Locally Constant Constructive Functions and Connectedness of Intervals
aa r X i v : . [ m a t h . L O ] J un LOCALLY CONSTANT CONSTRUCTIVE FUNCTIONSAND CONNECTEDNESS OF INTERVALS
VIKTOR CHERNOV
Abstract.
We prove that every locally constant constructive func-tion on an interval is in fact a constant function. This answers aquestion formulated by Andrej Bauer. As a related result we showthat an interval consisting of constructive real numbers is in factconnected, but can be decomposed into the disjoint union of twosequentially closed nonempy sets. Introduction
Constructive Topology and Constructive Analysis deal with the studyof objects that can be computed by some algorithm, for example by aTuring machine.A constructive real number is a Cauchy sequence of rational numbers { r n } ∞ n =1 equipped with an algorithm that describes the convergence, i.e.given ε > M ∈ N such that for all m, n > M we have | r n − r m | < ε. A constructive function is an algorithm that transformsconstructive numbers to constructive numbers. All the functions andnumbers in this paper are assumed to be constructive.A complete separable constructive metric space can be given by spec-ifying an algorithmically enumerable set P and a constructive met-ric function on this set. Points of the space are algorithmically givenCauchy sequences, whose members are elements of P . The metric isnaturally extended to the points of this space.The subject of constructive mathematics was developed by Markov [6,7] and Shanin [9] and their mathematical school, see Kushner [5] fora nice exposition. A different but to some extend similar approachto constructive mathematics was developed by E. Bishop [2] and hisfollowers.In our proofs we use the so called Markov’s principle saying that: ifthe assumption that a decidable subset of the set of natural numbers isempty yields a contradiction, then one can produce an element of thisset. This assumption is broader than the constructivism assumptionsof Bishop. Mathematics Subject Classification.
Primary 03D78; Secondary 03F60.
Key words and phrases. computable function, constructive point-set topology.
The constructive counterparts of many classical results fail. For ex-ample the constructive versions of the Intermediate value theorem [4]and the Brower fixed point theorem [8] are false.On the other hand many surprising facts that are clearly false in thetraditional versions of the subjects are true in the constructive world.For example, every constructive function defined on real numbers iscontinuous [3]. 2.
Main Results
Theorem 1.
Let f be a constructive locally constant real-valued func-tion on an interval [ a, b ] whose points are constructive real numbers,then f is a constant function. The proof requires the following Lemma.
Lemma 1.
Let f be a constructive fucntion on a complete separablemetric space X which is not a constant function, then you can algo-rithmically find two points p, q with f ( p ) = f ( q ) . Proof.
We will generate points of an enumerable everywhere dense setand compute the values of f at them with better and better precisionuntil we find two points p, q with f ( p ) = f ( q ) . If we do not succeedfinding two such points, then f is a constant function by the Ceitin [3]continuity theorem. (cid:3) Now we use the Lemma to prove Theorem 1.
Proof.
We argue by contradiction and assume that f is not a constantfunction. Then by the Lemma 1 we find two points p, q with f ( p ) = f ( q ) . Without the loss of generality we assume that p < q.
Take r = p + q and compute f ( r ) with a precision that guarantees that one ofthe facts f ( r ) = f ( p ) and f ( r ) = f ( q ) is true. Take one of the twohalves [ p, r ] and [ r, q ] of the intveral for which the values at its ends aredifferent and continue the constuction in a similar fashion. We will geta decreasing sequence of nested intervals [ p n , q n ] of length ( q − p ) / n such that f ( p n ) = f ( q n ) . The sequences { p n } ∞ n =1 and { q n } ∞ n =1 define the same computablenumber d. Since f was locally constant, there is an open neighbor-hood of d on which f is a constant function and this neighborhood hasto contain some pair p k , q k . Thus f ( p k ) = f ( q k ) for some k and we gotthe contradiction. (cid:3) Remark 1.
The statement of the Theorem 1 and its proof hold forcomputable maps of complete separable path connected constructivemetic spaces.
Definition 1.
A subset of a constructive separable metric space is open if it can be realized by an enumerable set of open balls of rational radii
OCALLY CONSTANT CONSTRUCTIVE FUNCTIONS 3 with centers in the points of an enumerable everywhere dense set. Asusbet is closed if it is a complement of an open subset.We say that a susbet S of a constructive separable metric space X is connected if it is impossible to represent S as a disjoint union of twononempty open sets. Theorem 2.
An interval I = [ a, b ] consisting of computable real pointsis connected.Proof. We argue by contradiction and assume that
A, B are disjointnonempty open sets with A ∪ B = I. So every point of I is either in A or in B. We will generate the set of all rational numbers and for eachof the numbers we will decide if it is in A or in B. We get two sets ofnumbers A and B . If either of A = ∅ or B = ∅ then, since the sets A and B are open, we will either get that A = ∅ or that B = ∅ . Thiscontradicts our assumptions.So both sets A and B are nonempty. Take an interval with oneend point in A and the other end point in B . Separate the intervalinto two subintervals of equal length and choose the subinterval forwhich the two ends belong to the two different sets A , B . Iteratingthis construction we get a sequence of nested intervals [ p n , q n ] ∞ n =1 suchthat q n − p n = q − p n − and such that for each n we have p n ∈ A and q n ∈ B . The limit point s defined by these two sequences { p n } ∞ n =1 and { q n } ∞ n =1 has to be either in A or in B. However since both A and B areopen sets, the whole tail of both sequences belongs to that open set.So we have a contradiciton. (cid:3) Definition 2.
A set S is sequentially closed if given a converging se-quence { s n } ∞ n =1 of points in S the limit point also is in S. Theorem 3.
An interval I = [0 , can be subdivided into the union oftwo nonempty disjoint sequentially closed subsets.Proof. Consider a Specker sequence { s n } ∞ n =1 i.e. a strictly increasingsequence of rational numbers in the interval [0 , . This sequence doesnot have a constructive limit [10].Consider two sequences of sets { A n } ∞ n =1 , { B n } ∞ n =1 where A n = [0 , s n )and B n = [ s n , A = ∪ n A n and B = ∩ n B n . The set B is closedand hence sequentially closed.The set A is open but still is sequentially closed. Indeed considerany seqence of points { a n } ∞ n =1 , a n ∈ A and let a = lim n →∞ a n . We note that for every n there is m such that s m > a n . Since the sequence s n is strictly increasing for every n we have that s n < a or a < s n +1 . If s n < a for all n , then we would have that thesequence s n converges to a . This contradcits the fact that the Speckersequence does not have a constructive limit.Thus there is m such that a < s m and then we have a ∈ A m ⊂ A. So the set A is sequentially closed. (cid:3) V. CHERNOV
Acknowledgments.
The author is thankful to Andrej Bauer forposing the question answered in this paper. He is also grateful toVladik Kreinovich and Yuri Matiyasevich for communicating this ques-tion and to the anonymous referee for the suggested improvements andcorrections.
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