Re: Why are reals uncountable yet algorithms countable (long)?
From: fishfry (BLOCKSPAMfishfry_at_your-mailbox.com)
Date: 11/15/04
- Next message: Daniel W. Johnson: "Re: Cantor's diagonal proof wrong?"
- Previous message: Shmuel (Seymour J.) Metz: "Re: About separable Hilbert space"
- In reply to: Ray Whitmer: "Re: Why are reals uncountable yet algorithms countable (long)?"
- Next in thread: Hibernatus: "Re: Why are reals uncountable yet algorithms countable (long)?"
- Messages sorted by: [ date ] [ thread ]
Date: Mon, 15 Nov 2004 00:34:06 GMT
In article <cn8k65$pu3$1@news.xmission.com>,
Ray Whitmer <ray@xmission.com> wrote:
> On Sun, 14 Nov 2004 18:22:51 +0000, fishfry wrote:
>
> > In article <cn85rn$ecb$1@news.xmission.com>,
> > Ray Whitmer <ray@xmission.com> wrote:
> >
> > [snip]
>
> You could theorize that reals exist that are not representable as any
> specifiable sequence of digits,
I believe this is the underlying problem you're having. In fact, a real
is exactly identifiable with an infinite sequence of digits.
For example, consider 1/3 = .33333...
This is an infinite sequence. The sequence is 3, 3, 3, 3, 3, 3, 3, ...
Another way to think of sequences is that a sequence is a function whose
domain is the natural numbers. So we have a function f given by f(1) =
3, f(2) = 3, f(3) = 3, f(4) = 4, etc.
That is what we mean by an infinite sequence. Given any infinite
sequence, it corrsponds to some real between 0 and 1, by imagining a
decimal point in front of the numbers of the sequence. And we can make
that statement more precise: we interpret the sequence by turning it
into an infinite series of the elements of the sequence divided by the
corresponding power of 10:
.33333... = 3/10 + 3/100 + 3/1000 + 1/10000 + ...
Likewise pi = 3 + 1/10 + 4/100 etc.
Now the natural numbers are countable, by definition. A set is countable
if it can be put into 1-1 corrrespondence with the natural numbers, and
of course any set can be put into 1-1 correspondence with itself; so the
naturals N are countable
The claim is now that the reals are NOT countable. So to prove that
wrong, you'd have to exhibit a bijection, or 1-1 correspondence, between
the naturals and the reals.
If the preceding makes sense, you can now take another look at the
Cantor diagonal argument, and see if it's any clearer to you.
To show that the reals are not countable. we assume we have a complete
list (or sequence) that exhausts all the reals. Then we show that this
is false, there is some real that is not on the list.
- Next message: Daniel W. Johnson: "Re: Cantor's diagonal proof wrong?"
- Previous message: Shmuel (Seymour J.) Metz: "Re: About separable Hilbert space"
- In reply to: Ray Whitmer: "Re: Why are reals uncountable yet algorithms countable (long)?"
- Next in thread: Hibernatus: "Re: Why are reals uncountable yet algorithms countable (long)?"
- Messages sorted by: [ date ] [ thread ]
Relevant Pages
|
