Re: Cantor Confusion
- From: "Newberry" <newberry@xxxxxxxxxx>
- Date: 22 Dec 2006 12:44:25 -0800
Jesse F. Hughes wrote:
"Newberry" <newberry@xxxxxxxxxx> writes:
Virgil wrote:
In article <1166765631.154421.111470@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
"Newberry" <newberry@xxxxxxxxxx> wrote:
Virgil wrote:
In article <1166762377.524661.268400@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
"Newberry" <newberry@xxxxxxxxxx> wrote:
OK, and why exactly can't we map the edges onto the paths?There aren't enough edges ( or nodes).
Each node can be represented uniquely by a finite string of left/right
branchings which carries you from the root node to the node in question,
with the empty string being the root node itself, and each edge by a
finite non-empty sequence terminating at it terminal node.
It has been shown many times that there are only countably many such
strings.
In the same manner of representation, it is clear that every different
infinite sequence of such left/right branchings represents a different
infinite path in the tree.
It has been shown many times and in many ways that the set of such
strings is not countable, in the sense that there is no way of
surjecting the natural numbers onto that set.
Right. So if he edges can be mapped onto the paths we have a
contradiction.
Exactly.
If we have a contradiction then ZFC is inconsistent.
Yes, if we can map edges onto paths, then we have a contradiction and
if we have a contradiction, ZFC is inconsistent.
But we can't map edges onto paths.
Does each path pass through at least one edge?
Ax(Path(x) -> Ey(Through(x,y)))
--
Jesse F. Hughes
"This Trojan appears to utilize a function of the Windows Media DRM
designed to enable license delivery scenarios as part of a social
engineering attack." -- MS candidly explains the role of DRM licenses
.
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