• Security incident: ISF was recently accessed by intruders. Please change your password, and change it anywhere else you used it. Read more

Cont: Deeper than primes - Continuation 2

Status
Not open for further replies.
Last edited:
ctamblyn said:
Choose whatever set S you like, I don't mind.
Now choose any function you like which maps elements of your set S to elements of P(S), and call it f.

Now construct the set Tf = { all x in S such that x is not contained in f(x) }.

Now answer the question, yes/no: do you agree that there is no y in S such that Tf = f(y)?
Click to expand...
As clearly seen in the dialog between Hans Hüttel and me in http://math.stackexchange.com/quest...f-set-s-for-which-there-is-a-bijection-with-s , also he uses a fundamental notion that does not distinguish between the infinite and the finite, as follows:

"What is the general definition of sets A and B having the same cardinality? All we know about A and B is that they are sets.".

ctamblyn and Hans Hüttel share a fundamental notion of A and B that stands at the basis of "general definition of sets A and B having the same cardinality".

Since only infinite sets are bijective with their proper subsets (known also as Dedekind-infinite property), no definition of sets A and B having the same cardinality, can avoid the difference between infinite sets and finite sets.

In other words, one can't provide general definition of sets A and B having the same cardinality which is based of a notion that does not distinguish between infinite sets and finite sets.

Any attempt to force general definition of sets A and B having the same cardinality, by explicitly avoid the distinction between non-empty finite sets and non-empty infinite sets, is doomed to fail.

ctamblyn and Hans Hüttel, explicitly avoid the distinction between non-empty finite sets and non-empty infinite sets.

I'll agree with their notion that explicitly avoids the distinction between non-empty finite sets and non-empty infinite sets, in case that they show even a single case of non-empty finite set that is bijective with any of its proper subsets.
 
Last edited:
My argument is very simple.

Mathematics can't be used in order to define a general theorem of cardinality that does not distinguish between non-empty finite sets and infinte infinite, exactly because non-empty finite sets and non-empty infinite sets are essentially different of each other as follows:

Any given infinite set is bijective with at least one of its proper subsets, where no non-empty finite set is bijective with any of its proper subsets.

Any attempt to claim otherwise establishes a naïve version of set theory, and if ZFC supports a general theorem of cardinality that does not distinguish between non-empty finite sets and infinite sets (as done in case of Cantor's theorem) it is a naïve version of set theory.

More details are already given in http://www.internationalskeptics.com/forums/showpost.php?p=11595997&postcount=2305.
 
Last edited:
By traditional mathematics the cardinality of the set of all real numbers is bijective with the set of all positive real numbers that are > 0 AND < 1.

It is also well known that multiplying all the positive real numbers that are > 0 AND < 1 with each other, we always get a real number > 0 AND < 1, which is smaller than all the considered multiplied real numbers, AND > 0.

If this smaller number is added to the multiplied positive real numbers that are > 0 AND < 1, we get another real number, which is smaller than all the considered multiplied real numbers, AND > 0.

So the cardinality of all positive real numbers that are > 0 AND < 1 is not satisfied, and since the cardinality of the set of all real numbers is bijective with the set of all positive real numbers that are > 0 AND < 1, we can conclude that the cardinality of the set of all real numbers is not satisfied.

In other words, bijection may measure that two sets have a complete 1-to-1 matchup among their members, but it does not tell us if the measured sets have a satisfied cardinality (cardinality with an accurate value).
 
Last edited:
Here is a clearer version of my previous post.

By Traditional Mathematics the cardinality of the set of all real numbers is bijective with the set of all positive real numbers that are > 0 AND < 1.

It is also well known that by multiplying all the positive real numbers that are > 0 AND < 1 with each other, we always get real number > 0 AND < 1, which is smaller than all the considered multiplied real numbers, AND it is > 0.

If this smaller number is added to the multiplied positive real numbers that are > 0 AND < 1, we get another real number, which is smaller than all the considered multiplied real numbers, AND it is > 0, etc. ad infinitum.

So the exact cardinality of all positive real numbers that are > 0 AND < 1 is not satisfied, and since the cardinality of the set of all real numbers is bijective with the set of all positive real numbers that are > 0 AND < 1, we can conclude that the cardinality of the set of all real numbers is not satisfied.

In other words, bijection may measure that two sets have a complete 1-to-1 matchup among their members (no member is left out) but it does not tell us if the measured sets have a satisfied cardinality (cardinality with an accurate value like |R|).
 
doronshadmi said:
Here is a clearer version of my previous post.

By Traditional Mathematics the cardinality of the set of all real numbers is bijective with the set of all positive real numbers that are > 0 AND < 1.
Click to expand...

You are blending to two concepts and inventing a term. Did you mean to say that the set of real numbers, R, and the reals on the open interval, (0, 1), have the same cardinality (or, equivalently, that there exists a bijection between R and (0, 1))?

