aa r X i v : . [ m a t h . L O ] J a n A.Miller Recursion Theorem The Recursion Theorem and Infinite Sequences
Arnold W. Miller AbstractIn this paper we use the Recursion Theorem to show the exis-tence of various infinite sequences and sets. Our main result isthat there is an increasing sequence e < e < · · · such that W e n = { e n +1 } for every n . Similarly, we prove that there ex-ists an increasing sequence such that W e n = { e n +1 , e n +2 , . . . } forevery n . We call a nonempty computably enumerable set A self-constructing if W e = A for every e ∈ A . We show that everynonempty computable enumerable set which is disjoint from aninfinite computable set is one-one equivalent to a self-constructingset.Kleene’s Recursion Theorem says the following:For any computable function f there exists an e with ψ e = ψ f ( e ) .In this Theorem h ψ e : e ∈ ω i is a standard computable numbering of allpartial computable functions. For example, ψ e might be the partial functioncomputed by the e th Turing machine. The number e is referred to as a fixedpoint for f and this theorem is also called the Fixed Point Theorem.For a proof of Kleene’s Theorem see any of the standard references,Cooper [1], Odifreddi [3], Rogers [4], or Soare [6]. See especially Smullyan [5]for many variants and generalizations of the fixed point theorem. The Recur-sion Theorem applies to all acceptable numberings (in the sense of Rogers,see Odifreddi [3] p.215-221). All natural enumerations are acceptable. Weuse W e to denote the domain of ψ e and hence h W e : e ∈ ω i is a uniformcomputable listing of all computably enumerable sets.The proof of the Recursion Theorem is short but tricky. It can be uni-formized to yield what is called the Recursion Theorem with Parameters.The proof also yields an infinite computable set of fixed points by using thePadding Lemma. See Soare [6] pages 36-37.We will use the following version of the Recursion Theorem with Param-eters which includes a uniform use of the Padding Lemma: Mathematics Subject Classification 2000: 03E25; 03E99Keywords: Recursion Theorem, fixed points. .Miller Recursion Theorem Lemma 1
For any computable function f : ω × ω → ω there is a computablefunction h : ω → ω such that W h ( x ) is an infinite computable set for every x and for every y ∈ W h ( x ) we have that ψ f ( x,y ) = ψ y . The proof is left to the reader.It is an exercise (see Miller [2] section 13) to show (using Smullyan’sdouble recursion theorem or more particularly its n -ary generalization, see[5] Chapter IX) that for any n > e < e < · · · < e n such that W e i = { e i +1 } for i < n and W e n = { e } . It occurred to us to ask ifit would be possible to have an infinite sequence like this. We show that itis. Here is our main result: Theorem 1
There is a strictly increasing sequence: e < e < · · · < e n < · · · such that W e n = { e n +1 } for every n. ProofWe use W e,s to denote the set of all y < s such that ψ e ( y ) converges inless than s -steps. We use h x, y i to denote a pairing function, a computablebijection from ω to ω , e.g., h x, y i = 2 x (2 y + 1) − q ( e, x ) be a computable function such that for all x and e : W q ( e,x ) = (cid:26) { y } if ( ∃ s ∃ y ∈ W e,s y > x ) and h s, y i is the least such pair ∅ otherwise.Such a q is constructed by a standard argument using the s-m-n or Parame-terization Theorem. To see this one defines a partial computable function θ as follows: θ ( e, x, y ) = (cid:26) ∃ s ∃ y ∈ W e,s y > x ) and h s, y i is the least such pair ↑ otherwise. .Miller Recursion Theorem q such that ψ q ( e,x ) ( y ) = θ ( e, x, y ) for all e, x, y. Using Lemma 1 let h be a computable function such that for every e , theset W h ( e ) is an infinite set of fixed points for q ( e, · ), i.e., W x = W q ( e,x ) for all x ∈ W h ( e ) . Let e be a fixed point for h , so W e = W h ( e ) .Let x be any element of W e . Then x ∈ W h ( e ) so W x = W g ( x,e ) = { y } where y > x and y ∈ W e . Hence, starting with any e ∈ W e we get aninfinite increasing sequence as required.QEDNote that to obtain a sequence with W e n +1 = { e n } is trivial. Also notethat the sequence in Theorem 1 must be computable. This is not necessarilytrue for our next result: Theorem 2
There exists a computable strictly increasing sequence h e n : n <ω i such that for every n W e n = { e m : m > n } . ProofUsing the s-m-n Theorem find q a computable function such that for every x and e : W q ( e,x ) = { max( W e,s ) : s ∈ ω } \ { , , . . . , x } . As in the above proof, let h be a computable function such that for every e , the set W h ( e ) is an infinite set of fixed points for q ( e, · ), i.e., W x = W q ( e,x ) for all x ∈ W h ( e ) . Let e be a fixed point for h , so W e = W h ( e ) .Note that W e is infinite and let { e < e < e < . . . } = { max( W e,s ) : s ∈ ω } . For any x ∈ W e = W h ( e ) we have that W x = W q ( e,x ) = { max( W e,s ) : s ∈ ω } \ { , , . . . , x } . Hence for any n we have that W e n = { e m : m > n } . .Miller Recursion Theorem e < e < · · · such that ψ e n ( m ) = e n + m +1 for every n, m < ω. The proof of this is left as an exercise for the reader.Usually the first example given of an application of the Recursion The-orem is to prove that there exists an e such that W e = { e } . We say that anonempty computably enumerable set A is self-constructing iff for all e ∈ A we have that W e = A . So W e = { e } is an example of a self-constructing set.Our next result shows there are many self-constructing sets. Theorem 3
