Re: Cantor Confusion

In article <1172253288.305584.116320@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> mueckenh@xxxxxxxxxxxxxxxxx writes:
On 23 Feb., 14:35, "*** T. Winter" <***.Win...@xxxxxx> wrote:
....
> Every element of P(oo) is a path and as such a subset of U(T(n)), as
> you say.
> U(T(N)) has only finite paths as subsets (as N has only finite initial
> segments --- both U(T(n)) and N have infinitely many such subsets,
> but these subsets are finite).

Eh? Above you state (properly) that p(oo) is a subset of U(T(N)).

I said: "as you say". But how is it possible that a finite tree
contains an infinite path? Every tree T(n) contains only finite paths,
namely such with n nodes. There is no tree containing an infinite
path.

Oh. But the union of all finite trees contains infinite paths. Or are you
now of the opinion that 1/3 is *not* a path in your infinite tree?

Are
you contradicting that now? And if we consider: N subset N, N contains
one infinite initial subset. Or do you mean something different with
U(T(N)) from U(T(n))? If so, what do you mean with it?

I mean that U(T(n)) is the union of all finite trees with n levels and
hence with paths with n nodes. I can't believe that there is an
infinite path in this union.

But when I argued before that 1/3 was not in your tree, you contradicted me.
(I thought so because there was a lack of a proper definition.) But now
you apparently have switched to a completely different view.

> There are countably many unions of finite subsets of U(T(n)).

Prove it. There are countably many *finite* unions of finite subsets,
not countably many arbitrary unions of finite subsets. Consider N.
Each subset of N is a union of finite subsets of N, namely the union
of all singleton sets that contain an element of that set.

> There are countably many unions like U(p(n)).

Prove it.

Every set of finite subsets of N is countable. The set of all finite
subsets of N is countable.

Ignoring infinite subsets.


> There are countably many unions like U(p(n)).

Prove it.

If A is countable, then the set of all finite sequences of elements of
A is countable.

Ignoring infinite sequences.

If you want to prove inconsistency of ZF with the axiom of infinity you
should allow infinite subsets in your proofs. Disallowing them does not
prove inconsistency of ZF with the axiom of infinity, only inconsistency
with the way *you* think things should be. Straight out: denial of the
axiom of infinity.
--
*** t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~***/
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