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Topic: the continuum is not for babies
Replies: 13   Last Post: Jul 11, 2014 2:27 AM

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 johngabriel2009@gmail.com Posts: 354 Registered: 5/25/14
Re: 1.28 - The myth of the 'real' number line.
Posted: Jul 11, 2014 2:27 AM

On Friday, July 11, 2014 12:35:45 AM UTC+2, dull...@sprynet.com wrote:
> On Thu, 10 Jul 2014 10:19:11 -0700 (PDT), John Gabriel
>
> <johngabriel2009@gmail.com> wrote:
>
>
>

> >On Thursday, July 10, 2014 6:56:19 PM UTC+2, dull...@sprynet.com wrote:
>
> >
>
> >> >> >> No. Saying N=eps-|f(x)-L| is nonsense, because N is not allowed
>
> >> >> >> to depend on x.
>
> >
>
> >> >Nonsense. Allowing N=eps-|f(x)-L| proves that N works for all x>N.
>
> >
>
> >> N is not allowed to depend on x. That's simply how the logic of "Ae EN (Ax ...) works.
>
> >
>
> >There is nothing that says N cannot be stated in terms of x.
>
> >
>
> >The definition is: "for every eps > 0 there exists N such that |f(x) - L| < eps for all x > N".
>
>
>
> _That_ is exactly what says N cannot depend on x. First "there exists
>
> N such that", and then something about x. N is chosen _first_,
> then given N it's suppoed to be true that something happens
> for every x > N.

> Look. Is the following true or false?
> (*) There exists N such that every x is larger than N.

> I hope you agree that (*) is false. No N is larger than
> _every_ x.

Agreed.

> But by your (wrong) logic it's easy to show that
> (*) is true: Just let N = x - 1.

Noooo, nooo, noooo. You are NOT understanding this problem correctly. Setting
N=eps-|f(x)-L| helps to prove the result. x is NOT fixed, therefore, you are not constraining the choice of N. You seem to miss the fine detail here, which is not surprising!

> If N = x - 1 then x > N. True, Does that say
> that there exists N which is larger than every x?
> No. It seys for every x there exists N which
> is larger than x. Very different statements.

> But if N is allowed to depend on x then the
> clear false statment (*) becomes true.
> N is simply not allowed to depend on x.

Yes, it is. There is NOTHING that prevents me from setting N=eps-|f(x)-L|.

> For the same reason as in the definition of limit.