Re: Cantor Confusion

Newberry <newberry@xxxxxxxxxx> wrote:

stephen@xxxxxxxxxx wrote:
Newberry <newberry@xxxxxxxxxx> wrote:

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)))

Every path passes through an infinite number of edges.

So the answer is yes. If every path passes through at least one edge
does it follow that the number of paths is less or equal than the
number of edges?

P = {x|Path(x)]; E = {y|Edge(y)}
|P| <= |E| ?

No. As I said, every path passes through one of the two
edges incident to the root. There are only two edges incident
to the root, but there are an infinite number of paths.
So it does not follow that if every path passes through at
least one edge in some set of edges, that the set of
edges is not larger than the set of paths.

Stephen

.

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