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Topic: looking for example of closed set that is *not* complete in a metric space
Replies: 26   Last Post: Feb 3, 2013 11:06 AM

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 Shmuel (Seymour J.) Metz Posts: 3,473 Registered: 12/4/04
Re: looking for example of closed set that is *not* complete in a metric space
Posted: Feb 2, 2013 9:41 PM

In <GbydnfG1Xq3Nu5HMnZ2dnUVZ_jqdnZ2d@giganews.com>, on 02/01/2013
at 02:32 PM, fom <fomJUNK@nyms.net> said:

>So, for example, there are "gaps" in the system of rational
>numbers. One can, assuming completed infinities, define infinite
>sets of rational numbers corresponding to the elements of a
>Cauchy sequence.

What would be the point? You're introducing extra machinery into what
is a very simple construction.

>When the limit of the sequence is, itself, a rational number,
>that infinite set becomes a representation of that rational number
>in the complete space whose "numbers" are equivalence classes of
>Cauchy sequences sharing the same limit.

No; the set of values taken by the sequence is irrelevant. If a Cauchy
sequence in the rationals converges then the *sequence* is a
representative of its limit.

>When the limit of a Cauchy sequence does not exist as a rational
>number, that Cauchy sequence becomes a representative of the
>equivalence class of Cauchy sequences that cannot be
>differentiated from that representative using the order relation
>between the rational numbers of the underlying set.

What are you trying to say? The definition of the equivalence relation
is the same whether the Cauchy sequences converge or not; two
sequences are equivalent if their difference converges to zero. Any
Cauchy sequence is a representative of its equivalence class, by
definition.

>These "numbers" have no corresponding rational number as a limit
>and are, therefore, distinguished as a different logical type in
>the *new*, completed space.

Non sequitor, and false. There is nothing logically special about
equivalence classes of Cauchy sequences that converge.

>To call a subset of a complete space a dense subset is to say that
>such a logical type construction could be made from that subset
>to recover the original space.

Completeness only applies to metric spaces[1], while denseness applies
to subsets of arbitrary topological spaces. You can't in general
reconstruct a topological space from only a dense subset, not even if
the space is compact.

[1] Well, slightly more general.

--
Shmuel (Seymour J.) Metz, SysProg and JOAT <http://patriot.net/~shmuel>

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Date Subject Author
2/1/13 Achimota
2/1/13 Paul
2/1/13 Paul
2/1/13 fom
2/1/13 fom
2/2/13 Shmuel (Seymour J.) Metz
2/3/13 fom
2/3/13 Shmuel (Seymour J.) Metz
2/2/13 Achimota
2/2/13 Butch Malahide
2/2/13 quasi
2/2/13 Butch Malahide
2/2/13 Achimota
2/2/13 quasi
2/3/13 Achimota
2/3/13 Paul
2/3/13 Achimota
2/1/13 Butch Malahide
2/1/13 J. Antonio Perez M.
2/1/13 William Hughes
2/2/13 J. Antonio Perez M.
2/1/13 Butch Malahide
2/1/13 William Elliot
2/2/13 Butch Malahide
2/2/13 William Elliot
2/2/13 Butch Malahide
2/2/13 Shmuel (Seymour J.) Metz