Deeper than primes

jsfisher · May 26, 2009

doronshadmi said:
The straightforward meaning of the word "no gaps" is "no interval".

I see you jump from one guessed meaning for the word gap to another. Perhaps you should consult a dictionary, because you are now zero for two trying to get this one right.

The mathematical meaning for no gaps between real numbers is precisely inline with the common English usage: the space between any two numbers is completely filled by other numbers (leaving no gaps).

In that case there is no room for z between x and y, if we deal with the non-finite collection of all members of set X.

Now we are back to you making baseless assertions. There is no immediate predecessor because there is always another real number between any two real numbers.

You have moved from drawing a false conclusion from a false premise to drawing the false conclusion from no premise at all. This is not helping to support your assertion that every real number has an immediate predecessor or an immediate successor.

doronshadmi · May 26, 2009

jsfisher said:
I see you jump from one guessed meaning for the word gap to another. Perhaps you should consult a dictionary, because you are now zero for two trying to get this one right.

The mathematical meaning for no gaps between real numbers is precisely inline with the common English usage: the space between any two numbers is completely filled by other numbers (leaving no gaps).

Completely filled by other numbers (leaving no gaps), means that by Standard Math, no single R member is missing in any collection of all non-finite R members of a given interval.

In other words, Y has an immediate predecessor by Standard Math, and it cannot be defined by a any collection of finite members of that interval, exactly because any finite collection is not the all R members of the given interval.

Once again jsfisher, Standard Math simply guesses that something that is based on a finite collection holds also in the case of a complete non-finite collection (where complete means that all non-finite elements of the given interval are included).

If you think that by using a categorical expression style in some formal mathematical definition makes the difference between right and wrong notions, then you are living in fantasy.

jsfisher said:
Now we are back to you making baseless assertions. There is no immediate predecessor because there is always another real number between any two real numbers.

jsfisher, you are using a baseless assertion taken from a finite collection, and force it on a complete non-finite collection.

This is a fact that you cannot deny, and because of this fact you avoid any detailed reply to http://www.internationalskeptics.com/forums/showpost.php?p=4749611&postcount=3158 .

The Man · May 26, 2009

doronshadmi said:
They are synonyms as long as they are related to the same mathematical case, which deals with what exists (or not) between a collection of all non-finite ordered elements.

EDIT:

As for "no gap" in the case of x < z < y.

If there is really no gap, then x = z = y.

Since we deal with x < z < y, then "no gap" is meaningless in the real sense.

No Doron the word gap is not synonymous with ‘interval’, but it is defined as an ‘empty interval’. There is a distinction, one that you seem to be unable to make with your notions that you claim have distinction as a ‘fist order property’. A gap does require difference between the boundaries of that gap, but it is specifically where something could be between those boundaries and is not. When there are no gaps between those boundaries that extent is said to be continuous, or in other words can continue from one boundary to the other without gaps. When there are gaps the extent is said to be discontinuous or discrete.

The interval (1,2) in the real numbers has no gaps (or is continuous). A set representing that interval would contain an infinite number of elements, each with a finite value greater then 1 and less then 2. However in the integers that same interval is a gap (it is an empty interval or would result in the empty set) as both of those discrete and discontinuous boundaries (1 and 2) are excluded and there are no integers between those boundaries.

The Man · May 26, 2009

doronshadmi said:
Completely filled by other numbers (leaving no gaps), means that by Standard Math, no single R member is missing in any collection of all non-finite R members of a given interval.

In other words, Y has an immediate predecessor by Standard Math, and it cannot be defined by a any collection of finite members of that interval, exactly because any finite collection is not the all R members of the given interval.

Once again jsfisher, Standard Math simply guesses that something that is based on a finite collection holds also in the case of a complete non-finite collection (where complete means that all non-finite elements of the given interval are included).

If you think that by using a categorical expression style in some formal mathematical definition makes the difference between right and wrong notions, then you are living in fantasy.

What the heck is a “non-finite R member” or “an immediate predecessor”, these are your expressions Doron you need to clearly define them first before you can claim anything about them other then they simply lack definition.

doronshadmi said:
In other words, Y has an immediate predecessor by Standard Math, and it cannot be defined by a any collection of finite members of that interval, exactly because any finite collection is not the all R members of the given interval.

