Deeper than primes

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The definitions given do not violate any tenants of standard math.




Any interval of the reals contains some element or elements that other portions of that interval are greater then and less then thus and interval is non-local by your ascriptions, since it dose represent a line segment. Are you now claiming that a line segment is not your non-local element?



Done, that definition you cited was specifically about the reals, if you are insisting on an example no problem.

In the reals the interval (-∞,1) is the immediate and no-local predecessor, while the interval (1, ∞) is the immediate non-local successor of the local and finite value 1.

Nonsense.

-∞ or ∞ are not real numbers (do not forget that we are talking here only about standard real analysis).
 
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What do you think [3, 5) means?
Good question.

The answer is (according to Standard Math):

[3,5) is an ordered collection of infinitely many R members, where this collection does not have the largest element.

As a result 5 (that is greater than any element of this collection) is not an immediate successor of this collection exactly because
(5 is not an element of this collection) AND (this collection does not have the largest element).
 
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Nonsense.

-∞ or ∞ are not real numbers (do not forget that we are talking here only about standard real analysis).

Do not forget that (-∞, ∞) is an interval of all real numbers. Don't like them then replace them with some numbers more to your preferance.
 
Good question.

The answer is (according to Standard Math):

[3,5) is an ordered collection of infinitely many R members, where this collection does not have the largest element.

As a result 5 (that is greater than any element of this collection) is not an immediate successor of this collection exactly because
(5 is not an element of this collection) AND (this collection does not have the largest element).

5 is the immediate successor of this collection exactly becouse it is the upper boundary of that caollection AND "is not an element of this collection".
 
So each element of B is a successor of A.

But no element of B is an immediate successor of A.

For example:

Interval A = [3,5]

Interval B = any real number > 5

No number of interval B is an immediate successor of A.

There is no requirement any number in the interval B be an immediate successor of any number in interval A. There's only two conditions that need to be satisfied:

1. A < B, and
2. There is no C such that A < C < B.

For A = [3, 5] and B = (5, +inf), both conditions are met, so B is an immediate successor to A.
 
Doron,

Since we have told you what we mean by successor with respect to intervals and what we mean by immediate successor, and since you seem to be using those terms completely differently, I think is is well past time you tell us your usage.

What do you mean by successor with respect to real intervals?

What do you mean by immediate successor?

It would be great if you could provide this information in a precise, comprehensive way, for example, using first-order predicates.
 
Nonsense.

-∞ or ∞ are not real numbers (do not forget that we are talking here only about standard real analysis).

Oh, geez. You aren't even acquainted with standard interval notation. [K, ∞) is a perfectly fine interval of all the numbers <= K.
 
There is no requirement any number in the interval B be an immediate successor of any number in interval A. There's only two conditions that need to be satisfied:

1. A < B, and
2. There is no C such that A < C < B.

For A = [3, 5] and B = (5, +inf), both conditions are met, so B is an immediate successor to A.

Pure nonsense.

< has a meaning only if it is used between things that can be compared with each other.

[3,5) cannot be compared with 5, unless the comparison is between 5 and some element of [3,5).

Do you want to play, so let us play this game:

[3,5) = [3,…[5,…

As we see [3,… is an open interval of real numbers that has no largest element.

Following [3,… there is [5,… that is another interval of real numbers that has no largest element.

No number of [5,… is an immediate successor of [3,… because [3,… does not have the largest element.

End of game.


EDIT:


jsfisher Please avoid your twisted games (now you enter to the game A and B intervals).

In http://www.internationalskeptics.com/forums/showpost.php?p=4721582&postcount=2864 you clearly say:

"Y is in fact an immediate successor to [X,Y)".

Well, Y (where Y is a real number) is not an immediate successor to [X,Y) interval , as clearly shown in the example above.

(Also you can't use "<" relation between A and B intervals, because any use of "<" relation has a meaning only if it used between the elements of A and B intervals).
 
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Doron,

Since we have told you what we mean by successor with respect to intervals and what we mean by immediate successor, and since you seem to be using those terms completely differently, I think is is well past time you tell us your usage.

What do you mean by successor with respect to real intervals?

What do you mean by immediate successor?

It would be great if you could provide this information in a precise, comprehensive way, for example, using first-order predicates.

Nothing is different here, immediate successor has a one and only one meaning in Standard Math which is:

If a and b are two whole numbers and b immediately follows a such that a<b, then b is called the immediate successor of a.

(you can't take a and b as two intervals, because any use of "<" relation has a meaning only if it used between the elements of a and b intervals).

Any other use of the term 'immediate successor' under Standard Math is a load of crap.

Please see http://www.internationalskeptics.com/forums/showpost.php?p=4790953&postcount=3530 for more details, about this subject.
 
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Do not forget that (-∞, ∞) is an interval of all real numbers. Don't like them then replace them with some numbers more to your preferance.
So what if (-∞, ∞) is the interval of all real numbers?

We are taliking about the relation "<" that has a meaning only if it used between the numbers of this interval.
 
