|July 19th, 2016, 06:44 PM||#21|
Joined: Dec 2013
Math Focus: Mainly analysis and algebra
To me, the most elegant formulation of the proof uses explicit indices in both the enumeration and the sequences and does away entirely with the proof by contradiction which is the root of many misunderstandings in favour of a subsidiary theorem that no countable set of sequences/real numbers contains every such sequence/real number. The contrapositive of this is the statement that the reals are not countable.
So his "proof" fails not because there is no mapping with the naturals: because it is a list the mapping exists and the set is countably infinite; but because the set he creates is not the set he believes it to be: the real numbers. $\frac13$, $(\pi-3)$ and $\sqrt2 -1$ are not in it. He is unable or unwilling to see this fact.
Last edited by v8archie; July 19th, 2016 at 06:51 PM.
|decimal, fractions, numbers, ordering, real|
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