Re: Question about a minor recasting of Dedekind cuts.

On Jul 20, 3:15 am, riderofgiraffes <mathforum.org...@xxxxxxxxxxxxxx>
wrote:
I've been explaining various constructions to a
maths club. I've constructed the integers as
equivalence classes of pairs of natural numbers,
where (a,b) == (c,d) if a+d=b+c, and I've
constructed the rationals similarly with ad=bc.
Now I'm constructing the reals in two ways.

Firstly I've used Cauchy sequences, equivalent
if their difference goes to zero. Easy enough.
With Cauchy sequences the limit doesn't have
to exist in the set from which the elements
are drawn - easy to prove.

Then I've used Dedekind cuts, but it's proving
really, really messy. I want to define the
negative, and subtraction, but having the limit
point (if it exists) in only one set means all
sorts of nasty conditions. Try it. Try to
define the quotient of two Dedekind cuts.

Why not simply say this:

A D-cut of the rationals is a pair of sets A,B
such that:

(using
u for union,
n for intersection,
e for element
)
a in A, b in B => a<=b
AuB = Q
(q e AnQ) & (a e A => a<=q) => q in B
(q e BnQ) & (b e B => q<=b) => q in A

Using shorter notation:

A <= B
AuB = Q
A<=q & q e A => q e B
q e B & q<=B => q e A

Now everything follows simply from the concept
that if A and B intersect then the intersection
is the rational they define, and if not, then
the irrational they define is "between" them.
Now all the arithmetic operations go through
without messy case analyses.

Thoughts?

I am sorry to say there is a SiNGLE way to state all reals.

A cut may NOT BE cut.

A set to define all rationals is not each cut as allowed. Your
rational cut intersects and does not include all real.

Why?

And you gave an answer. Division!

A set divisor as a Dedekind cut must obey a single rule.

1/0=not allowed!

Your sets allow may non disallowed answers.

.

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