Re: Question

In article <e2rqgp$1uf6$1@xxxxxxxxxxxxxxxxxx>,
magidin@xxxxxxxxxxxxxxxxx (Arturo Magidin) wrote:

In article <vmhjr2-F2A105.16310927042006@xxxxxxxxxxxxxxxxxxxxxx>,
Virgil <vmhjr2@xxxxxxxxxxx> wrote:
In article <1146165177.094108.274750@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
"zuhair" <zaljohar@xxxxxxxxx> wrote:

Arturo Magidin wrote:

The definition states that if the difference converges into zero then
the two Cauchy sequences are equivalent, I disagree with that.

One cannot "disagree" with a definition in mathematics. One either
accepts it or rejects it.

They should be equivalent only when the difference REACHES zero.

Using that as your equivalence relation, what you would get is only
another copy of the rational numbers, not the usual real numbers at all.

I don't think you get "the rational numbers". Cauchy sequences that
converged to a rational number but were not eventually constant would
not be equivalent to the constant sequence corresponding to that
rational; and any such sequence would be non-equivalent to any shift
on that sequence. So the sequences

(1, 1, 1, ...)
(1/2, 3/4, 7/8, ..., (2^n-1)/2^n, ...)
(0, 1/2, 3/4, 7/7, ..., 2^{n-1}-1/2^{n-1}, ... )

would lie in distinct equivalence classes. Your equivalence classes
are just the classes of all sequence that are eventually equal, so I
think you would get essentially just the set of all Cauchy sequences
back (essentially, mind you).

I spoke too quickly. The zero equivalence class would be the set of
sequences with finite carrier (zero for all but finitely many terms).

An equivalence class in general would be the set of all sequences which
differed from a given sequence at only finitely many terms.

The result would not even be a field, since there are obvious zero
divisors. e.g., the product of a sequence zero for all odd terms times a
sequence zero at all odd positions would always be the zero sequence.
.

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