Re: Cardinality of Real Numbers
- From: The Ghost In The Machine <ewill@xxxxxxxxxxxxxxxxxxxxxxx>
- Date: Mon, 29 Aug 2005 00:30:20 GMT
In sci.math, jswimr3@xxxxxxxxx
<jswimr3@xxxxxxxxx>
wrote
on 28 Aug 2005 10:54:01 -0700
<1125251641.615664.93900@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>:
> I've been thinking about cardinality proofs lately, and I've run into
> something that's been bothering me. I thought of what seems like a
> mapping from the set of integers to the set of real numbers. Now, of
> course, this can't exist, so there must be something wrong with my
> mapping, but I can't see what it is.
The cardinality proof works against *any* mapping, be it yours,
mine, Aunt Millie's, |-|erc's ... anyone's.
Usually, the range is [0,1) but if one allegedly has a
mapping M from N to [0,1) the mapping can be extended to
(-oo, +oo) by various means -- such as M'() from M():
M'(3n) = M(n)
M'(3n+1) = 1/(1 - M(n))
M'(3n+2) = 1 - 1/(1 - M(n))
(there's a few problems at M(n) = 0 but you get the idea).
Or one can use M'(n) = tan(pi * M(n) - pi/2), if one doesn't
mind transcendentals. Any 1-1 onto mapping from [0,1) -> R will
do for composition.
>
> The mapping works like this: for each integer, map it onto all the
> reals you can get by putting a decimal point anywhere in it. For
> example, 123 would map to:
>
> 123
> 12.3
> 1.23
> .123
This is not a mapping from N to R, but a mapping from N to a
subset of R (or, more realistically, a subset of Q or a set I
like to call T_10 = {k/10^n: k, n in J}, which is a proper subset
of Q (it doesn't contain 1/3, for example).
One can of course work around this problem without too much trouble;
if you're familiar with the N -> Q mapping, a similar technique
resolves this issue.
>
> It seems like this would cover the full set of real numbers.
Not even close. :-) For starters, where's 1/3?
[rest snipped]
--
#191, ewill3@xxxxxxxxxxxxx
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