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Re: Matheology ?019
Posted:
Jun 4, 2012 2:37 PM
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On 4 Jun., 19:42, "dilettante" <n...@nonono.no> wrote:
> >The limit was exactly the same if all your intervals started at -1 or
> >+1, respectively. The difference is that a truely connected component
> >has its limit independent of the finite definition of a sequence: If
> >there are intervals like [0, 1-1/n], then, independent of any
> >definition, the first possibility for an uncovered point would be the
> >0.
>
> Of course the limit is the same. There is no dispute about that. But the
> connected components, your "clusters" are not the same. And there are whole
> intervals uncovered arbitrarily close to 0, as well as whole intervals
> covered.
>
Your example differs somewhat from mine. In addition 0 is disconnected
from all these intervals by extended intervals. But certainly there is
no point connected to 0.
And finally another idea: If the ends of my intervals are multiples of
sqrt(2) and if I dont use overlapping intervals, but am satisfied if
every rational is covered once only. Then there cannot exist any
uncovered irrational that is incommensurable with sqrt(2). How about
that?
> >> Thus it makes no sense to say that 0 is the endpoint of
> >> a cluster, but not of any particular interval.
> >For the proof it does not make a difference whether the intervals are
> >connected or could be connected, as long as the limit is defined.
> >Important for my proof is only that every neighbourhood of the limit
> >contains points of intervals.
>
> Okay. In my example also, every neighborhood of 0 contains removed
> intervals, as well as intervals in C. My question is, why do you see a
> contradiction in your example, but none here?
The contradiction would occur if 0 was not the one and only limit
point. Every neighborhood of every of my cluster endpoints can contain
further cluster endpoints. But there must not be uncovered irrationals
without a cluster, and be it as small as you like, between them.
Regards, WM
>
>
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