For any nonempty computably enumerable set B the followingare equivalent:1. B is disjoint from an infinite computable set.2. There is a computable permutation π of ω such that A = π ( B ) and A is self-constructing. Proof(2) → (1)Let E be an infinite computable set such that for every e ∈ E we havethat W e = ∅ . Any self-constructing set is disjoint from E .(1) → (2)Given any e consider the following computably enumerable set Q e . Let { c < c < c < . . . } = { max( W e,s ) : s ∈ ω } which may be finite or even empty. Then put Q e = { c n : n ∈ B } . By the s-m-n Theorem we can find a computable q such that for every e : W q ( e ) = Q e . .Miller Recursion Theorem h such that for every e the set W h ( e ) is infinite and for all x ∈ W h ( e ) we have that W x = W q ( e ) . Now let e bea fixed point for h so that W h ( e ) = W e . Let A = Q e . Then for all x ∈ A wehave that W x = A . So A is self-constructing.To get π let D and E be two infinite pairwise disjoint computable setsdisjoint from B . Take one-one computable enumerations of them: D = { d n : n < ω } and E = { e n : n < ω } . Note that W e = W h ( e ) is infinite and let C be the infinite computable set: C = { c < c < c < . . . } = { max( W e,s ) : s ∈ ω } . Since W e = W h ( e ) is a set of fixed points, it is coinfinite and hence C ⊆ W e is coinfinite. Take a one-one computable enumeration of the complement C of C : C = { c n : n < ω } . Now we can define π : π ( c n ) = d n if n ∈ Dd n +1 if n ∈ En otherwise π ( c n ) = e n . Note that π bijectively maps C to E and C to E . Furthermore if n ∈ B then π ( c n ) = n , and since A = { c n : n ∈ B } we have that π ( A ) = B .QEDAs a corollary we get that there are self-constructing sets of each finitecardinality and there is a self-constructing set which is not computable, infact, there is a creative self-constructing set.It is not hard to show that S = { e : W e is self-constructing } is Π -complete. To see this first note that it is easy to show that S is Π .We can get a many-one reduction f ofTot = def { e : W e = ω } .Miller Recursion Theorem S as follows. Fix an infinite self-constructing set A with one-to-one com-putable enumeration A = { a n : n ∈ ω } . By the s-m-n Theorem construct acomputable f so that for every e : W f ( e ) = { a n : n ∈ W e } . Hence, e ∈ Tot iff f ( e ) ∈ S . Since Tot is Π -complete (see Soare [6] page66), S is too. References [1] Cooper, S. Barry
Computability theory . Chapman & Hall/CRC,Boca Raton, FL, 2004. x+409 pp. ISBN: 1-58488-237-9[2] Miller, Arnold W.
Lecture Notes in Computability Theory ∼ miller/res[3] Odifreddi, Piergiorgio Classical recursion theory. The theory offunctions and sets of natural numbers.
With a foreword by G.E. Sacks. Studies in Logic and the Foundations of Mathematics, 125.North-Holland Publishing Co., Amsterdam, 1989. xviii+668 pp. ISBN:0-444-87295-7[4] Rogers, Hartley, Jr.
Theory of recursive functions and effec-tive computability.
McGraw-Hill Book Co., New York-Toronto, Ont.-London 1967 xx+482 pp.[5] Smullyan, Raymond M.
Recursion theory for metamathematics.
Oxford Logic Guides, 22. The Clarendon Press, Oxford University Press,New York, 1993. xvi+163 pp. ISBN: 0-19-508232-X[6] Soare, Robert I.
Recursively enumerable sets and degrees. Astudy of computable functions and computably generatedsets. ∼ miller .Miller Recursion Theorem ppendix Electronic on-line version only Theorem 1
There exists an infinite strictly increasing sequence h e n : n < ω i such that for every n W e n = { e m : m > n } and the sequence is not computable. ProofThis is a combination of the proof of Theorem 2 and Theorem 3. Fix B acomputably enumerable but not computable set. For any e let { c < c < c < . . . } = { max( W e,s ) : s ∈ ω } By the s-m-n Theorem find computable q such that for any x and eW q ( e,x ) = { c n : n ∈ B and c n > x } . Take h computable so that W h ( e ) is an infinite set of fixed points for q ( e, · ),i.e., x ∈ W h ( e ) implies W x = W q ( e,x ) . Take e to be a fixed point for h , so W h ( e ) = W e . Then putting { e n : n < ω } = { c n : n ∈ B } is not computable and have the required property.QED Theorem 2
There is a computable strictly increasing sequence e < e < · · · such that ψ e n ( m ) = e n + m +1 for every n, m < ω. ProofAs in the proof of Theorem 2 given any e let { e < e < e < . . . } = { max( W e,s ) : s ∈ ω } ppendix Electronic on-line version only q a computable function such that forevery x , m , and e : ψ q ( e,x ) ( m ) = e n + m +1 where n is minimal so that e n ≥ x } . Let h be a computable function such that for every e , the set W h ( e ) is aninfinite set of fixed points for q ( e, · ), i.e., ψ x = ψ q ( e,x ) for all x ∈ W h ( e ) . Let e be a fixed point for h , so W e = W h ( e ) .Now for any n we have that e n ∈ W e = W h ( e ) and so ψ e n = ψ q ( e,e n ) andhence for every m : ψ e n ( m ) = e n + m +1 ..