Search All of the Math Forum:

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

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: Cantor's first proof
Replies: 31   Last Post: Nov 9, 2012 1:04 AM

 Search Thread: Advanced Search

 Messages: [ Previous | Next ]
 rt servo Posts: 19 Registered: 2/1/05
Re: Cantor's first proof
Posted: Nov 8, 2012 1:33 PM
 Plain Text Reply

On 08/11/2012 2:12 PM, Zuhair wrote:
> On Nov 8, 8:15 pm, gus gassmann <g...@nospam.com> wrote:
>> On 08/11/2012 12:00 PM, Zuhair wrote:
>>
>>
>>
>>
>>
>>
>>
>>
>>

>>> On Nov 8, 2:10 pm, "LudovicoVan" <ju...@diegidio.name> wrote:
>>>> "Zuhair" <zaljo...@gmail.com> wrote in message
>>
>>>> news:dac67778-bb43-4846-996b-0cd984b03922@s12g2000vbw.googlegroups.com...
>>
>>>>> From what I knew after surfing about this proof is that Cantor
>>>>> presented it to Dedekind first containing infinite number of gaps, but
>>>>> what was published was the customary one.

>>
>>>>> A nice proof that I saw online is this one:
>>
>>>>> Quote:
>>
>>>>> Cantor's 1874 proof. To show that the real numbers are uncountable, we
>>>>> must show that given any countable sequence of distinct real numbers,
>>>>> there exists another real number not in the sequence.

>>
>>>>> Like the diagonalization argument, we will do so by providing
>>>>> an explicit algorithm which produces such a number; unlike the
>>>>> diagonalization argument, we will employ not decimal expansions but
>>>>> order properties of the real numbers.

>>
>>>>> Let (a_n) be a countable sequence of distinct real numbers. Suppose
>>>>> that there are two distinct terms a_j and a_k such that no term a_l
>>>>> lies strictly between a_j and a_k in other words, suppose that (a_n)
>>>>> does not possess the Intermediate Value Property. Let L be any real
>>>>> number strictly between a_j and a_k, for example (a_j + a_k) /2 . Then
>>>>> L is not in the sequence (a_n).

>>
>>>>> Now suppose that (a_n) does have the Intermediate Value Property.
>>
>>>> Which is equivalent to assuming that (a_n) is an enumeration of (all) the
>>>> real numbers, is it not?

>>
>>> I already explained it is not, but because this is an important point,
>>> I'll expand on showing that it is not!

>>
>>> Take any countable enumeration of all rationals, it does have the
>>> Intermediate Value Property, yet it is not an enumeration of (all)
>>> real numbers.

>>
>> It seems you should define what you mean by "Intermediate Value
>> Property". In analysis one talks about the intermediate value theorem
>> for continuous functions, but this is not what you mean here.

>
> Here it is about enumeration of reals, so such an enumeration would be
> said to possess an Intermediate Value Property iff for every two
> distinct reals it enumerates there is a real that is strictly between
> them that it enumerates. Actually it is obvious from the quote.

It clearly was not obvious to LV.

Cheers

© The Math Forum at NCTM 1994-2018. All Rights Reserved.