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

Replies: 3   Last Post: Feb 5, 2014 4:36 PM

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Ben Bacarisse

Posts: 1,972
Registered: 7/4/07
Re: § 424 Actual Infinity: We never get it - but
we get it!

Posted: Feb 5, 2014 2:07 PM
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WM <> writes:

> Am Mittwoch, 5. Februar 2014 17:48:42 UTC+1 schrieb Ben Bacarisse:
>> WM <> 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 numbers 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?

> Of course there is a bijection between |N and all finite
> definitions. The set of paths is a subset of the latter.

So another question you won't answer. You are not fooling anyone with
this tactic.

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

> The simplest construction to show that every antidiagonal is in a
> bijection is the set of all rational numbers.

So no, you can't. You just say stuff like this and hope the everyone
else goes "oh, that must be a bijection -- Prof Mueckenheim say so -- no
need for an actual proof".

>> 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!

> Definition: Every path *is* a finite definition.

<flashing lights; sound effect: woop! woop! woop!>

Almost an actual definition. So there is a bijection between N and some
set of paths that meet this rather vague criterion. I'll buy that
(despite the issues raised by William Hughes). Let's call them the

>> >> 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.

> I have defined three sets of paths, namely leading from the root node
> to each node and thereafter being completed by a tail of 000..., or
> 111..., or 010101...

Why are you answering yourself?

> You can add any of your favourite completeions by using any of a
> countable set of finite definitions.

Yes, I know how to make the set of WMpaths.

>> No, I have defined the set of paths
> That has cardinal number 1.

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

> Please write every path in as simple terms as you can. Perhaps that
> will remove your block.

Funny. Do you still claim not to understand the definition of the set
of paths or do you want me to write it out in some other form?


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