Deeper than primes

[doronshadmi]

That's simply what the notation [X, Y) means.

Notation means nothing without a notion at the basis of it.

Please show the notion at the basis of [X,Y) notation.

 

[doronshadmi]

That's a circular definition you have for complete, or didn't you notice? Also, in the link you provide, I provide no definition for completeness.

I wrote "your determination" , exactly because you are using words like "complete" without define them.

EDIT:

Ok, so not only do you prove, again, you don't understand what complete means ...

Since you have used this word, then please show that you also understand it (something that you did not do, yet).

 

[doronshadmi]

Yes. If the successor concept is extended to include intervals, then Y is an immediate successor.

If the word "immediate" is used, it does not matter if an (as a non-particular case) is used.

Furthermore, by using an we actually determine that any arbitrary R member has an immediate successor.

 

[zooterkin]

In other words, you do not follow your own words.

Erm, are you misreading one of those occasions where jsfisher has written "it does matter" as being "it doesn't matter"? If not, what you are saying makes no sense (not much change there, then).

 

[zooterkin]

Notation means nothing without a notion at the basis of it.

Please show the notion at the basis of [X,Y) notation.

The finite interval starting with X, and including everything up to, but not including, Y.

 

[jsfisher]

Are you seriously claiming these two statements are inconsistent?

And it does matter if we deal with [X,Y] or [X,Y).

Yes, it does matter. [X,Y) has immediate successors while [X,Y] does not.

(Hi-lighting added.)

 

[doronshadmi]

Erm, are you misreading one of those occasions where jsfisher has written "it does matter" as being "it doesn't matter"? If not, what you are saying makes no sense (not much change there, then).

Are you seriously claiming these two statements are inconsistent?

You are right, my mistake, I really wrongly read "it does matter" as "it doesn't matter".

My apologies to you jsfisher, and thank you zooterkin for open my eyes to this case.

This mistake of mine has no influence on my argument (if you think otherwise, then please explicitly show why I am wrong also in this case) ,so let us focused now on the difference between [X,Y] and [X,Y).

You say that [X,Y) is:

The finite interval starting with X, and including everything up to, but not including, Y.

Now, according to jsfisher's determination, Y is an immediate successor of "up to".

How can Y be an immediate successor of "up to", but "up to" is not an immediate predecessor of Y?

 

[integral]

[...] Standard Math [...]

You keep using this term; what do you mean by it? The entire body of mathematics that is derived using deductive principles? 'Standard', as opposed to what?

 

[jsfisher]

I wrote "your determination" , exactly because you are using words like "complete" without define them.

I didn't use any word like complete. And there is no reason for me to define my terms when the usage is in complete agreement with accepted definitions.

EDIT:

Since you have used this word, then please show that you also understand it (something that you did not do, yet).

See above. Also, what's the point? You'll go off on some tirade about mathematicians hijacking things as a cover for your own ignorance. If you really want to know what something means, look it up. Stop wasting everyone else's time, here, with your abuse of the language.

 

[doronshadmi]

I didn't use any word like complete. And there is no reason for me to define my terms when the usage is in complete agreement with accepted definitions.



See above. Also, what's the point? You'll go off on some tirade about mathematicians hijacking things as a cover for your own ignorance. If you really want to know what something means, look it up. Stop wasting everyone else's time, here, with your abuse of the language.

the space between any two numbers is completely filled by other numbers (leaving no gaps).

Please explain what do you mean by "completely filled" or "leaving no gaps"?.

There is no immediate predecessor because there is always another real number between any two real numbers.

Do you understand that that you can show that "there is always another real number between any two real numbers" exactly because you always based on a finite case?

 

[jsfisher]

You are right, my mistake, I really wrongly read "it does matter" as "it doesn't matter".

My apologies to you jsfisher, and thank you zooterkin for open my eyes to this case.

Apology noted and accepted.

This mistake of mine has no influence on my argument (if you think otherwise, then please explicitly show why I am wrong also in this case) ,so let us focused now on the difference between [X,Y] and [X,Y).

