Re: Dedekind Cuts, Fundamental Sequences: why?

On Jun 4, 4:39 am, Hatto von Aquitanien <a...@xxxxxxxxxxxxxx> wrote:
Glen Wheeler wrote:
"Hatto von Aquitanien" <a...@xxxxxxxxxxxxxx> wrote in message
news:TKidnbd-d8d7Rf7bnZ2dnUVZ_jmdnZ2d@xxxxxxxxxxxxxxxx
Virgil wrote:

In article <Vf-dnV0lgNg_4v7bnZ2dnUVZ_qmpn...@xxxxxxxxxxxxx>,
Hatto von Aquitanien <a...@xxxxxxxxxxxxxx> wrote:

Bob Kolker wrote:

Hatto von Aquitanien wrote:

What is the step of logic which leads one to seek an extention of
the rational numbers to the real numbers?

Very simple. You want every set of numbers bounded from below to have
a
greatest lower bound and every set of numbers bounded from above to
have a least upper bound. While rational numbers are dense in their
ordering they lack the closure of boundedness, hence real numbers are
invented to extend the rationals.

For example, the set of positive rationals whose squares are less than
or equal to 2 is bounded above but has no least upper bound.

Similarly, the set of positive rationals whose squares are greater than
or equal to 2 is bounded below but has no greatest lower bound.

Isn't that just a fancy way of saying x^2=2, x>0 has no solutions in the
rational numbers? The example helps me understand how a set of rational
numbers could have no least upper bound in the rational number.

No, I'd say that is not the best way to think about it. Think of the
set

S = {x \in \Q : x^2 > 2}.

Now think of a lower bound on the set. You can do better than what you
thought of, try again. An elementary proof will show that the greatest
lower bound is not in \Q.

I'm not sure what you intend for me to prove. If I prove that sqrt(2) is
not in \Q, then I don't have definitions for sqrt(2) < q and sqrt(2) = q.
All the definitions given up to this point apply to rational numbers.


Consider this different but similar situation: It is also true that
x^2 = -1 has no solution in the rationals; but the set

S' = {x in Q : x^2 > -1}

has no lower bound at all.

The difference here is that the set

S = {x \in \Q+ : x^2 > 2}

does have a /lower/ bound (in fact, it has infinitely many lower
bounds); it just has no /greatest/ lower bound.

From the perspective of already knowing about real numbers, the assertion is
blatantly obvious, and easily proven. That is, if I also assume that I can
use the ordering of the real numbers in the proof.


But given some arbitrary function f and the set:

S = {x in Q : f(x)>0}

where S has a /lower/ bound, we don't know whether in fact S has a /
greatest/ lower bound (in Q) or not.

It might, or it might not. Sometimes, as in the case of sqrt(2), we
can easily prove that there is no element of Q which is a greatest
lower bound; but sometimes it may not be obvious (or even possible) to
prove this.

When we extend Q to form R, we no longer have this problem: if S is a
subset of Q, and S has a lower bound; then it perforce has a (unique)
greatest lower bound in R (possibly in Q and possibly not). It even
has the stronger property: if S is a subset of R, and S has a lower
bound, then it has a unique greatest lower bound in R.

And that makes various theorems much easier to conceptualize and prove
(e.g., the mean value theorem), which satisfy various notions of
continuity. If f is any continuous function, and we have rational
numbers a < b, and f(a) < 0, and f(b) > 0, then we can talk about "the
number c" with a < c < b and f(c) = 0, without knowing whether c is
rational or not.

Cheers - Chas

.

Relevant Pages

  • Re: Dedekind Cuts, Fundamental Sequences: why?
    ... You want every set of numbers bounded from below to have a greatest lower bound and every set of numbers bounded from above to have a least upper bound. ... While rational numbers are dense in their ordering they lack the closure of boundedness, hence real numbers are invented to extend the rationals. ...
    (sci.math)
  • Re: Dedekind Cuts, Fundamental Sequences: why?
    ... ordering they lack the closure of boundedness, ... the set of positive rationals whose squares are less than ... or equal to 2 is bounded below but has no greatest lower bound. ... Pickert and Görke call fundamental sequences "Cantor sequences". ...
    (sci.math)
  • Re: Dedekind Cuts, Fundamental Sequences: why?
    ... ordering they lack the closure of boundedness, ... the set of positive rationals whose squares are less than ... or equal to 2 is bounded above but has no least upper bound. ... or equal to 2 is bounded below but has no greatest lower bound. ...
    (sci.math)
  • Re: Dedekind Cuts, Fundamental Sequences: why?
    ... Bob Kolker wrote: ... greatest lower bound and every set of numbers bounded from above to have ... they lack the closure of boundedness, hence real numbers are invented to ... extend the rationals. ...
    (sci.math)
  • Re: Dedekind Cuts, Fundamental Sequences: why?
    ... they lack the closure of boundedness, hence real numbers are invented to ... the set of positive rationals whose squares are less than ... the set of positive rationals whose squares are greater than ... or equal to 2 is bounded below but has no greatest lower bound. ...
    (sci.math)
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