Re: Infinte strings of digits

From: Mike Oliver (mike_lists_at_verizon.net)
Date: 09/19/04


Date: Sun, 19 Sep 2004 16:48:37 -0500

N. Silver wrote:
> Mike Oliver wrote:
>
>
>>But that isn't what you get, from the "intermediate
>>step"! What you get is a bijection between the
>>set of *representatives* for equivalence classes,
>>and R. Not the same thing at all.
>
>
> Since the sequences in each equivalence class
> are a countable set, yes it is.

No it isn't. It does *imply* the existence
of such a bijection, using the Axiom of Choice,
but it doesn't let you *construct* one, in
the sense I suspect Peter was looking for.
In fact no "reasonably definable" such bijection
exists. That's the interesting mathematical
point to be seen here, and it can be quite well-formulated
at the level of formality from Peter's original
post.



Relevant Pages

  • Re: Infinte strings of digits
    ... > Since the sequences in each equivalence class ... > are a countable set, ... In fact no "reasonably definable" such bijection ...
    (sci.logic)
  • Re: Infinte strings of digits
    ... Mike Oliver wrote: ... What you get is a bijection between the ... Since the sequences in each equivalence class ... are a countable set, yes it is. ...
    (sci.logic)
  • Re: Infinte strings of digits
    ... Mike Oliver wrote: ... What you get is a bijection between the ... Since the sequences in each equivalence class ... are a countable set, yes it is. ...
    (sci.math)
  • Re: Cantor Confusion
    ... By definition a Cauchy sequence denotes a real number. ... Why then do you require a definable bijection between ... are a subset of a definable countable set. ... You do not even need a bijection, an injection is ...
    (sci.math)
  • Re: How many bijections are there from A to A?
    ... There is an injection from this into all sequences comprising natural ... there is an injection from permutations over N to R. ... As there is an injection in both directions, it must be a bijection. ...
    (sci.math)