You say that [X,Y) is:

The finite interval starting with X, and including everything up to, but not including, Y.

Now, according to jsfisher's determination, Y is an immediate successor of "up to".

No. I said that if we extend the concepts of successor and predecessor to include intervals (along with the reals), then Y (a real) is an immediate successor to [X,Y) (a half-open interval).

How can Y be an immediate successor of "up to", but "up to" is not an immediate predecessor of Y?

The phrase, up to, is neither a real number nor an interval, so it has no part in this discussion. On the other hand, [X,Y) is an immediate predecessor of Y, in case you were interested.

 

[doronshadmi]

EDIT: Please answer to http://www.internationalskeptics.com/forums/showpost.php?p=4752487&postcount=3190 .

On the other hand, [X,Y) is an immediate predecessor of Y, in case you were interested.

What if Y is the least upper bound of [X,Y)?

I claim that Standard Math uses here a notion that is based on a finite case of R members, and concludes something about the non-finite case of R members, which is something that cannot be done in the framework of Standard Math (this is exactly my argument).

If you disagree with me then please explicitly show how one can use a finite amount of R members in order to conclude something about a non-finite collection of all R members of some interval.

X and Y of [X,Y] of what is called "finite interval" do not give any knowledge (accept a the title "finite interval") of what is really going in the "completely filled" collection of all R members, of what is called "finite interval".

The best we can do is to use a finite collection of R members, and say that any arbitrary R member that is not X or Y is > X and < Y.



EDIT:

http://en.wikipedia.org/wiki/Supremum

In mathematics, given a subset S of a partially ordered set T, the supremum (sup) of S, if it exists, is the least element of T that is greater than or equal to each element of S.

The least upper bound axiom, also abbreviated as the LUB axiom, is an axiom of real analysis stating that if a nonempty subset of the real numbers has an upper bound, then it has a least upper bound. It is an axiom in the sense that it cannot be proven by the other axioms within the system of real analysis.

Following the wiki quotes above, we get the follows:

1) Upper bound is an axiom under real analysis (it is a categorical determination that must be accepted or rejected).

2) jsfisher accepts this axiom, and avoids any research about its validity.

3) doronshadmi rejects this axiom because he researches its validity and claims that this axiom is based on a notion taken from a finite case and used on a non-finite case, which is something that cannot be done.

4) In this case, jsfisher has to show that it is valid to take a finite case and use it also in a non-finite case.



Let us focused on "greater than OR equal to ..." because (as I get it) this is the heart of this discussion, as follows:

Let us focused only on Standard Math (OM is not used at this part of the discussion).

By using only Standard Math, we follow your own determinations along the discussion, as follows:

1) By standard Math, any interval of non-finite R members is (as jsfisher says) "completely filled" (all R members of that interval are included, without any exceptions (otherwise the interval is incomplete)) (here http://www.internationalskeptics.com/forums/showpost.php?p=4749909&postcount=3161 we can define your own determination to "completely filled", which is related to this case).

2) By Standard Math "completely filled" of a non-finite collection AND no gap (where "no gap" by Standard Math means "there is a room for more R members") is a contradiction, because something can't be (R "completely filled") AND (enables a room for more R members).


Now, let us examine again why Standard Math is derived to this contradiction.

Standard Math is derived to this contradiction, because it tries to understand the non-finite by notions and examples that are based on the finite.

Let us see how Standard Math doing it:

First, we shell examine the case of two integers a and b, such that a is an immediate predecessor of b.

It is easy to get that a is an immediate predecessor of b, because we deal here with a finite case of distinct a and distinct b (this is exactly the construction of the integers).

Now let us examine the case of R members.

By using a finite amount of R members, it is easy to show that there is a room for more R members between distinct a and distinct b.

This room exists exactly because we are using a finite amount of R members, and we cannot conclude that this room still holds when we deal with the non-finite case of all R members.