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Prove it.

Did you read the following?
You seem to want to totally ignore the interval in favor of its elements. I, on the other hand, ignore neither the interval nor its membership. The object under consideration is the interval, but its properties are shaped by its elements.

Here's what I (and everyone else here) mean by "interval A precedes interval B". What do you mean?

[latex]$$ (A \prec B) \, \Leftrightarrow \, (\forall x \, \forall y \, (x \in A \wedge y \in B) \Rightarrow (x \prec y)) $$[/latex]​
 
Did you read the following?

You know what ?

If we wish to extend the use of "<" relation beyond numbers, then this relation must be used between things of the same type.

For example: if A and B are intervals, then B is indeed the immediate successor of A, as long as we ignore A and B elements.

But jsfisher claims that Y (which is a real number) is an immediate successor of [X,Y) interval.

By doing this, jsfisher mixes between [X,… (which is an interval that has no largest element) and Y, where Y is not another interval but it is the first element of [Y,… interval, where [Y,… interval is not Y element.

A = [X,…

B = [Y,…

B immediately follows A, and B is an immediate successor of A exactly because A and B are of the same type (both of them are intervals, in this case).

This is defiantly not the case in [3,5) < 5, where two different types are mixed with each other, and as a result "<" relation has no meaning in "[3,5) < 5" gibberish expression.


EDIT:


In http://www.internationalskeptics.com/forums/showpost.php?p=4721582&postcount=2864 jsfisher clearly says:

"Y is in fact an immediate successor to [X,Y)".

By doing that he mixes between an interval and an element of an interval, and the result is an utter gibberish.
 
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You know what ?

If we wish to extend the use "<" relation beyond numbers, then this relation must be used between things of the same type.

For example: if A and B are intervals, then B is indeed the immediate successor of A, as long as we ignore A and B elements.

But jsfisher claims that Y (which is a real number) is an immediate successor of [X,Y) interval.

By doing this, jsfisher mixes between [X,… (which is an interval that has no largest element) and Y, where Y is not another interval but it is the first element of [Y,… interval, where [Y,… interval is not Y element.

A = [X,…

B = [Y,…

B immediately follows A, and B is an immediate successor of A exactly because A and B are of the same type (both of them are intervals, in this case).

This is defiantly not the case in [3,5) < 5, where two different types are mixed with each other, and as a result "<" relation has no meaning in "[3,5) < 5" gibberish expression.


EDIT:


In http://www.internationalskeptics.com/forums/showpost.php?p=4721582&postcount=2864 jsfisher clearly says:

"Y is in fact an immediate successor to [X,Y)".

By doing that he mixes between an interval and an element of an interval, and the result is an utter gibberish.



So, did you read it but not understand it, or not read it?

ETA:
Do you think jsfisher is making this up as he goes along? There is only one person in this thread doing that.
 
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So, did you read it but not understand it, or not read it?

ETA:
Do you think jsfisher is making this up as he goes along? There is only one person in this thread doing that.

No, you quated it without undertand it.

Now let us see how you explain it in details, since you are the one who used it in order to support your claims.

Please do it, the stage is yours, I am wating ...
 
Write what down?

This:


[latex]$$ (A \prec B) \, \Leftrightarrow \, (\forall x \, \forall y \, (x \in A \wedge y \in B) \Rightarrow (x \prec y)) $$[/latex]​


Write it in plain English, 10 minutes are over, you have only 10 minutes to do that, before I do thet.
 
This:


[latex]$$ (A \prec B) \, \Leftrightarrow \, (\forall x \, \forall y \, (x \in A \wedge y \in B) \Rightarrow (x \prec y)) $$[/latex]​


Write it in plain English, 10 minutes are over, you have only 10 minutes to do that, before I do thet.

A is less than B if for all x and y, where x is a member of A and y is a member of B, x is less than y.


Where's your version?
 
You seem to want to totally ignore the interval in favor of its elements. I, on the other hand, ignore neither the interval nor its membership. The object under consideration is the interval, but its properties are shaped by its elements.

Here's what I (and everyone else here) mean by "interval A precedes interval B". What do you mean?

[latex]$$ (A \prec B) \, \Leftrightarrow \, (\forall x \, \forall y \, (x \in A \wedge y \in B) \Rightarrow (x \prec y)) $$[/latex]​

This is quite pathetic jsfisher, you are using any possible trick in order to not admit that you made a fundamental mistake by claim that Y number is an immediate successor of [X,Y) interval.

Furthermore, let us write
[latex]$$ (A \prec B) \, \Leftrightarrow \, (\forall x \, \forall y \, (x \in A \wedge y \in B) \Rightarrow (x \prec y)) $$[/latex]​
in simple English and show how you are using the elements of A and B in order to show that "interval A precedes interval B".

(A < B) iff (for all x and for all y (such that x is a member of B AND y is a member of B) implies (x < y))

It is clearly seen that A<B because we compare between each element of A (called x) and each element of B (called y).

