Hi list,

I have a question I've been thinking about for a while... It may seems stupid, 
but I need to understand where I'm wrong.

So here it is... Does the set N contains infinite number ?

I ask this because Cantor prove with the diagonalisation argument that the set 
R is uncountable and cannot be map to N, so that the cardinality of R is 
higher than the cardinality of N.

But if N contains infinite numbers then with the diagonalisation argument I 
can do something like this, a mapping of N to N :

N       
1       D1 = d11d12d13d14 ... d1k ...
2       D2 = d21d22d23d24 ... d2k ...
3       D3 = d31d32d33d34 ... d3k ...
...      
n       Dn = dn1dn2dn3dn4 ... dnk ...
... till infinity

So DX is any infinite natural number, dxx is a digit between 0 and 9. No I 
have the number X = x1x2x3x4x5 ... xk ... which is an infinite natural number 
with x1 != d11, x2 != d22 ... xn != dnn. So this natural number is not in the 
list, so it means N is not countable but it cannot be !

Also there is an infinity of finite length natural number... but also an 
infinity of infinite natural number...

Also if infinite natural number exists how can we conserver the well order of 
the natural number ?

How can we find the successor of the natural number  
1111111111111.......111111..... till infinity ?

I get lost, I think there is a problem with how infinity is handled in the 
argument.

Regards,
Quentin

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