Re: Cantor's wrong! (doubt it)

• Previous message: Paul Chapman: "Re: Cantor's wrong! (doubt it)"
• In reply to: remlaps_at_despammed.com: "Cantor's wrong! (doubt it)"
• Next in thread: Will Twentyman: "Re: Cantor's wrong! (doubt it)"
• Messages sorted by: [ date ] [ thread ] [ subject ] [ author ]

Date: Wed, 16 Jun 2004 00:26:14 +0100

Hi Steve,

<remlaps@despammed.com> wrote in message
news:2b2845e7.0406151239.6726bfbe@posting.google.com...
> Has this web page been accepted or discredited?
>
>
http://www.andyland.com/education/rensselaer/01999spring/paradox/cantorswron
g.html
>
> I came across it several weeks ago and did many web searches trying to
find
> someone who had discredited it, but could not. Could it be correct?
> I don't think so, but I don't trust me. I would like to see someone who
> knows a little about set theory explain why it's wrong (or right?).

A general comment - there are many people out on the web who think they've
disproved some well known mathematical result, and generally they simply do
not understand the proof they are disproving. There's no reason to be
worried that you can't find anyone who's contradicted what they say, as
mathematicians have better things to do than scour the net looking for false
arguments to debunk! You should have more faith in your own judgement, as
you are (more or less) correct in what you say below... :-)

>
> I'll take a stab, but can some one do a better job?
>
> 1.) The proposed algorithm (in section 3) will enumerate every real of
> finite length and exactly 1 real of infinite length, not all reals.
> The reason the rationals (which can be also be expressed as infinite
> length) can be enumerated is that they can also be expressed as
> fractions, eliminating the need to list an infinite number of decimal
> places. Contrarily, transcendantals cannot be similarly expressed.

The algorithm constructs every decimal of finite length. (It doesn't seem
to construct even one of infinite length, so I'm not sure which decimal of
infinite length you thought it would construct?)

So for example, the algorithm does not construct 1/3 or Pi. Sure, it
constructs numbers arbitrarily close to these numbers, but it never
constructs these numbers themselves - if it did, then it would construct
them *on some particular (finite) step*, but this never happens as 1/3 and
Pi do not have finite decimal expansions.

Your remark about the rationals being countable is correct but not really
relevent to the argument on the web page...

>
> 2.) Cantor's diagonalization is in fact right, not wrong as claimed (in
> section 4), because it shows not that the constructed number HAS not
> been included, but rather (and more strongly) that it CAN not be
included.
> The argument presented to the contrary is fallacious. The argument
> presented in contradiction to cantor is invalid when it compares an
> enumeration of integers because it only shows that an integer HAS not
> been included, not that it CAN not be included.

not sure what you're getting at with HAS/CAN etc.

Cantor's argument shows that *any* attempt you make to enumerate the real
numbers will fail, because there will be (at least) one number missing from
your enumeration.

Obviously, if you produce enumeration attempt #1, and then Cantor produces
number z_1 that is missing from #1, you could try and do better and come up
with another attempt #2 which includes z_1 and all the numbers that were in
#1. But if someone tries to go down this route that just shows they don't
understand the structure of Cantor's proof because the new list will have a
new missing number z_2 and so on.

You can think of the structure of Cantor's proof as:
a) you claim to have enumerated the set? give me the enumeration then!
b) aha, look you were wrong - you missed out number XXX.
c) (and since (a)+(b) applies for any enumeration you might come up with,
     there aren't any such enumerations...)

Sometimes people claim they have produced an enumeration which contains
numbers that are "infinitely far down the list", but this just shows they
don't understand what enumerable means - every number that appears in the
enumeration appears at some finite position in the list! (There are no
numbers in the list that are "infinitely far down the list", although of
course there is no finite upper bound to how far down the list numbers
appear).

Someone was posting a similar argument here a few months ago, claiming that
the power set of the natural numbers was countable, but they were making
essentially the same mistake as on this web site - they were constructing an
enumeration which included every single subset of N of finite size, and
concluding that "in the limit" this would include every subset of N.
(Whatever "in the limit" means... it's probably trying to suggest that there
are subsets appearing infinitely far down the list, which is wooly
thinking.)

>
> 3.) Where in does this come from (the bottom of section 3)?
> "For all computational procedures P, and for all computational
procedures
> C, if P is an infinite loop whose contents are exactly C, and C is
> computable in finite time, then P is computable in infinite time."
>
> Am I supposed to just take the author's word for it?

Better ask the author exactly what he means here.

>
> That's the best I can put into words. This is all formatted real pretty
> before the web form has it's way with it. Let's see how it looks when
> google gets through with it :-).
>
> At least now if someone searches groups.google.com for andyland and
cantor,
> they'll find one person who is skeptical of the validity of this argument.

phew! <grins>

>
> Any one else want to support it or dispute it? for the archive?
>
> -Steve-
>
> Steve Palmer

Regards,
Mike.

• Previous message: Paul Chapman: "Re: Cantor's wrong! (doubt it)"
• In reply to: remlaps_at_despammed.com: "Cantor's wrong! (doubt it)"
• Next in thread: Will Twentyman: "Re: Cantor's wrong! (doubt it)"
• Messages sorted by: [ date ] [ thread ] [ subject ] [ author ]