Re: Countability of real numbers

On 6 Nov 2006 23:14:48 -0800, "Ajeet" <asgrewal@xxxxxxxxx> wrote:


David R Tribble wrote:
David C. Ullrich wrote:
A number is an element of a set or not. It cannot be a
non-element and then somehow _become_ an element "when
n -> infinity".


Ajeet wrote:
I see your argument.

I meant it as a construction of S

so sqrt(2) would be added to S, when n (length of string being
considered) -> oo


David C. Ullrich wrote:
Saying that something is added to S "when n -> infinity is
considered" still doesn't make much sense.


Ajeet wrote:
why not? It makes sense to me. What I am trying to construct is the set
of all reals. However, given the mode of construction, reals that are
irrationals or recurring decimals would not be added till the length of
the string tends to infinity.

"Tends to infinity" or "is infinite"? Which one?

You can't consider the length of strings to "be infinite" if they only
"tend to" infinity.

The set of strings with a finite number of digits, where the number of
digits tends to infinity, is a countable set.

The set of strings with an infinite number of digits, where the number
of digits is infinite, is an uncountable set.

They are two different sets.

So you are saying the set I am constructing is countable? At any point
in the construction there are only a finite number of digits, however
since the construction goes on endlessly, it would contain all reals.

<PROOF>
Let S_i denote the set with all strings of length i. Therefore the set
that I am talking about is
S = U S_i (0<=i<infinity)

Suppose a real number "r" does not exist in my set. let the decimal
representation of "r" be d_0.d_1d_2d_3......(to infinity) where d_i is
the ith digit.
+Since r does not belong to S, there will be some digit k which will be
off. More precisely

the prefix d_0.d_1d_2d_3....d_k-1 belongs to S but,
the prefix d_0.d_1d_2d_3....d_k-1d_k does not belong to S

No. That's simply not so.

But this is not possible because that prefix would have been added in
all sets S_j with j>=k+1.

Therefore no such k exists.

Therefore r belongs to S.
</PROOF>


************************

David C. Ullrich
.

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