Re: Should oo+a=oo be abandoned?

On 6 Jul., 19:08, Eckard Blumschein <blumsch...@xxxxxxxxxxxxxxxxxxx>
wrote:
On 7/5/2007 10:34 PM, hagman wrote:

I use N,Z,Q,R,C for naturals, integers, rationals,reals, complex
numbers.
I see that you try to imitate blackboard bold in ASCII arts, but I'm
not
fond of that.

I just alluded a bit to Dedekind who wrote R for the body of rationals
bur IR for reals.

Which is a bit obsolete by todays standard (btw. the English
translation
for "Körper" is field).


The illusion that real numbers are numbers of the same sort as rational
numbers. They are uncountable. They do not have a direct and complete
numerical identity.

What is a numerical identity?
The positive real number x such that x*x=2 is quite well defined
and is unique and thus has an identity.

Imagine a meter. Where is x on it?

What does a physical body as a meter have to do with numbers of a
field?


Then what distinguishes qualities from quantities?

Quantities can be expressed numerically, i.e. they somehow relate to a
unity via multiplcation or division.
Zero as well as infinity fail to do so in a reproducible manner.

Thus 5 is a quantity because 5*1=5?

Yes.

Here's news to you: 0*1=0, oo*1=oo.

I wrote in a reproducible manner. Maybe, I should get more precise.
I meant unambiguously invertible.

I imagine the continuum like something fictitious and already compact
that does neither require nor allow any compactification.

Exclusion of oo from IR is the usual tenet but not convincing to me
because oo is not much different from any real number in that it is
numerically not accessible.

For the decision of what is in and what is not in e.g. R, allow me
to introduce a concept that seems to be new for you: rigorous
definition.
One possible definition is that R is the set of equivalent classes
of Cauchy sequences of rationals modulo zero sequences.
Thus if oo were in R, there would exist a Cauchy sequence of rationals
converging to it.

This way, oo is excluded while 1/oo is included. Therefore I am not sure
whether or not this definition is the best choice.

oo excluded: That's what I intended (especially to make R a field)
1/oo included? If you mean something different than 0, it is excluded
as well


I do not see any reason for compactification because uncountable numbers
must be compact. I argue: Otherwise the were not uncountable.

Absolutely nonsense.
Or else present a finite subcover of the countable open cover of the
reals I gave in my previous post.

I guess, this matter deserves careful elucidation prior to arbitrary
definitions.

You use mathematical terms in a mathematical discussion that are in
use with a very clear meaning.


I hope it is not too much if I introduce another concept that
has turned out to be quite useful in mathematics: proof.

Construction of an insulated network of useless proffs is not my goal.

Consider the infinitely many open intervals of the form (x-1,x+1).
They cover R.

Really?

If x is a real number the it is contained in (x-1,x+1).

Admittedly, I intended to let x run over the naturals only in the
original post,
and here I changed my mind to let it run over all reals just in order
to avoid a discussion whether or not each real number has an integer
at distance less than one.


There is no hole in N, but you were referring to the continuum.

Is there a finite subcover?

Just a sieve with holes without an end.

Is that a "yes" or a "no"?

I cannot get the point. What do you mean by subcover?

*Sigh* If X is a set, then a family {A_i | i in I} of subsets of X
is a COVER if the union of all A_i is X.
It is a FINITE COVER if the index set I is finite.
A second cover {B_j | j in J} is a SUBCOVER if JcI and B_j=A_j
for all j in J.
A topological sace is called COMPACT if every cover by open sets
has a finite subcover.
You said R were compact.
I specified an open cover (you may take it for granted that the
open intervals are open sets).
I asked you to describe a finite subcover of it.


Each single point of continuum is fictitious. Peirce wrote a mere
potentiality.

Then so is the point 1

Yes in IR, no within rational numbers.

If *all* points of R are fictitious (on-existet)

Perhaps you meant no-existent.

Indeed, my fingers were faster than my keyboard, sorry.


, then

R is not uncountable, it is empty.

Not existing within the rationals and as rationals does not exclude
existence as reals within the reals.

So they are existent but fictitious?


Especially, the power set of such a set A could be
shown to be at most as big as A.

No.

Why not?

Because there is no quantitative basis for such comparison.

Since you claim that A "cannot be enlarged" you must explain
in what manner Power(A) (or maybe rather B := Power(A) union A?)
is not an enlargement of A.
Strictly speaking, B is clearly an enlargement because AcB
and AeB but not AeA.
I was willing to soften your claim by assuming you merely
meant card(B)=card(A) but you refute that (it would be wrong
anyway).


Oooookay.
You accept that
- the rationals are countable

Of course.

- the reals are uncountable (somewhat)

Absolutely uncountable.

but not
- no map from R to Q is injective; no map Q->R is surjective; there
exists a surjection R->Q; there exists an injection Q->R ?

Today I understood that border-lines may make continuous areas not
isotropic but homeotropic (or the like). However, to my understanding
the continuum does not have an approachable inner structure. We just may
attribute a rational approximation to it as good as we like.

I begin to think that Kurt Schwitters is your favourite mathematician.

Having an incomplete A does not bother much because I am able to get as
many points as I like. However, if a complete A is the precondition for
getting P(A), then I do not have anything available until A is complete.

I can get as many points of P(A) as I like as well.

Really? Every point of P(A) must be represented by means of infinitely
many numbers. This does not even work in principle until A is completely
available.

Definitely not. Indeed I can present a "point" (usually called
element) of P(A)
even before you start giving me elements of A: {}

Defining "There are at least as many A as there are B"
as "There exists an injective map B->A" would be
sufficient.

Just this is justified only for finite, i.e., genuine sets, for settled
ones.

It is provably a transitive relation on sets which
coincides with your finite notion of size comparison etc.
Such observations are what I call justification.

Call it an unjustified justification. It is just a guess based on analogy.


Can you come up with a consistent finer equivalence relation of sets
together with a linear order of the equivalence classes?

On what grounds should f(x)=anything make sense for a function like *?

I lost the reference to *

The "*" is multiplication.

f(x)= * ? Hm.

Of course * is a function of two arguments and is usually written in
infix notation. I merely wanted to stress that your claim about
multiplication should be valid more generally if it were justified at all.

Regards,
EB, admiring all who really understand the fundamentals of mathematics

At least you are not narcistic.

I am tired, did not understand all details and made perhaps several
mistakes. If you read German, you might find in de.sci.mathematik what I
just wrote in two threads:
- Bitte keine Ontologie sondern Konsequenz in den Grundlagen
and
- kein Schnellschuss zur LT (Terhardt+LMT)

Regards,
EB

Nö, hab keinen Bock, den Kram nochmal auf Deutsch durchzukauen.

.

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