This is exactly where Standard Math fails in its own framework, because what can be concluded about integers (by using a finite case) cannot be concluded about R (by using a finite case).



Let us return once more to:

http://en.wikipedia.org/wiki/Supremum

In mathematics, given a subset S of a partially ordered set T, the supremum (sup) of S, if it exists, is the least element of T that is greater than or equal to each element of S.

"greater than or equal to" is understood here by using a notion taken from the finite case of R members, therefore it is meaningless in the non-finite case.

 

[The Man]

EDIT: Please answer to http://www.internationalskeptics.com/forums/showpost.php?p=4752487&postcount=3190 .



What if Y is the least upper bound of [X,Y)?

I claim that Standard Math uses here a notion that is based on a finite case of R members, and concludes something about the non-finite case of R members, which is something that cannot be done in the framework of Standard Math (this is exactly my argument).

If you disagree with me then please explicitly show how one can use a finite amount of R members in order to conclude something about a non-finite collection of all R members of some interval.


EDIT:




Following the wiki quotes above, we get the follows:

1) Upper bound is an axiom under real analysis (it is a categorical determination that must be accepted or rejected).

2) jsfisher accepts this axiom, and avoids any research about its validity.

3) doronshadmi rejects this axiom because he researches its validity and claims that this axiom is based on a notion taken from a finite case and used on a non-finite case, which is something that cannot be done.

4) In this case, jsfisher has to show that it is valid to take a finite case and use it also in a non-finite case.



Let us focused on "greater than OR equal to ..." because (as I get it) this is the heart of this discussion, as follows:

Let us focused only on Standard Math (OM is not used at this part of the discussion).

By using only Standard Math, we follow your own determinations along the discussion, as follows:

1) By standard Math, any interval of non-finite R members is (as jsfisher says) "completely filled" (all R members of that interval are included, without any exceptions (otherwise the interval is incomplete)) (here http://www.internationalskeptics.com/forums/showpost.php?p=4749909&postcount=3161 we can define your own determination to "completely filled", which is related to this case).

2) By Standard Math "completely filled" of a non-finite collection AND no gap (where "no gap" by Standard Math means "there is a room for more R members") is a contradiction, because something can't be (R "completely filled") AND (enables a room for more R members).


Now, let us examine again why Standard Math is derived to this contradiction.

Standard Math is derived to this contradiction, because it tries to understand the non-finite by notions and examples that are based on the finite.

Let us see how Standard Math doing it:

First, we shell examine the case of two integers a and b, such that a is an immediate predecessor of b.

It is easy to get that a is an immediate predecessor of b, because we deal here with a finite case of distinct a and distinct b (this is exactly the construction of the integers).

Now let us examine the case of R members.

By using a finite amount of R members, it is easy to show that there is a room for more R members between distinct a and distinct b.

This room exists exactly because we are using a finite amount of R members, and we cannot conclude that this room still holds when we deal with the non-finite case of all R members.

This is exactly where Standard Math fails in its own framework, because what can be concluded about integers (by using a finite case) cannot be concluded about R (by using a finite case).

Again your misunderstanding is about the number of members in a set representing an interval in the real numbers. As I and others have said before their would be an infinite number of members in that set. Again if you think otherwise then please list all the members of a set representing the interval (1,2) in the real numbers. You seem to have completely ignored when I challenged you to do that before and explained the fact that a finite interval like (1,2) in the reals has an infinite number of members. Again the term ‘finite’ in ‘finite interval’ refers to its expanse (meaning the difference between the boundaries) not the number of elements in that interval. Having been explained to you so many times you seem determined to ignore the infinite aspect of a finite interval in the reals in order to profess your failure in understanding as some failure of math in general. The interval [1,2] is a finite interval in the reals, the rationales and the integers because it has a finite expanse or finite difference between the boundaries. In the reals and the rationales there would be a infinite number of elements in that interval, however in the integers there would be only two members and thus a finite number of members. Your whole ‘argument’ is based on this misunderstanding and conflation of the applicability of the word ‘finite’ to one aspect, the difference between the boundaries of an interval, and another aspect, the number of members in that interval, the former can be finite when the latter is infinite. However the reverse is not true if the interval has an infinite expanse it must also have a infinite number of members even in the integers. A finite interval in the integers always has a finite number of members while a finite interval in the reals or rationals always has a infinite number of members.