It is also well known that by multiplying all the positive real numbers that are > 0 AND < 1 with each other, we always get real number > 0 AND < 1, which is smaller than all the considered multiplied real numbers, AND it is > 0.
Click to expand...

Well known? Got a reference for that? Just so we are clear, here, are you suggesting multiplying all of the numbers on the interval, (0, 1), together to get a single product? If so, then you will need to (a) prove the product exists and (b) prove the product is greater than zero.
 
doronshadmi said:
[snip]

So the exact cardinality of all positive real numbers that are > 0 AND < 1 is not satisfied, and since the cardinality of the set of all real numbers is bijective with the set of all positive real numbers that are > 0 AND < 1, we can conclude that the cardinality of the set of all real numbers is not satisfied.

In other words, bijection may measure that two sets have a complete 1-to-1 matchup among their members (no member is left out) but it does not tell us if the measured sets have a satisfied cardinality (cardinality with an accurate value like |R|).
Click to expand...

Please define " satisfied cardinality " when you can't even have a number value for |R|.

You do know that there are sets that don't have a cardinality that can be expressed as a number.
 
jsfisher said:
Did you mean to say that the set of real numbers, R, and the reals on the open interval, (0, 1), have the same cardinality (or, equivalently, that there exists a bijection between R and (0, 1))?
Click to expand...
That is indeed the Traditional Mathematics' claim, but this claim is not enough in order to also conclude that bijection between two sets of infinitely many R members, have, for example, an accurate cardinality like |R|.

In other words, Traditional Mathematics has to prove that bijection between two sets of infinitely many R members also provides, for example, an accurate cardinality like |R|.


jsfisher said:
Well known? Got a reference for that? Just so we are clear, here, are you suggesting multiplying all of the numbers on the interval, (0, 1), together to get a single product? If so, then you will need to (a) prove the product exists and (b) prove the product is greater than zero.
Click to expand...
The reference is already in your mind jsfisher, all you need is to multiply all of the numbers on the interval (0, 1) in order to immediately realize that the result is always smaller than each one of the multiplied numbers, and it can't be 0 or 1 by the very definition of the interval (0, 1).

In other words, the term "all", in case of infinitely many numbers, is not satisfied, and as a result no accurate cardinality exists for set R, as explained in my previous post.
 
Last edited:
Little 10 Toes said:
Please define " satisfied cardinality " when you can't even have a number value for |R|.

You do know that there are sets that don't have a cardinality that can be expressed as a number.
Click to expand...
|R| is called transfinite number, and it is considered as an accurate mathematical value (which means that it is a satisfied cardinality) by Traditional Mathematics.

In http://www.internationalskeptics.com/forums/showpost.php?p=11674901&postcount=2312 it is shown that Traditional Mathematis' reasoning that is based on bijection between two infinite sets, does not provide the needed proof for accurate |R|.
 
Last edited:
doronshadmi said:
The reference is already in your mind jsfisher, all you need is to multiply all of the numbers on the interval (0, 1) in order to immediately realize that the result is always smaller than each one of the multiplied numbers, and it can't be 0 or 1 by the very definition of the interval (0, 1).
Click to expand...

What you think I may be thinking is not relevant.

You have falsely claimed something to be well-known. It isn't. You have also used your own incredulity to justify your alleged facts. They aren't.

Your argument is stuck unless and until you provide proof that (a) the product of all of the numbers on the interval (0, 1) exists and that (b) the product is greater than zero.
 
Last edited:
jsfisher said:
Your argument is stuck unless and until you provide proof that (a) the product of all of the numbers on the interval (0, 1) exists and that (b) the product is greater than zero.
Click to expand...
I'm not a mathematician, but my reasoning is as follows:
IF the product exists, it must be greater than 0, since all the numbers that are multiplied are greater than zero.
However, if we call the product x, x can not be greater than zero, since if it was, infinitely many of the numbers that are multiplied would be smaller than x.

Therefore, the product can not exist.

How did I do?
 
jsfisher said:
(a) the product of all of the numbers on the interval (0, 1) exists
Click to expand...

What about: the product of all the numbers in an uncountable set exists if every[*] sequence of multiplications of numbers in that set converges to the same number?

(b) the product is greater than zero.
Click to expand...

It's zero though in this case.

* ETA: actually it depends on the sequences you consider, so not every sequence but every sequence which converges. Differently, let S be an uncountably infinite subset of R. Let U be the set of sequences in S which converge to an element of S. Then the product of all numbers in S exists if there exists some r in R such that for every u in U the corresponding sequence of partial products converges to r. The product of all numbers in S is then equal to r.
 
Last edited:
Status
Not open for further replies.

Share this page

ISF - Join now!

Every member here is approved by hand. No bots, no spam, just people who care about evidence and honest debate.

Membership is free!

Create your free account

Back
Top Bottom