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I was recently reading a Ph.D. thesis in which the following result was cited:

(Jockusch-Mohrherr) Let $A$ be any r.e. set except for $\mathbb{N}$. $A$ is creative iff for every r.e. set $B$ disjoint from $A$, it follows that $A \equiv_{1} A \cup B$.

I found this interesting and was trying to track down the source of the theorem, but nothing was given in the bibliography. Does anyone know where I could find this result? Or, I guess, alternatively provide a sense of how to prove this? Thanks.

2 Answers 2

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You can use the fact that being creative is the same as being $1$-complete.

Assume $A$ is creative and fix some r.e. $B \subset \overline{A}$. If we show that $A \cup B$ is also creative then $A \equiv_1 A \cup B$ follows, since both these sets are $1$-complete. Let $W_x \subset \overline{A \cup B}$, then $W_x \cap B$ is an r.e. subset of $\overline{A}$. Using productive function for $A$ we can effectively find an element $y \in \overline{A} \setminus (W_x \cap B)$, so $x \in \overline{A \cup B} \setminus W_x$.

To show the converse, note that since $A \neq \mathbb{N}$ its complement must be infinite. Assume that $\overline{A}$ contains an infinite r.e. set $B$. Take an infinite recursive $R \subseteq B$ and a creative $C \subseteq R$. We then have $A \equiv_1 A \cup C$, and since $A \cup C$ is creative (as in the proof above) $A$ is also creative. Now let us show that $\overline{A}$ contains an infinite r.e. subset. Assume the converse and pick some $n \notin A$. Set $B = A \cup \{n\}$ and fix an infinite recursive $R \subseteq A$. There is a recursive permutation $p$ mapping $R \cup \{n\}$ onto $R$. Then $B \leqslant_1 A$ via $$ g(x) = \begin{cases} x,& \text{if } x \notin R \cup \{n\},\\ p(x),& \text{otherwise.} \end{cases} $$ Now assume that $A \leqslant_1 B$, hence $A \equiv_1 B$ and hence they are recursively isomorphic by Myhill's theorem. Let $h$ be a recursive permutation mapping $A$ onto $B$. Then $n, h(n), hh(n), \dots$ is an infinite r.e. subset of $\overline{A}$, a contradiction.

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    How are you sure about the existence of the creative set C in R? – 2017-02-15
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    @Wore, it is true even for r.e. sets. Take some creative set $K$, now let $f$ be a 1-1 enumeration of $R$ and consider r.e. set $C = f(K) \subset R$. We have $K \leqslant_1 C$ and hence $C$ is also creative. – 2017-02-15
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The result is stated as exercise 4.17 in Soare's Recursively Enumerable Sets and Degrees, and the only Jockusch-Mohrherr publication in the reference list of that book is:

C.G. Jockusch, Jr. and J. Mohrherr (1985). Embedding the diamond lattice in the r.e. tt-degrees, Proceedings of the AMS 94, pp. 123-128.