 

[doronshadmi]

Again your misunderstanding is about the number of members in a set representing an interval in the real numbers. As I and others have said before their would be an infinite number of members in that set. Again if you think otherwise then please list all the members of a set representing the interval (1,2) in the real numbers. You seem to have completely ignored when I challenged you to do that before and explained the fact that a finite interval like (1,2) in the reals has an infinite number of members. Again the term ‘finite’ in ‘finite interval’ refers to its expanse (meaning the difference between the boundaries) not the number of elements in that interval. Having been explained to you so many times you seem determined to ignore the infinite aspect of a finite interval in the reals in order to profess your failure in understanding as some failure of math in general. The interval [1,2] is a finite interval in the reals, the rationales and the integers because it has a finite expanse or finite difference between the boundaries. In the reals and the rationales there would be a infinite number of elements in that interval, however in the integers there would be only two members and thus a finite number of members. Your whole ‘argument’ is based on this misunderstanding and conflation of the applicability of the word ‘finite’ to one aspect, the difference between the boundaries of an interval, and another aspect, the number of members in that interval, the former can be finite when the latter is infinite. However the reverse is not true if the interval has an infinite expanse it must also have a infinite number of members even in the integers. A finite interval in the integers always has a finite number of members while a finite interval in the reals or rationals always has a infinite number of members.

You did not get my previous post.

My argument is very simple:

You cannot use a finite amount of R members, in order to say any meaningful thing about a collection of all R members (and if we use jsfisher's phrase then the space between any two R numbers is completely filled by other R numbers (leaving no gaps), and we cannot use a finite amount of R members, to say some meaningful thing, in this case).

EDIT:

Please look at http://www.internationalskeptics.com/forums/showpost.php?p=4752487&postcount=3190 , I think it can help to understand my argument.

 

[jsfisher]

You cannot use a finite amount of R members

Exactly what "finite amount of R members" are you referring to? Surely not the two numbers that establish the bounds for an interval. After all, the interval includes an infinite supply of numbers regardless of which two distinct bounds are chosen.

...in order to say any meaningful thing about a collection of all R members

It is not a collection of all R members, just the ones between the two bounds.


Be this all as it may, your statement is false anyway. For example, {X : 0 < X < 2} uses two real numbers (0 and 2) to say something meaningful about all real numbers (and I do mean all real numbers).

 

[doronshadmi]

Ok jsfisher it is about time to expose your profile about this discussion, which is based only on the technical character of definitions, and not on the will to understand the fundamental notions that stand at the basis of any formal definition (I wish to add that this post is related only to Standard Math).

Again, we are in a philosophical forum that uses questions like "what is this?" , "why it is?" or "how it is?" etc. ... about fundamental concepts, where your basic approach is limited to "how to define and\or use?" questions.

Here is your basic attitude about this dialog:

In http://www.internationalskeptics.com/forums/showpost.php?p=4746256&postcount=3103 your reply is:

It is obvious that we can't used notions taken from a finite collection and use them in order to conclude something about a non-finite collection.

It is neither obvious nor true. This is just another baseless allegation offered without evidence. It is a fantasy of doronetics, formed out of ignorance, inconsistency, and contradiction.

Doron, you really need to stop just making stuff up.

No details are provided by you, in order to support your claim that what I say "is neither obvious nor true" or " It is a fantasy of doronetics" , etc …


In http://www.internationalskeptics.com/forums/showpost.php?p=4748865&postcount=3148 your reply is:

...that cannot be found by using any particuler and finite case of pair of R members, as used in your "proof".