Have you been missing the parts where we have been telling you that a set or collection representing an interval in the real numbers would not have a finite number of members? The members themselves would be finite (meaning each having a finite value) but there would be an infinite number of them in that collection.

jsfisher · May 26, 2009

doronshadmi said:
Completely filled by other numbers (leaving no gaps), means that by Standard Math, no single R member is missing in any collection of all non-finite R members of a given interval.

No, it doesn't mean that at all. Why would you say such a thing? It is trivially false.

In other words, Y has an immediate predecessor by Standard Math

Not that that actually follows from your premise, but since your premise is false, your conclusion has no merit.

...
jsfisher, you are using a baseless assertion taken from a finite collection, and force it on a complete non-finite collection.

Yawn. You are the one with the baseless assertion that all real numbers have immediate predecessors. Care to try again to give your assertion some basis?

This is a fact that you cannot deny.

It is neither a fact nor impervious to denial.

doronshadmi · May 26, 2009

The Man said:
No Doron the word gap is not synonymous with ‘interval’, but it is defined as an ‘empty interval’. There is a distinction, one that you seem to be unable to make with your notions that you claim have distinction as a ‘fist order property’. A gap does require difference between the boundaries of that gap, but it is specifically where something could be between those boundaries and is not. When there are no gaps between those boundaries that extent is said to be continuous, or in other words can continue from one boundary to the other without gaps. When there are gaps the extent is said to be discontinuous or discrete.

The interval (1,2) in the real numbers has no gaps (or is continuous). A set representing that interval would contain an infinite number of elements, each with a finite value greater then 1 and less then 2. However in the integers that same interval is a gap (it is an empty interval or would result in the empty set) as both of those discrete and discontinuous boundaries (1 and 2) are excluded and there are no integers between those boundaries.

(let us use the term "ordered pair" for X<Y)

The Man, thank you for the accurate explanation (given by using Standard Math) about the difference between an empty interval (in the case of an ordered pair of integers) and a non-empty interval (in the case of an ordered pair of Q or R members).

Let us examine it very carefully.

The ordered pair of integers is finite, and because it is finite it is easy to show that there is no another integer between the ordered pair (what you call empty interval).

It is also easy to show that between any finite ordered pair of Q or R members, there are more Q or R members (what you call a non-empty interval), and it is easily shown exactly because we are using a finite amount of members.

But there is no way to use the finite Q or R case, in order to determine something about the non-finite case of all Q or R members of some complete and non-finite interval.

Standard Math is the framework that defines the complete non-finite collection of a given interval, and the same Standard Math tries to determine things about the complete non-finite collection of a given interval, by using terms that are based on a finite collection (which is something that cannot be done).

As a result, there must be an immediate predecessor to Y of [X,Y) or [X,Y] (because the interval is a complete non-finite collection), but it cannot be defined by any finite amount of Q or R members.

In other words, by Standard Math there is an immediate predecessor to Y, but is cannot explicitly be defined or disproved by the same Standard Math, because no finite amount of elements can be used in order to conclude anything about a complete collection of non-finite elements.

Once again:

We are in the same state of Godel's incompleteness theorems, where things must be true but cannot be proved or disproved within the deductive framework that deals with the non-finite.

jsfisher · May 27, 2009

doronshadmi said:
The ordered pair of integers is finite

Finite? That's an odd way to use the term. An ordered pair is a single thing, so, in the sense of "how many", yeah, it is finite.

...and because it is finite it is easy to show that there is no another integer between the ordered pair (what you call empty interval).

Between the ordered pair? What's the difference between an orange? Did you mean to say, "between the members of the ordered pair"?

Double nonsense, at any rate. Your conclusion doesn't follow from the premise. And your conclusion is either gibberish or is patently false.

It is also easy to show that between any finite ordered pair of Q or R members, there are more Q or R members (what you call a non-empty interval), and it is easily shown exactly because we are using a finite amount of members.

Finite ordered pair? You seem hung up on the word, finite, because you use it in so many inappropriate places. Wouldn't it have been far more direct to just say "between any two real numbers there is another real number" and "between any two rational numbers there is another rational number"? All the rest is just complication that serves no purpose.

But there is no way to use the finite Q or R case, in order to determine something about the non-finite case of all Q or R members of some complete and non-finite interval.

