X$ is complete, there is a reduction from $\emptyset'$ to $X$, and so by composing these we see that $X$ is uniformly many-one complete (this argument works for any complete set).
Now, we'll make a $\Sigma^0_1$ set $V$, and by a standard argument with the recursion theorem we can assume we already know an $n$ with $V = W_n$.  Using the uniformity, this means we know the index of a computable reduction $f$ from $V$ to $X$.
Begin by computing $f(0), f(1)$ and $f(2)$.  These are rational numbers, and so are arranged in some order.  Without loss of generality, let's assume that $f(0) \le_\mathbb{Q} f(1) \le_\mathbb{Q} f(2)$.  Then enumerate $0$ and $2$ into $V$, but leave $1$ out.  By assumption, this means that $f(0)$ and $f(2)$ are in $X$, but $f(1)$ is not.  But this contradicts the convexity of $X$.