Now, this part is just illogical and silly. If it exists, then we can give it a name and explore its properties. If not, well, then this is just another Doron fantasy.

Also in this case no details are provided by you, in order to support your claim (also in this case your basic approach is limited to "how to define and\or use?" questions)

Your claim in this case is limited to the "how to define and\or use?" question as follows:

"If it exists, then we can give it a name and explore its properties.", and you don't ask questions like "how it exists?" , "what is this?" , "why it exists like this?" .

You really believe that "how to define and\or use?" questions really give you all the needed knowledge in order to successfully explore something that you never bothered to understand by other questions.

As a result you have no idea with what you deal, because only "how to define and\or use?" questions are not sufficient enough if you deal with the non-finite.

Furthermore, you also provide the reason of why to avoid any further research of fundamental mathematical notions by saying (please be aware that I used the word "define" and not "understand", in order to communicate with you):

I wrote "your determination" , exactly because you are using words like "complete" without define them.

I didn't use any word like complete. And there is no reason for me to define my terms when the usage is in complete agreement with accepted definitions.

You used the phrase "completely filled" or "leaving no gaps" instead of complete, but this is not the main point here.

The main point here that you explicitly say:

"And there is no reason for me to define my terms when the usage is in complete agreement with accepted definitions."

So jsfisher, you have no motivation to understand with what you deal, and you limit your mathematical activity to the agreed and accepted answers, that were given by others that used only "how to define and\or use?" questions.

As a result you do not understand, for example, this fundamental question:

There is no immediate predecessor because there is always another real number between any two real numbers.

Do you understand that you can show that "there is always another real number between any two real numbers" exactly because you always based on a finite case?

Here is your last phrase:

Be this all as it may, your statement is false anyway. For example, {X : 0 < X < 2} uses two real numbers (0 and 2) to say something meaningful about all real numbers (and I do mean all real numbers).

that clearly shows that you do not ask the needed questions in order to really understand the meaning of words like "all" that are used by you here, but you simply do not get them like any person that its framework is limited only to "how to define and\or use?" questions (your last phrase shows again that you don't understand the question that I asked you above (the bolded one)).

By the way, I already covered your last phrase by this answer:

X and Y of [X,Y] of what is called "finite interval" do not give any knowledge (accept the title "finite interval") of what is really going in the "completely filled" collection of all R members, of what is called "finite interval".

The best we can do is to use a finite collection of R members, and say that any arbitrary R member that is not X or Y is > X and < Y.

And I wish to add, that any exploration that is based only on "how to define and\or use?" questions cannot understand the non-finite.

 

[jsfisher]

Ok jsfisher it is about time to expose your profile about this discussion....

Here is the statement at issues:

It is obvious that we can't used notions taken from a finite collection and use them in order to conclude something about a non-finite collection.

You must mean something by it, because you continually repeat it, but it is not at all clear what you mean. A couple of us have asked you a couple of times in a couple of different ways what you mean to no avail.

If I recall correctly, it first arose as an objection to my proof the set of reals {X : X<Y} for some number, Y, has no largest element. You posed your statement as a summary dismissal of the proof after your several other attempts to discredit it had failed.

Yes, {X : X<Y} is an infinite set determined by a single value (single = finite number of). Ok, so what? And what do you mean by your statement relative to this set?

Given just that one value, Y, I can make determinations about an infinite number of reals, namely whether each is a member of the set, above.

Is that what you are saying is impossible? Clearly it is not impossible, so what are you really saying?

I can prove the formula N(N+1)/2 gives the sum of the integers between 1 and N using just two example cases. I first show that it is true for the case N=1, then I show that if it is true for N=X, then it must be true for N=X+1.

Is that a case of a finite collection (two cases) being used to draw a conclusion about an infinite collection (the set of all integers)? The proof is perfectly valid, so what are you really saying?

 

[doronshadmi]

Here is the statement at issues:



You must mean something by it, because you continually repeat it, but it is not at all clear what you mean. A couple of us have asked you a couple of times in a couple of different ways what you mean to no avail.