Nonsense. You continue to allege things without support. The mere fact an interval can be identified by it's two boundaries (a finite number of boundaries) doesn't prevent conclusions being drawn about it's infinite membership.

Standard Math is....

As demonstrated again and again, standard mathematics is something you know nothing about, so please spare us your misinformed lecture.

As a result, there must be an immediate predecessor to Y of [X,Y) or [X,Y] (because the interval is a complete non-finite collection)

I'll just add "complete" to the list of terms you don't understand. That aside, I see you again are using a irrelevant premise (infinite collection) to reach a false conclusion (existence of immediate predecessors).

...but it can never be defined by any finite amount of Q or R members.

Are you really saying an immediate predecessor is an infinite collection. How bizarre of you. If an immediate predecessor actually existed in R for some number, I'd need only one thing (a finite number of things) to name it.

doronshadmi · May 27, 2009

jsfisher, in my previous post I explicitly wrote that "ordered pair" means X<Y.

doronshadmi said:
Completely filled by other numbers (leaving no gaps), means that by Standard Math, no single R member is missing in any collection of all non-finite R members of a given interval.

jsfisher said:

No, it doesn't mean that at all. Why would you say such a thing? It is trivially false.

Click to expand...

jsfisher said:
I will use a simple proof by contradiction. As with all such proofs, it begins with an assumption then proceeds to construct a contradiction, thereby showing the assumption to be false.

Assume the set {X : X<Y} does have a largest element, Z.

For Z to be an element of the set, Z < Y.
Let h be any element of the interval (Z,Y).
By the construction of h, Z < h < Y.
Since h < Y, h is an element of the set {X : X<Y}.
Since Z < h, the assumption Z was the largest element of the set has been contradicted.

Therefore, the set {X : X<Y} does not have a largest element.

If your interval is [Z,Y) such that Y is not the least upper bound of [X,Z] then Z is the largest member of [X,Z] but it is not the immediate predecessor of Y.

If your interval is [Z,Y) such that Y is the least upper bound of [X,Z] then Z is the largest member of [X,Z] and the immediate predecessor of Y, and you cannot define Z by using a finite amount of R members.

doronshadmi · May 27, 2009

jsfisher said:
Finite? That's an odd way to use the term. An ordered pair is a single thing, so, in the sense of "how many", yeah, it is finite.

Between the ordered pair? What's the difference between an orange? Did you mean to say, "between the members of the ordered pair"?

Double nonsense, at any rate. Your conclusion doesn't follow from the premise. And your conclusion is either gibberish or is patently false.

Finite ordered pair? You seem hung up on the word, finite, because you use it in so many inappropriate places. Wouldn't it have been far more direct to just say "between any two real numbers there is another real number" and "between any two rational numbers there is another rational number"? All the rest is just complication that serves no purpose.

Nonsense. You continue to allege things without support. The mere fact an interval can be identified by it's two boundaries (a finite number of boundaries) doesn't prevent conclusions being drawn about it's infinite membership.

As demonstrated again and again, standard mathematics is something you know nothing about, so please spare us your misinformed lecture.

I'll just add "complete" to the list of terms you don't understand. That aside, I see you again are using a irrelevant premise (infinite collection) to reach a false conclusion (existence of immediate predecessors).

Are you really saying an immediate predecessor is an infinite collection. How bizarre of you. If an immediate predecessor actually existed in R for some number, I'd need only one thing (a finite number of things) to name it.

You did not understand this post.

Please refreash your screen and read all of it, before you air your view on any part of it.

If you do not read all of it before you you air your view on any part of it, I promise you that you again and again going to not get it.

jsfisher · May 27, 2009

doronshadmi said:
jsfisher, in my previous post I explicitly wrote that "ordered pair" means X<Y.

(a) You added it after the fact in one of your legendary post edits.
(b) So what? What did I write that didn't take that into account?

If your interval is [Z,Y)

It wasn't.

...such that Y is not the least upper bound of [X,Z]

It wasn't, and what's with this X?

...then Z is the largest member of [X,Z] but it is not the immediate predecessor of Y.

Since the premise is false, the conclusion is irrelevant.

If your interval is [Z,Y)

Still wasn't.

