Let be a function that takes as input a set of nonnegative integers A and return

Question

Let be a function that takes as input a set of nonnegative integers A and returns the set of indices of all recursively enumerable subsets of A. i.e.

(A) = {z | Wz A}.

Let f be a one-to-one total recursive function. The center Cf of f is the smallest set of nonnegative integers such that

1. Every index of the empty set is in Cf and

2. If A Cf , then f((A)) Cf . (f((A)) is the f-image of (A).)

We will partition Cf into levels: The lowest level is the set of indices of the empty set; i.e. z is in the lowest level of Cf if, and only if, Wz = . Given a level L of Cf , there is a uniquely determined next level defined to be f((L)). Given any set of levels of Cf there is a uniquely determined limit level of this collection defined to be the union of the levels in the collection.

Suppose we give names to the levels such that no two levels have the same name. What the names specifically are doesn’t matter. Let be the collection of names. Given a name , let L be the level with name . Let V . We then get a set of levels {Lv | v V }. Denote the limit level {Lv | v V } by lim vV Lv.

Theorem: (DO NOT PROVE THIS THEOREM: take it as given.)

1. Every element of Cf is in some level.

2. In any nonempty set of levels of Cf there is a least level with respect to set-inclusion.

3. Let z Cf . Every element n of Wz is a member of a level lower than the least level in which z is a member.

We can play the chicken game on Cf . Play alternates by starting with an arbitrary element of Cf and successively choosing elements in lower and lower levels: if z is chosen then the next player must choose an element of Wz. If there is nothing in Wz to choose, the last player must have chosen an index of the empty set and wins.

Prove that (1) play must terminate. (2) The number of dollars won has no upper bound. (3) It is undecidable whether play has terminated. HINT: If play does not always terminate, then there must be a starting integer from which it is possible to make an infinite sequence of plays. Consider the least level in which there is such a starting integer.

Explanation / Answer

supSsupS is shorthand for the least upper bound of SS in some set TSTS with respect to an order on TT. However, we often omit TT and when they are obvious from the context.

Consider the case when T=T=Q and is the usual order. Since sup{x:x2<2}sup{xQ:x2<2} has an upper bound but no least upper bound (in T=T=Q), this supremum does not exist. Hence, Q does not satisfy the least upper bound property (a.k.a. Dedekind completeness).

However, if instead we had T=T=R, the above supremum is 22.

Another example is given by @Hagen von Eitzen, where we look for supsup. Regardless of whether T=T=Q or T=T=R, in both of these spaces, 00 is an upper bound of . Furthermore, if xx is an upper bound of the empty set, so too is x1x1. Therefore, there exists no least upper bound. We can write this as sup=sup= (in fact, if T=T=R¯, the extended reals with the obvious order, this is a precise statement).

Note that this does not contradict Dedekind completeness of the reals because Dedekind completeness only requires bounded nonempty sets to have supremums.

Related Questions

Navigate

• Browse (All)
• Browse L
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.