If I recall correctly, it first arose as an objection to my proof the set of reals {X : X<Y} for some number, Y, has no largest element. You posed your statement as a summary dismissal of the proof after your several other attempts to discredit it had failed.

Yes, {X : X<Y} is an infinite set determined by a single value (single = finite number of). Ok, so what? And what do you mean by your statement relative to this set?

Given just that one value, Y, I can make determinations about an infinite number of reals, namely whether each is a member of the set, above.

Is that what you are saying is impossible? Clearly it is not impossible, so what are you really saying?

I can prove the formula N(N+1)/2 gives the sum of the integers between 1 and N using just two example cases. I first show that it is true for the case N=1, then I show that if it is true for N=X, then it must be true for N=X+1.

Is that a case of a finite collection (two cases) being used to draw a conclusion about an infinite collection (the set of all integers)? The proof is perfectly valid, so what are you really saying?

Jsfisher,

You are talking about a proof by induction.

It is valid with integers.

It is invalid with reals.

Again, I am talking here only about Standard Math (OM is not used, in this case).

Is that what you are saying is impossible? Clearly it is not impossible, so what are you really saying?

Jsfisher, please, please first try to answer very carefully to the following question (it is essential to the discussed subject):

Jsfisher, do you understand that you can show that "there is always another real number between any two real numbers" exactly because you are always based on a finite case?

 

[jsfisher]

Jsfisher,

You are talking about a proof by induction.

It is valid with integers.

It is invalid with reals.

Don't move the goal posts. There is nothing in your declaration that restricts it to the reals. Here is your statement again. Notice all it mentions are finite collections and infinite collections; there are no restrictions or exclusions other than that:

It is obvious that we can't used notions taken from a finite collection and use them in order to conclude something about a non-finite collection.

I have also now given you several examples which contradict your statement. The statement is clearly bogus, so either try again to say what you really mean, clearly and succinctly, or just retract the whole thing.

 

[The Man]

You did not get my previous post.

My argument is very simple:

You cannot use a finite amount of R members, in order to say any meaningful thing about a collection of all R members (and if we use jsfisher's phrase then the space between any two R numbers is completely filled by other R numbers (leaving no gaps), and we cannot use a finite amount of R members, to say some meaningful thing, in this case).

EDIT:

Please look at http://www.internationalskeptics.com/forums/showpost.php?p=4752487&postcount=3190 , I think it can help to understand my argument.

Again you simply do not understand the invalid assertion that opens your argument. As has been repeatedly explained the interval [1,2] in the reals is not “a finite amount of R members”. Again if you think it is then list all those members or simply answer jsfishers direct question.

Exactly what "finite amount of R members" are you referring to? Surely not the two numbers that establish the bounds for an interval. After all, the interval includes an infinite supply of numbers regardless of which two distinct bounds are chosen.

Ok jsfisher it is about time to expose your profile about this discussion, which is based only on the technical character of definitions, and not on the will to understand the fundamental notions that stand at the basis of any formal definition (I wish to add that this post is related only to Standard Math).

Again, we are in a philosophical forum that uses questions like "what is this?" , "why it is?" or "how it is?" etc. ... about fundamental concepts, where your basic approach is limited to "how to define and\or use?" questions.

Instead of answering this simple and direct question you obfuscate with a dissertation about the “will to understand the fundamental notions that stand at the basis of any formal definition” while apparently lacking the will to clearly explain what “finite amount of R members” you are talking about.

So to put it in the format you are requesting above; “what is this” “finite amount of R members” you are referring to, “why it is” “a finite amount of R members” and not the infinite amount of members in any interval of the reals and "how it is” being used “in order to say any meaningful thing about a collection of all R members”? Only you can explain yourself and what you mean by “a finite amount of R members”, no amount of philosophical rambling, finite or infinite, can be substituted for a simple and direct explanation by you as to what “finite amount of R members” you are referring to.