Re: Another set with cardinality |Z|

From: Eray Ozkural exa (erayo_at_bilkent.edu.tr)
Date: 09/29/04 

Date: 28 Sep 2004 20:26:10 -0700

stephen@nomail.com wrote in message news:<cjc2af$1l0o$1@msunews.cl.msu.edu>...
> In comp.theory Eray Ozkural exa <erayo@bilkent.edu.tr> wrote:
> : stephen@nomail.com wrote in message news:<cjahpf$cld$1@msunews.cl.msu.edu>...
>
> : That's very interesting indeed, but a path does not have to be named
> : like that to exist, in my opinion.
>
> In order for a path to exist it must connect to vertices that exist.
> Every vertex has a finite depth.

Existence does not necessarily imply nameability. (According to
standard set theory) Every vertex does indeed has finite depth in my
paragraph below.

> : I can show inductively that I can always find a path of length one
> : more than you can (stating your choice as a base case), e.g. the
> : agreed upon definition of countable infinity, hence there is at least
> : one simple path that has infinite cardinality. You can view the path
> : as a set of pairs (u,i) where u is a vertex and i is its a position in
> : the sequence, starting from 1. All i's are finite, but the cardinality
> : of the set is |N|. Does that make sense?
>
> No. It makes as mush sense as claiming that there is an infinite
> integer because you can always find an integer larger than any
> integer I name. There are no infinite integers.

Excuse me, but that is a broken analogy. In the above, I make it
explicit that the depth of individual vertices in the path sequence
are finite, but the cardinality of the path set {(u,i}} is Aleph-0.

Likewise, for integers, there is no "infinite" integer by definition,
but the set of all integers has cardinality Aleph-0.

If you do not trust proofs by induction, then I cannot show you why my
argument above holds. You might be beating the wrong straw man. The
one you would be interested in is different, see my latest response to
Robert Low. (a bijection that works in finite case, but is supposedly
defunct in infinite case)

Again, I would welcome corrections, e.g. why I cannot apply the
induction above to have a simple path set {(u,i) : u in V, i index of
vertex in path} with cardinality Aleph-0. This would imply that the
level set L has cardinality Aleph-0 as well. Disproof for either would
be acceptable.

Thanks for all your responses, they've been really helpful. I now have
a much better idea of what should go into an essay on constructive
set theory. It seems that such a theory is incommensurable with the
received set theory, but that is no big deal as I see it. (e.g. in the
constructivist set theory, we might not be able to distinguish among
cardinalities exactly in the same way, but there are ways of
distinguishing among kinds of infinities which are just as powerful,
via measures of computational complexity)

Regards, 

--
Eray Ozkural

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