Re: Non-countability of R

On Sun, 3 Aug 2008 newsgr.mail@xxxxxxxxx wrote:

prove that "the set R of real numbers is uncountable".

PROOF. It is enough to show that the set I of all real numbers r such
that 0 <= r <= 1 is uncountable: this is because |I| <= |R|.
Assume that I is countable, so that it can be written in the form
{r_1,r_2,r_3,...}. Write each r_i as a decimal, say

r_i = 0.r_{i1} r_{i2} ...

where 0 <= r_{ij} <= 9. We shall get a contradiction ny producing a
number in the set I which does not equal any r_i. Define

s_i = 0 if r_{ii} <> 0; 1 if r_{ii} = 0

and let s be the decimal 0.s_1 s_2 ...; then certainly s \in I. Hence
s=r_i for some i, so that s_i=r_{ii}; but this is impossible by the
definition of s_i. QED

I think that this method won't work because imho the author should
have spoke about the fact that the decimal representation of reals is
not unique (as other authors do).

For example, consider if we list (as first number) 0.1000... and
eventually 0.09999... at the same position; in the second case we
would get 0.1 even though it was already present in the list as
0.09999... .

You are correct. To correct for this, require that the
reals in [0,1] be expressed without ending in 9999....

Now however to write 1 in the form of 0.d1 d2 d3 ...
requires 1 = 0.9999.... So we divest ourselves of 1
by just listing the reals in [0,1) expressed as decimals
not ending in 999....

Now, nothing excludes the fact that there could be listed
other numbers in the second, third etc. positions that are represented
in terms of a 9 expansion, determining a final number that is already
present in the list.
Any comment?


.

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