Re: Cantor won, but one final question anymore thnk u
From: Dave Seaman (dseaman_at_no.such.host)
Date: 09/26/04
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Date: Sun, 26 Sep 2004 00:55:27 +0000 (UTC)
On Sun, 26 Sep 2004 00:59:36 +0300, Keckman wrote:
> On Sat, 25 Sep 2004 21:13:37 +0000 (UTC), Dave Seaman
><dseaman@no.such.host> wrote:
> --clik --
> Suppose a set R is
> 1) linearly ordered, and
> 2) densely ordered, i.e., between any two members there is another, and
> 3) has no "endpoints", i.e., smallest or largest members, and
> 4) has no gaps, i.e., if it is partitioned into two nonempty sets A and B
> in such a way that every member of A is less than every member of B, then
> there is a boundary point c, so that every point less than c is in A and
> every point greater than c is in B.
> ->Then R is not countable.
> --clik --
>>> The theorem says that if there is no gaps then then R is uncountable.
>>> But there was gap in A U B concerning real numbers when c=srt(2).
>> No, there is no gap. Point 4) explains what it means for a set to have
>> no gaps. To show a gap in R, you have to provide suitable sets A and B
>> such that no boundary point c exists in R. In the case mentioned,
>> c=sqrt(2) is the required boundary point, and c is in R. Therefore, you
>> have not demonstrated a gap in R.
> There is no gap _in_ _R_ allthough there _is_ gap in A U B because
> "every point less than c is in A and every point greater than c is in B"
> (there is not word equal). c does not belongs to A and it doesn't belong
> to B. Of course the is then gap in A U B because c doesn't belong to A
> either B.
No, there is no gap in A U B, since A U B = R and the required boundary
point c does belong to R. Notice that 4) begins, "if [R] is partitioned
into two nonempty sets A and B...", and a "partition" in mathematics
means a decomposition of a set into pairwise disjoint subsets whose union
is the original set. Therefore A U B = R, and since c belongs to R, it
follows that c must belong either to A or to B. We don't know which, and
it doesn't matter.
>>> There was no gap concerning rational numbers i.e. A U B contains all
>>> rationals? So
>>> rationals are not countable but reals are?
>> On the contrary, there is a gap in the rationals, because in that case
>> c=sqrt(2) is not a member of Q, and therefore a gap exists.
> There is not gap when we stay in Q. c=sqrt(2) doesn't belong to Q, then A
> U B contains all the numbers than R.
The fact that c=sqrt(2) doesn't belong to Q means that 4) is not
satisfied with respect to Q, and therefore Q has a gap. Remember that 4)
is the definition of what it means to have no gaps. There is no c in Q
such that every member of Q that is less than c is in A and every member
of Q that is greater than c is in B.
>We didn't not suppose anything more
> than that R is a set in the first hand.
Yes, we did. We supposed 1), 2), 3), and 4).
> R is not said to be the hole one
> "dimensional line - system of co ordinates x-axle" or something like that.
> If we then divede the R to two sets, and choose c out of that R of course
> that c doesn't belong to R because it doesn't belong there at the
> beginning.
I don't understand. If you choose c out of R, then c belongs to R.
>But there is no gap in A U B.
That depends on the original set R.
> but ok. You said = "To show a gap in R, you have to provide suitable sets
> A and B
> such that no boundary point c exists in R."
> Sorry i have to repeat to try to understand:
> "no boundary point c exists in R"? does this boundary point c exist or
> does it not exist in R? Of course that boundary point c exist in R.
> sqrt(2) belongs to R. It belongs to R but it does'nt belong to A U B.
Yes, it does belong to A U B, since A U B = R.
> Ok,
> there is no boundary point in R alltohough it could be divede to two sets
> A and B...ok now i maybe made it clear tomyself.
You have it backwards. There *is* a boundary point in the reals, but not
in Q.
-- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. <http://www.commoncouragepress.com/index.cfm?action=book&bookid=228>
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