Re: Non-countability of R

On Mon, 4 Aug 2008, David C. Ullrich wrote:
<marsh@xxxxxxxxxxxxxxxxxx> wrote:
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).

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....

No, that's not sufficient to fix the proof. Think again
about what the problem with the proof is.

It should work because the diagonal decimal is entirely of 0's and 1's.
Now if perhaps, it ends in d 0000..., (0 < d) then it should be in the
list because decimals ending in (d-1) 9999... have been omitted as a
representation of a real in [0,1).

What do you see wrong with the method? Yes, if one used 9's instead of
1's in the diagonal decimal definition, then it wouldn't work. Also
this method won't work in binary decimals but it will for all other
bases b in N - {1,2}.

----
.

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