Drexel dragonThe Math ForumDonate to the Math Forum



Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.


Math Forum » Discussions » sci.math.* » sci.math.independent

Topic: § 424 Actual Infinity: We never get it - but
we get it!

Replies: 3   Last Post: Feb 5, 2014 4:27 PM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
Ben Bacarisse

Posts: 1,281
Registered: 7/4/07
Re: § 424 Actual Infinity: We never get it - but
we get it!

Posted: Feb 5, 2014 11:48 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

WM <wolfgang.mueckenheim@hs-augsburg.de> writes:

> Am Mittwoch, 5. Februar 2014 15:49:43 UTC+1 schrieb Ben Bacarisse:
>

>> There either is or there is not a bijection between N and the
>> set of paths.

>
> If we find that an uncountable set like all real nunbers is a subset
> of a countable set like all definitions of numbers, then we have a
> contradiction.


Is or there is not a bijection between N and the set of paths?

>> You said you prove to your students that the set of paths is
>> countable.

>
> Yes. The set of all finite definitions is countable and I have proved
> that there is no infinite definition.


But not that there is a bijection between N and the set of paths.

> The latter is that part of my
> proof that you erroneously think would construct a bijection of all
> paths and |N.


So there is not a bijection between N and the set of paths? Which is
it? Can you construct one or not?

>> You've agreed that that means nothing more than that there is a
>> bijection between N and this set. You have not show this, because, of
>> course, there is no such bijection. In fact, elsewhere, you have agreed
>> that this set is uncountable

>
> and is a subset of a countable set. Why do you always tend to "forget"
> that decisive addition?


I never forget it. You have no proof that the set of paths is a subset
of anything since you have not yet even defined it! Despite being asked
repeatedly.

>> Here is the definition of the path set to which you have agreed:
>>
>> || [...] The infinite binary tree, B = (N, T) is a graph
>> || whose noes are the natural numbers (1, 2, 3...) and whose edges are the
>> || pairs T = { (i, j) | j = 2i or j = 2i+1 }. The set if infinite rooted
>> || paths in B are the set of sequences p(n) with p(1) = 1 and p(n+1) =
>> || 2p(n) or p(n+1) 2p(n)+1. Is this set of paths countable? No.

>
> You have not defined any path.


No, I have defined the set of paths that, until you supply your own, is
the set that you are claiming is countable. It's properties can be
determined by finite arguments based on the set's finite definition.

> A path contains no "or". The set of all
> paths that you could define is a subset of the countable set of all
> definitions.


Don't play silly games. You know what that "or" means. Do you want to
be write it out in simpler terms for you?

--
Ben.



Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© Drexel University 1994-2014. All Rights Reserved.
The Math Forum is a research and educational enterprise of the Drexel University School of Education.