Re: Cantor / Abbildung N auf R

On Jan 12, 7:00 pm, WM <mueck...@xxxxxxxxxxxxxxxxx> wrote:

Der binäre Baum erlaubt einen ebenso korrekten Schluss wie der
Diagonalbeweis, nur ist das Ergebnis ein anderes. Zwar wird keine
Bijektion zwischen |N und |R erzeugt, aber es wird gezeigt, dass |R
nicht mehr Elemente besitzt, als |N.

Zitat aus Beitrag in NG "Aspects of the infinite",

From: WM <mueck...@xxxxxxxxxxxxxxxxx>
Date: Sun, 13 May 2007 07:38:26 -0700
Local: Sun, May 13 2007 3:38 pm
Subject: Re: Binary Tree

<recht gute Ausführung gesnippt>
...
The union U(T(n)) of all finite trees T(n) with n levels (n in N)
yields an infinite tree with a countably infinite cardinal number of
nodes and a countably infinite cardinal number of finite paths. When
extending to the complete infinite tree T, the set of nodes remains
the same whereas the paths become infinite and their cardinal number
becomes uncountable. So considering the nodes we have T = U(T(n))
whereas considering the paths T =/= U(T(n)).

Kritisch wird nur ob eine unendliche Menge die keine
Bijektion mit IN erlaubt vielleicht doch "mehr" Elemente
besitzt als IN. Sicher möchte eine naive Auffassung es
so haben.

Grüße, kluto
.

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