...such that Y is the least upper bound of [X,Z]

Still wasn't

...then Z is the largest member of [X,Z] and the immediate predecessor of Y

Still a false premise trailed by an irrelevant conclusion.

...and you cannot define Z by using a finite amount of R members.

One real number should be sufficient. That's a finite number of members, right?

jsfisher · May 27, 2009

doronshadmi said:
You did not understand this post.

And you evade mine.

The Man · May 27, 2009

doronshadmi said:
(let us use the term "ordered pair" for X<Y)

The Man, thank you for the accurate explanation (given by using Standard Math) about the difference between an empty interval (in the case of an ordered pair of integers) and a non-empty interval (in the case of an ordered pair of Q or R members).

No problem, but if you actually understood the ‘standard math’ you are claiming to supplant I would not have to explain it to you.

doronshadmi said:
Let us examine it very carefully.

Well let’s be more careful in that examination then your usual ‘research’.

doronshadmi said:
The ordered pair of integers is finite, and because it is finite it is easy to show that there is no another integer between the ordered pair (what you call empty interval).

The interval [1,3] in the integers would meet your qualification of an “ordered pair” as 1<3. Although the number of members in a set representing this interval would be finite (just 3 members that we can list as 1, 2 and 3) it is easy to show that there is another integer (2 in this case) between that “ordered” pair. Also this is a finite interval as it spans a finite difference between the boundaries (2 in this case)

doronshadmi said:
It is also easy to show that between any finite ordered pair of Q or R members, there are more Q or R members (what you call a non-empty interval), and it is easily shown exactly because we are using a finite amount of members.

Again the interval [1,3] in the rationals or reals still meets your qualification for an ordered pair. However in this case a set representing that interval would not have a finite number of members. If you think otherwise it should be a simple matter to list those members as in the above example. So again the interval is finite that it spans a finite difference between the boundaries (again 2 in this case), but it certainly does not have a “finite amount of members”, again if you think so then list those members.

doronshadmi said:
But there is no way to use the finite Q or R case, in order to determine something about the non-finite case of all Q or R members of some complete and non-finite interval.

Again a finite interval does not infer or require a finite number of members in that interval it simply refers to the finite expanse (or finite difference between the boundaries) of that interval, this seems to be the fact that you can not accept.

doronshadmi said:
Standard Math is the framework that defines the complete non-finite collection of a given interval, and the same Standard Math tries to determine things about the complete non-finite collection of a given interval, by using terms that are based on a finite collection (which is something that cannot be done).

Well again if it is a “non-finite collection” as you claim above then it is not “a finite collection” as you claim above. The only misuse of terms is yours as usual.

doronshadmi said:
As a result, there must be an immediate predecessor to Y of [X,Y) or [X,Y] (because the interval is a complete non-finite collection), but it cannot be defined by any finite amount of Q or R members.

Well again you have to define what constitutes your ‘immediate predecessor’ in order for anyone (including you) to have any basis to say what meets or does not meet your definition. The [1,3] interval in the integers before we can say that 2 is a predecessor of 3 in that interval and no other integer precedes 3 that is greater then 2. However that claim does not hold when that interval includes the rationals or reals.

doronshadmi said:
In other words, by Standard Math there is an immediate predecessor to Y, but is cannot explicitly be defined or disproved by the same Standard Math, because no finite amount of elements can be used in order to conclude anything about a complete collection of non-finite elements.

Once again:

We are in the same state of Godel's incompleteness theorems, where things must be true but cannot be proved or disproved within the deductive framework that deals with the non-finite.

Once again the claim that there is an “immediate predecessor” in an interval of the reals or rationales is your claim so it is incumbent on you to explicitly define what would constitute an “‘immediate predecessor”. You are the only one in the ‘state’ of thinking what you can not prove or disprove “must be true” so much so that you do not seem much concerned with actually trying to do either.

jsfisher · May 27, 2009

Is it worth asking what's the point of this latest tangent? Other than Doron wanting to claim some moral superiority for something he's imaged in his fictitious OM, where is this all going?

On the other hand, will Doron understand that his invented immediate predecessor for some particular real number isn't a real number and therefore isn't an immediate predecessor? And once he wiggles into the shelter of denial over that, will we respond favorably to a question about what's the immediate predecessor to the immediate predecessor?

Train wrecks are such fascinating events.

