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: Matheology S 162
Replies: 7   Last Post: Nov 30, 2012 5:00 PM

Advanced Search

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

Posts: 9,012
Registered: 1/6/11
Re: Matheology S 162
Posted: Nov 30, 2012 12:43 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

In article <87pq2vbx8v.fsf@phiwumbda.org>,
"Jesse F. Hughes" <jesse@phiwumbda.org> wrote:

> WM <mueckenh@rz.fh-augsburg.de> writes:
>

> > On 28 Nov., 19:46, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> >

> >> So, your conclusion is that, for every n, the set {n,n+1} is finite?
> >
> > My conclusion is that for every set {1, ..., n} also the set {1, ...,
> > n, n+1} is finite!

>
> Boy, are you incoherent. Let's refresh your memory, since you
> accidentally snipped the context.
>
> ,----
> | >> >and the set containing
> | >> > both, n and n + 1 ist finite too.
> | >>
> | >> There is of course no such thing as
> | >>  "the set containing both n and n+1".
> | >
> | > Here it is: {n, n+1}
> `----
>
> As everyone can see, what you seem to have prove is that the pair
> {n,n+1} is finite.
>
> Of course, it's also true that every proper initial segment of N is
> finite, but if you'll simply read the above, that wasn't what you
> argued.
>

> >
> >> If so, surely we agree.  And from this, we infer that every set of
> >> natural numbers is finite, er, how?

> >
> > Every set, that is formed by induction beginning with {1}, is finite.
> > For every set of natural numbers we can prove that all numbers are
> > finite, hence the set is finite (for completed infinity an infinite
> > number would be required), and, moreover we can prove that there are
> > (potentially) infinitely many numbers not in that set.

>
> Yes, yes, same ol' silly disregard for what the principle of induction
> actually says. I'm not interested in covering this well-trod ground.
>

> > But a real crackpot stamping with feet and shouting "there is the set
> > containing all naturals" will impress some other crackpots. No one
> > else.

>
> Sure. Aside from the fact, you know, that ZFC proves there is a set of
> natural numbers.


And note that neither WM nor anyone else has yet been able to show that
ZFC contains any self-contradictions or internal inconsistencies.
--





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.