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On two types of infinite sets of infinite sequences.

by the Tuesday Afternoon Club.

In the following x will stand for the infinite sequence (x(0), x(1), x(2), ...........) where the x(i) are taken from a finite alphabet of at least two characters, say {0, 1}

Let us consider the sets of sequences S1 and S2 , defined as the solution sets of the equations P1(x) and P2(x), respectively, with

 Pl(x): (E i: i ≥ 0: (A j: 0 ≤ j < i: x(j) = 0) and (A j: i ≤ j: x(j) = 1)) P2(x): (A i: i ≥ 0: x(i) = 0 or (A j: j ≥ i: x(j) = 1)) .

Because P1(x) ⇒ P2(x) , S1 is a subset of S2 , even a proper subset: there exists one solution of P2(x) and non P1(x) , viz. the sequence with x(1) = 0 for all i ≥ 0.

The set S2 has the property that for any non-member of S2 —i.e. any solution of non P2(x)— non-membership can be established on account of an initial segment of it. As a matter of fact, in this particular case only two elements suffice for this evidence, as follows from non P2(x):

 (E i: i > 0: x(i) ≠ 0 and (E j: j > i: x(j) ≠ 1)) .

The set S1 doesn’t have this property because there exists a non-member —viz. x(i) = 0 for all i ≥ 0— with the property that any initial segment of it is also the initial segment of a member of S1 . We call S2 “closed” and S1 “non-closed”, that is:

 “The set S is closed” means “any non-member of S has an initial segment that is not the initial segment of any member of S”. “The set S is non-closed” means “there exists a non-member of S such that any initial segment of it is also the initial segment of some member of S ”.
From the above it follows
1)     that a set of sequences is either closed or non-closed,
2)     that any finite set —in particular the empty set— is closed,
3)     that the (uncountably infinite) universe of all possible sequences is closed.

*              *
*

The notion of a closed set is of significance in connection with non-deterministic infinite computations. Consider the set of possible output sequences y corresponding to

 “initialize; i:= 0; (1) do true → compute; print(y(i)); i:= i + 1 od”
Here “initialize” and “compute” stand for terminating computations not affecting the value of i . If “initialize” and “compute” stand for deterministic computations, the set of possible output sequences consists of a single element; if “initialize” and (in particular) “compute” are non-deterministic, the corresponding set of possible output sequences can be infinite.
 “Program (1) is a continuous machine” means “in program (1) ”initialize“ and ”compute“ are both of bounded non-determinacy”.

We can now prove the following

Theorem 1. The possible output sequences of program (1) form a closed set or program (1) is not a continuous machine.

Proof. Let S be the set of possible output sequences of program (1). Either S is closed —in which case Theorem 1 holds— or S is non-closed. In the latter case, let x be a non-member of S such that each initial segment of x is also an initial segment of some member of S . Consider now the following program:

 “initialize; i=: 0; (2) do ”y(0),...,y(i-1) is an initial segment of x “ → compute; print(y(i)); i:= i + 1 od; print(i)”

Non-termination of (2) is excluded because x is not a member of S Because each initial segment of x is also the initial segment of some member of S , the final value of i is unbounded. Hence, program (2) is a (weakly) terminating program of unbounded non-determinacy; hence “initialize” or “compute” is of unbounded non-determinacy, i.e. program (1) is not a continuous machine. (End of Proof.)

Theorem 2. A set of sequences is non-closed or is the set of possible output sequences of some continuous machine.

Proof. Let S be the set of sequences. Either set S is non-closed —in which case Theorem 2 holds— or it is closed. In the latter case consider program (1) with “initialize” deterministic and “compute” only constrained by the requirement that the next y(i) leads to an initial segment of some element of S . Because the alphabet is finite, “compute” need not be of unbounded non-determinacy. By virtue of its construction program (1) may then generate any member of S , and, S being closed, it cannot generate any non-member of S . (End of Proof.)

Having identified closed sets of possible output sequences with continuous machines, we propose to ignore for the time being non-deterministic infinite computations to which non-closed sets correspond.

 Plataanstraat 5 19th February 1980 5671 AL NUENEN prof.dr.Edsger W.Dijkstra The Netherlands Burroughs Research Fellow

Last revision 2015-02-28 .