The Man · May 27, 2009

jsfisher said:
Is it worth asking what's the point of this latest tangent? Other than Doron wanting to claim some moral superiority for something he's imaged in his fictitious OM, where is this all going?

On the other hand, will Doron understand that his invented immediate predecessor for some particular real number isn't a real number and therefore isn't an immediate predecessor? And once he wiggles into the shelter of denial over that, will we respond favorably to a question about what's the immediate predecessor to the immediate predecessor?

Train wrecks are such fascinating events.

Indeed, as observed before with Dorn’s claims that he is ‘bridging’ logic with ethics; moral superiority seems to be one of his goals or fantasies. However, logically and ethically one would need to first understand and communicate effectively ‘standard math’ before one claims to be able to supplant or even argue against that standard. Thus the train wreck continues.

doronshadmi · May 27, 2009

Let us go back to the beginning of the last discussion of immediate successors or predecessors to some R member.

At post http://www.internationalskeptics.com/forums/showpost.php?p=4721582&postcount=2864 jsfisher writes this:

jsfisher said:
Assuming X < Y < Z, and any reasonable definition for "immediate successor", then, like it or not, doron, Y is in fact an immediate successor to [X,Y). So is [Y,Z]. So is ....

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

So jsfisher, by your own words you determine that "Y is in fact an immediate successor to [X,Y)".

Moreover, by your own words you determine that " … it does matter if we deal with [X,Y] or [X,Y)".

Now, at post http://www.internationalskeptics.com/forums/showpost.php?p=4736076&postcount=2974 you determine that "No real number has an immediate predecessor or immediate successor.

These two determinations of yours clearly contradict each other.

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

Jsfisher, (by ignoring the contradiction that is derived from your two determinations above) since you explicitly say (in your first determination above) that Y is an immediate successor of [X,Y] or [X,Y) , then Y is an immediate R member to some another R member, which is not Y.

In this case, and by following your own determination, this another R member (which is not Y) must be the immediate predecessor of Y.

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 complete (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 completeness, which is related to this case).

2) By Standard Math completeness 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 complete) 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 the immediate predecessor of b.

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

Now let us examine the case of Q or R members.

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

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

zooterkin · May 27, 2009

doronshadmi said:
So jsfisher, by your own words you determine that "Y is in fact an immediate successor to [X,Y)".
.

He doesn't determine it, that's what it means. Do you understand the phrase, 'by definition'?

By definition, the successor of [X, Y) is Y.

ETA:

Y is not the immediate successor of [X, Y].

jsfisher · May 27, 2009

doronshadmi said:
Let us go back to the beginning of the last discussion of immediate successors or predecessors to some R member.

At post http://www.internationalskeptics.com/forums/showpost.php?p=4721582&postcount=2864 jsfosher writes this:

So jsfisher, by your own words you determine that "Y is in fact an immediate successor to [X,Y)".

Yes. If the successor concept is extended to include intervals, then Y is an immediate successor. So is [Y,Z] for any Z>Y.

Moreover, by your own words you determine that " … 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.

Now, at post http://www.internationalskeptics.com/forums/showpost.php?p=4736076&postcount=2974 you determine that "No real number has an immediate predecessor or immediate successor.

Yes, exactly right.

These two determinations of yours clearly contradict each other.

Nope. Not in the slightest. Within the context of the real numbers (or the rational numbers), no number has an immediate predecessor or and immediate successor. If you extend that to include intervals as well, then things change.

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

Jsfisher, (by ignoring the contradiction that is derived from your two determinations above)

There is no contradiction, just a lack of comprehension by you.

...since you explicitly say (in your first determination above) that Y is an immediate successor of [X,Y] or [X,Y)

More lack of comprehension by you. I did not ever say Y was a successor, immediate or otherwise, to [X,Y].

...then Y is an immediate R member to some another R member, which is not Y.

No, not true. [X,Y), for example, is not a real number (i.e. not a member of R).

In this case, and by following your own determination, this another R member (which is not Y) must be the immediate predecessor of Y.

Your faulty premises have led you to a faulty conclusion.

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 complete (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 completeness, which is related to this case).

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.

2) By Standard Math completeness 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 complete) AND (enables a room for more R members).

Ok, so not only do you prove, again, you don't understand what complete means, you have also reverted back to misrepresenting the phrase, no gap. You also assert a conclusion that doesn't follow from your premises (even if they weren't faulty).

