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Topic: Discussion with WM - Frustration reaches boiling point (What
is not clear?)

Replies: 9   Last Post: Jul 5, 2014 8:30 PM

 Messages: [ Previous | Next ]
 ross.finlayson@gmail.com Posts: 2,720 Registered: 2/15/09
Re: Discussion with WM - Frustration reaches boiling point (What
is not clear?)

Posted: Jul 5, 2014 12:18 PM

On 7/5/2014 8:14 AM, PotatoSauce wrote:
> On Saturday, July 5, 2014 10:55:51 AM UTC-4, muec...@rz.fh-augsburg.de wrote:
>> On Saturday, 5 July 2014 16:25:47 UTC+2, PotatoSauce wrote:
>>
>>
>>
>>
>>

>>>
>>
>>> Right, so you could've just looked at s_n = Q \ {q_1,..., q_n} to arrive at the same conclusions.
>>
>>>
>>
>> But the problem would not have been so obvious. Just this example has been accepted without further thinking by many.
>>

>>>
>>
>>>
>>
>>> In fact, you were using a needlessly convoluted example.
>>
>>
>>

>>> For instance: s_n = {n,... , 2n}
>>
>>
>>
>> s_n = {n+1, ..., 2n} would be more systematic. I have used it frequently.

>>>>
>>
>>
>>

>>> lim s_n = { }
>>
>>>
>>
>>> lim |s_n | = infinity
>>
>>>
>>
>>
>>

>>> Applying your interpretation, we get the ridiculous conclusion that there is no bijection between a set to itself.
>>
>>
>>
>> That is only a ridiculous conclusion if you forget that there is no infinite set itself.
>>

>>>
>>
>>> But given any set, the identity map is a bijection.
>>
>>
>>
>> Who has given the set? God?
>>

>>>
>>
>>
>>

>>
>>>
>>
>>> s_n = {1/n , ... , 1/2n}
>>
>>
>>
>> Same a sabove.
>>
>>
>>
>> Please do not believe that you can remove an internal contradiction of a theory by quoting arguments infavour of one side.
>>

>>>
>>
>>
>>

>>> And yet, there is a bijection n -> 1/n between natural numbers and their reciprocals.
>>
>>
>>
>> There is not even a bijection between all natural numbers. Don't confuse this with every natural number. Every bijection involves numbers n that have a finite number of predecessors but an infinite number of successors. That is the basic feature of every natural number. All natural numbers would spoil this, because none could follow. Really ridiculous!
>>
>>
>>
>> Regards, WM

>
> If you are assuming from the start N doesn't exist to prove that there is no bijection between N and Q, then your logic is entirely off.
>
> You want lim card(s_n) to represent the cardinality of the sequence s_n "at infinity."
>
> But you have failed to provide a definition of s_n "at infinity."
>
>
>
>

Cardinals with their arithmetic are
not so suited to this as ordinals are.

lim_oo

Here 'oo' fits as an ordinal, not a
cardinal, then later in terms of
products of ordinals as placeholders
of the notation.

You are using the wrong tool.

Ordinals maintain reciprocals that
cardinals do not.

Cardinals are particular to a numeric
purpose, here to kite the ceiling of
powersets.

No, cardinal arithmetic isn't used in
areas and in terms of, for example,
products of omega, for oo, sees the
usual methods of real analysis hold
more directly with notions of an oo
as ordinal than cardinal.

Of course, this is with usual notions
of infinity in the numbers as there are.

Heh, calculus is perfect, and cardinals
(as a form) are irrelevant to it.

Date Subject Author
7/5/14 ross.finlayson@gmail.com
7/5/14 ArtflDodgr
7/5/14 Virgil
7/5/14 ArtflDodgr
7/5/14 Virgil
7/5/14 Virgil
7/5/14 Virgil