Let A be a set of real numbers, and define a=sup A. Is there any way to proof the existence of a sequence {a_n} with a_n in A and a_n-->a without using the Axiom of Countable Choice?
Frank Lovelace wrote: > Let A be a set of real numbers, and define a=sup A. Is there any way > to proof the existence of a sequence {a_n} with a_n in A and a_n-->a > without using the Axiom of Countable Choice?
A plan that might work is:
Let B be the set of real numbers that are limits of sequences of points in A. Show that B is closed. B obviously contains A. So B contains the closure of A, including sup A.
In article <vzqIm.5300$Xf2.5...@newsfe12.iad>, Eric Schmidt <eric41...@comcast.net> wrote:
>Frank Lovelace wrote: >> Let A be a set of real numbers, and define a=sup A. Is there any way >> to proof the existence of a sequence {a_n} with a_n in A and a_n-->a >> without using the Axiom of Countable Choice? >A plan that might work is: >Let B be the set of real numbers that are limits of sequences of points >in A. Show that B is closed. B obviously contains A. So B contains the >closure of A, including sup A.
Without using countable choice, how do you show that B is closed? -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hru...@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
Frank Lovelace (frank.lovel...@gmail.com) writes: > Let A be a set of real numbers, and define a=sup A. Is there any way > to proof the existence of a sequence {a_n} with a_n in A and a_n-->a > without using the Axiom of Countable Choice?
Paul Cohen's first ~AC model in _Set Theory and the Continuum Hypothesis_ provides a counterexample to this.
Namely, Cohen's model has T, an infinite subset of P(omega), such that T has no countable infinite subsets in the model.
By considering characteristic functions of subsets of omega, and in turn considering these as binary digits, we can surject P(omega) onto the real interval [0, 1], so let A be the range of this surjection on T.
The methods of analysis of Cohen's model show A has sup 1, and 1 is not a member of A.
So if there were an a_n sequence in A approaching a, it would have to have infinite range.
But A has no infinite countable subsets, by the corresponding property of T.
Herman Rubin wrote: > In article <vzqIm.5300$Xf2.5...@newsfe12.iad>, > Eric Schmidt <eric41...@comcast.net> wrote:
>>Frank Lovelace wrote:
>>>Let A be a set of real numbers, and define a=sup A. Is there any way >>>to proof the existence of a sequence {a_n} with a_n in A and a_n-->a >>>without using the Axiom of Countable Choice?
>>A plan that might work is:
>>Let B be the set of real numbers that are limits of sequences of points >>in A. Show that B is closed. B obviously contains A. So B contains the >>closure of A, including sup A.
> Without using countable choice, how do you show that > B is closed?
Yeah, I couldn't work that out. And apparently it's not possible to do. Oh well.
> Frank Lovelace (frank.lovel...@gmail.com) writes: > > Let A be a set of real numbers, and define a=sup A. Is there any way > > to proof the existence of a sequence {a_n} with a_n in A and a_n-->a > > without using the Axiom of Countable Choice?
> Paul Cohen's first ~AC model in _Set Theory and the Continuum Hypothesis_ > provides a counterexample to this.
> Namely, Cohen's model has T, an infinite subset of P(omega), such that > T has no countable infinite subsets in the model.
> By considering characteristic functions of subsets of omega, and in turn > considering these as binary digits, we can surject P(omega) onto the real > interval [0, 1], so let A be the range of this surjection on T.
> The methods of analysis of Cohen's model show A has sup 1, and 1 > is not a member of A.
> So if there were an a_n sequence in A approaching a, it would have to > have infinite range.
> But A has no infinite countable subsets, by the corresponding property of T.
Thanks for the reference and the sketch of the counterexample, which is clear to the level of detail you have provided; I'll search out the book and learn some model theory.
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