Re: Another set with cardinality |Z|

From: Eray Ozkural exa (erayo_at_bilkent.edu.tr)
Date: 09/28/04 

Date: 28 Sep 2004 08:52:49 -0700

stephen@nomail.com wrote in message news:<cjahpf$cld$1@msunews.cl.msu.edu>...
> In comp.theory Eray Ozkural exa <erayo@bilkent.edu.tr> wrote:
> : "Tim Peters" <tim.one@comcast.net> wrote in message news:<R5adnTP0EPYGCMXcRVn-uw@comcast.com>...
> :> It's unclear what you mean by "infinite path" (I realize you don't think
> :> it's unclear <wink>). Perhaps part of what you intend is that by "path" you
> :> mean "simple path" (a path that never visits a vertex twice). It's also
> :> unclear what you mean by "pick".
>
> : I indeed meant simple path. It's been a while since I took the graph
> : theory course, so some imprecision may have found its way into my
> : terms.
>
> : By "pick", I mean the ordinary graph theoretic usage of "choose", such
> : as in "choose any vertex u of G", or "choose a minimal A-B path".
>
> If you pick any two vertices u and v, then the depth of u, d(u), will
> be finite and the depth of v, d(v), will be finite. The length
> of the path between u and v will be either d(u)+d(v) or |d(u)-d(v)|,
> both of which are finite. It seems pretty intuitive to me that
> there are no infinite paths in your tree.

That's very interesting indeed, but a path does not have to be named
like that to exist, in my opinion.

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?

Regards, 

--
Eray Ozkural
PS: I see you don't trust your intuition, but to say that there are no
infinite paths in the limiting tree would be a stretch of imagination
as I see it.
PS2: I am not suggesting that the discussion goes anywhere. Please see
this as merely an exercise.
PS3: I am referring to the second simpler limiting tree with vertex
labels chosen from {0,1}, and not the first one which was labelled
with (rational) binary expansions ...