Cantor Confusion
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From: Virgil on 23 Dec 2006 00:35 In article <1166845904.426550.122020(a)48g2000cwx.googlegroups.com>, "Newberry" <newberry(a)ureach.com> wrote: > stephen(a)nomail.com wrote: > > Newberry <newberry(a)ureach.com> wrote: > > > > So which edge is mapped to the path that always goes left? > > Remember, you need to uniquely map each path to an edge. > > No, I don't. What I am showing here is that each path will accumulate > the weight of two edges as it approaches infinity. Then what fractional of the first left edge is assigned to each of the infinitely many paths passing through it? By my count, the pieces of any one edge have to add up to more than one edge in order to assign a piece to each path through it.
From: Virgil on 23 Dec 2006 00:39 In article <1166849530.679744.243900(a)a3g2000cwd.googlegroups.com>, "Newberry" <newberry(a)ureach.com> wrote: > stephen(a)nomail.com wrote: > > Newberry <newberry(a)ureach.com> wrote: > > > > > No, I don't. What I am showing here is that each path will accumulate > > > the weight of two edges as it approaches infinity. > > > > That is not a mapping. > Correct. > You need to construct a surjection. > No, I don't. You do if you want to claim that there are "as many" edges as paths, as that is what such a claim says. > > Showing that each path will accumalate the weight of two edges > > has nothing to do with constructing a mapping. > But it shows that in an infinite tree there are two edges per path. And the same two edges for infinitely many other paths as well, so your paths are being used more than once.
From: stephen on 23 Dec 2006 00:35 Newberry <newberry(a)ureach.com> wrote: > stephen(a)nomail.com wrote: >> Newberry <newberry(a)ureach.com> wrote: >> >> > stephen(a)nomail.com wrote: >> >> Newberry <newberry(a)ureach.com> wrote: >> >> >> >> >> >> > What is wrong with WM's mapping? You divide each edge into two halves >> >> > and pass one half to each branch. Then you divide the passed half again >> >> > and again. >> >> >> >> So which edge is mapped to the path that always goes left? >> >> Remember, you need to uniquely map each path to an edge. >> >> > No, I don't. What I am showing here is that each path will accumulate >> > the weight of two edges as it approaches infinity. >> >> That is not a mapping. > Correct. So that answers your question "what is wrong with WM's mapping?" It is not a mapping. > You need to construct a surjection. > No, I don't. Yes you do, if you want to show that there exists a surjection from edges to paths. >> Showing that each path will accumalate the weight of two edges >> has nothing to do with constructing a mapping. > But it shows that in an infinite tree there are two edges per path. No. It shows that in large finite trees there are approximately two edges per path. It says nothing about infinite trees. Stephen
From: Newberry on 23 Dec 2006 01:08 stephen(a)nomail.com wrote: > Newberry <newberry(a)ureach.com> wrote: > > > stephen(a)nomail.com wrote: > >> Newberry <newberry(a)ureach.com> wrote: > >> > >> > stephen(a)nomail.com wrote: > >> >> Newberry <newberry(a)ureach.com> wrote: > >> >> > >> >> > >> >> > What is wrong with WM's mapping? You divide each edge into two halves > >> >> > and pass one half to each branch. Then you divide the passed half again > >> >> > and again. > >> >> > >> >> So which edge is mapped to the path that always goes left? > >> >> Remember, you need to uniquely map each path to an edge. > >> > >> > No, I don't. What I am showing here is that each path will accumulate > >> > the weight of two edges as it approaches infinity. > >> > >> That is not a mapping. > > Correct. > > So that answers your question "what is wrong with WM's mapping?" > It is not a mapping. > > > You need to construct a surjection. > > No, I don't. > > Yes you do, if you want to show that there exists a surjection > from edges to paths. > > >> Showing that each path will accumalate the weight of two edges > >> has nothing to do with constructing a mapping. > > But it shows that in an infinite tree there are two edges per path. > > No. It shows that in large finite trees there are approximately > two edges per path. It says nothing about infinite trees. Does it converge to two edges per path as we approach infinity? > > Stephen
From: Newberry on 23 Dec 2006 01:11
Virgil wrote: > In article <1166849530.679744.243900(a)a3g2000cwd.googlegroups.com>, > "Newberry" <newberry(a)ureach.com> wrote: > > > stephen(a)nomail.com wrote: > > > Newberry <newberry(a)ureach.com> wrote: > > > > > > > > No, I don't. What I am showing here is that each path will accumulate > > > > the weight of two edges as it approaches infinity. > > > > > > That is not a mapping. > > Correct. > > You need to construct a surjection. > > > No, I don't. > > You do if you want to claim that there are "as many" edges as paths, as > that is what such a claim says. > > > > Showing that each path will accumalate the weight of two edges > > > has nothing to do with constructing a mapping. > > > But it shows that in an infinite tree there are two edges per path. > > And the same two edges for infinitely many other paths as well, so your > paths are being used more than once. Is it true that the ratio of edges over paths converges to two as we approach infinity? lim{n-->oo} (2*2^n - 2)/2^n = 2 |