Is it possible in ZF? - sci.math

Message from discussion Is it possible in ZF?

David Libert

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More options Nov 6 2009, 2:24 am

Newsgroups: sci.math

From: ah...@FreeNet.Carleton.CA (David Libert)

Date: 6 Nov 2009 02:24:02 GMT

Local: Fri, Nov 6 2009 2:24 am

Subject: Re: Is it possible in ZF?

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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.

--
David Libert ah...@FreeNet.Carleton.CA

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