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| Frank Lovelace... |
 Posted: Wed Nov 04, 2009 12:59 pm |
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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? |
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| Eric Schmidt... |
| Posted: Wed Nov 04, 2009 9:15 pm |
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Frank Lovelace wrote:
[quote]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?
[/quote]
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.
--
Eric Schmidt |
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| Herman Rubin... |
| Posted: Thu Nov 05, 2009 3:41 pm |
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In article <vzqIm.5300$Xf2.5283 at (no spam) newsfe12.iad>,
Eric Schmidt <eric41293 at (no spam) comcast.net> wrote:
[quote]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.
[/quote]
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
hrubin at (no spam) stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 |
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| David Libert... |
| Posted: Thu Nov 05, 2009 9:24 pm |
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Frank Lovelace (frank.lovelace at (no spam) gmail.com) writes:
[quote]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?
[/quote]
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.
--
David Libert ah170 at (no spam) FreeNet.Carleton.CA |
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| Eric Schmidt... |
| Posted: Sat Nov 07, 2009 1:21 pm |
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Herman Rubin wrote:
[quote]In article <vzqIm.5300$Xf2.5283 at (no spam) newsfe12.iad>,
Eric Schmidt <eric41293 at (no spam) 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?
[/quote]
Yeah, I couldn't work that out. And apparently it's not possible to do.
Oh well.
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Eric Schmidt |
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| Frank Lovelace... |
| Posted: Sat Nov 07, 2009 2:21 pm |
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On Nov 5, 7:24 pm, ah... at (no spam) FreeNet.Carleton.CA (David Libert) wrote:
[quote]Frank Lovelace (frank.lovel... at (no spam) 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.
--
David Libert ah... at (no spam) FreeNet.Carleton.CA
[/quote]
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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