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

Since you are completely wrong to this point, there is no need to continue.

doronshadmi · May 27, 2009

zooterkin said:
He doesn't determine it, that's what it means. Do you understand the phrase, 'by definition'?

By definition, the successor of [X, Y) is Y.

ETA:

Y is not the immediate successor of [X, Y].

And on what this definiton is based?

zooterkin · May 27, 2009

doronshadmi said:
And on what this definiton is based?

What do you mean, what is it based on? That's simply what the notation [X, Y) means. It refers to the finite interval starting with X, and including everything up to, but not including, Y.

(I despair, really. This is a notation I'd not encountered a week ago, yet here I am explaining it to someone who claims to know conventional maths well enough to state that it doesn't work.)

doronshadmi · May 27, 2009

jsfisher said:
More lack of comprehension by you. I did not ever say Y was a successor, immediate or otherwise, to [X,Y].

doronshadmi said:
Moreover, by your own words you determine that " … it does matter if we deal with [X,Y] or [X,Y)".

doronshadmi said:

At post http://www.internationalskeptics.com/forums/showpost.php?p=4721582&postcount=2864 jsfisher writes this:

jsfisher said:

Assuming X < Y < Z, and any reasonable definition for "immediate successor", then, like it or not, doron, Y is in fact an immediate successor to [X,Y). So is [Y,Z]. So is ....

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

Click to expand...

Click to expand...

jsfisher said:

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

Click to expand...

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

doronshadmi · May 27, 2009

zooterkin said:
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 · May 27, 2009

jsfisher said:
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:

jsfisher said:
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 · May 27, 2009

jsfisher said:
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 · May 27, 2009

doronshadmi said:
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 · May 27, 2009

doronshadmi said:
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 · May 27, 2009

Are you seriously claiming these two statements are inconsistent?

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

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

(Hi-lighting added.)

doronshadmi · May 27, 2009

zooterkin said:
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).

jsfisher said:
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 · May 27, 2009

doronshadmi said:
[...] 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 · May 27, 2009

doronshadmi said:
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 · May 27, 2009

jsfisher said:
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.

jsfisher said:
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"?.

jsfisher said:
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 · May 27, 2009

doronshadmi said:
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 · May 27, 2009

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

jsfisher said:
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:

wiki said:
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.

wiki said:
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:

wiki said:
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 · May 27, 2009

doronshadmi said:
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 · May 27, 2009

The Man said:
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 · May 27, 2009

doronshadmi said:
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 · May 27, 2009

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:

doronshadmi said:
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.

jsfisher said:

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.

Click to expand...

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:

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

jsfisher said:

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.

Click to expand...

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):

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

jsfisher said:

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.

Click to expand...

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:

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

doronshadmi said:

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?

Click to expand...

Here is your last phrase:

jsfisher said:
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:

doronshadmi said:
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 · May 27, 2009

doronshadmi said:
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 · May 28, 2009

jsfisher said:
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).

jsfisher said:
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 · May 28, 2009

doronshadmi said:
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 · May 28, 2009

doronshadmi said:
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.

jsfisher said:
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.

doronshadmi said:
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.

Deeper than primes

ETcorngods survivor

Penultimate Amazing

Unbanned zombie poster

Unbanned zombie poster

ETcorngods survivor

Penultimate Amazing

ETcorngods survivor

Penultimate Amazing

Penultimate Amazing

ETcorngods survivor

ETcorngods survivor

Unbanned zombie poster

ETcorngods survivor

Unbanned zombie poster

Penultimate Amazing

Nitpicking dilettante, Administrator

ETcorngods survivor

Penultimate Amazing

Nitpicking dilettante, Administrator

Penultimate Amazing

Penultimate Amazing

Penultimate Amazing

Penultimate Amazing

Nitpicking dilettante, Administrator

Nitpicking dilettante, Administrator

ETcorngods survivor

Penultimate Amazing

Student

ETcorngods survivor

Penultimate Amazing

ETcorngods survivor

Penultimate Amazing

Unbanned zombie poster

Penultimate Amazing

ETcorngods survivor

Penultimate Amazing

ETcorngods survivor

Penultimate Amazing

ETcorngods survivor

Unbanned zombie poster