So A<B cannot be found in this expression, independently of the elements of A and B.

Also in your expression above there is no mixing between interval and an element of an interval, as you do here:

In http://www.internationalskeptics.com/forums/showpost.php?p=4721582&postcount=2864 you clearly say:

"Y is in fact an immediate successor to [X,Y)".

Since [X,Y) is an interval and Y is an element of an interval, then "Y is in fact an immediate successor to [X,Y)" is an utter gibberish, because you mix between different types (between an interval and an element of an interval, in this case).

EDIT:

Since you show that A<B expression depends of the elements of A (called x) and the elements of B (called y), and since no y is an immediate successor of any x, then (by following this reasoning) you cannot use
[latex]$$ (A \prec B) \, \Leftrightarrow \, (\forall x \, \forall y \, (x \in A \wedge y \in B) \Rightarrow (x \prec y)) $$[/latex]​
in order to claim that B is an immediate successor of A, because by your expression above A<B has a meaning only according the relations of A and B elements (notated as x and y, where no y element of interval B is an immediate successor of any x element of interval A).
 
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A is less than B if for all x and y, where x is a member of A and y is a member of B, x is less than y.


Where's your version?

This is wrong, it has to be:

A is less than B iff (if and only if) for all x and y, where x is a member of A and y is a member of B, x is less than y.

EDIT:

But more importent, you do not understand what follows from this expression, about "[3,5) < 5" gibberish expression, as clearly explained in http://www.internationalskeptics.com/forums/showpost.php?p=4791330&postcount=3547 .
 
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Nothing is different here, immediate successor has a one and only one meaning in Standard Math which is:

If a and b are two whole numbers and b immediately follows a such that a<b, then b is called the immediate successor of a.

What ever gave that idea? There is nothing to limit the immediate successor construct to whole numbers.

(you can't take a and b as two intervals, because any use of "<" relation has a meaning only if it used between the elements of a and b intervals).

This is another misunderstanding on your part. Nothing restricts the ordering relation (symbolized by "<") to be just the conventional numeric less-than comparator. Moreover, I have been very clear what I am taking to be the ordering relation, so any confusion is of your own manufacturing.

Any other use of the term 'immediate successor' under Standard Math is a load of crap.

Ok, so you are failing at Order Theory, too. What branch will you be failing at next?
 
What ever gave that idea? There is nothing to limit the immediate successor construct to whole numbers.



This is another misunderstanding on your part. Nothing restricts the ordering relation (symbolized by "<") to be just the conventional numeric less-than comparator. Moreover, I have been very clear what I am taking to be the ordering relation, so any confusion is of your own manufacturing.



Ok, so you are failing at Order Theory, too. What branch will you be failing at next?

The revevant posts for this case are

http://www.internationalskeptics.com/forums/showpost.php?p=4791128&postcount=3537

and

http://www.internationalskeptics.com/forums/showpost.php?p=4791330&postcount=3547

Please reply to them.
 
In http://www.internationalskeptics.com/forums/showpost.php?p=4721582&postcount=2864 jsfisher clearly says:

"Y is in fact an immediate successor to [X,Y)".

By doing that he mixes between an interval and an element of an interval, and the result is an utter gibberish.

And I have explained how the ordering would apply to a domain of real numbers and real intervals combined. It would seem every has been able to follow my explanation except you, Doron.

No matter. I can restrict it entirely to intervals, still consistent with the already-presented ordering relation:

[Y, Y] is an immediate successor to [X, Y).

Is that better for you, Doron?
 
This is wrong, it has to be:

A is less than B iff (if and only if) for all x and y, where x is a member of A and y is a member of B, x is less than y.
Now explain to me in what way that is significantly different from what I said.
But more importent, you do not understand what follows from this expression, about "[3,5) < 5" gibberish expression, as clearly explained in http://www.internationalskeptics.com/forums/showpost.php?p=4791330&postcount=3547 .
One of us doesn't understand it, that much is clear.
 
EDIT: jsfisher, you are the one who supported [3,5) < 5 in http://www.internationalskeptics.com/forums/showpost.php?p=4788779&postcount=3458 .



And I have explained how the ordering would apply to a domain of real numbers and real intervals combined. It would seem every has been able to follow my explanation except you, Doron.

No matter. I can restrict it entirely to intervals, still consistent with the already-presented ordering relation:

[Y, Y] is an immediate successor to [X, Y).

Is that better for you, Doron?

As long as you ignore the elements of [X,Y) and [Y,Y].

Now please show us how [X,Y)< [Y,Y] by ignoring the elements of [X,Y) and the elements of [Y,Y].
 
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Now explain to me in what way that is significantly different from what I said.

One of us doesn't understand it, that much is clear.

EDIT: iff is = , (A = B)

If is a part of "if,then" , (B implies A)
 
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Now explain to me in what way that is significantly different from what I said.

EDIT: (A < B) = (for all x and for all y (such that x is a member of B AND y is a member of B) implies (x < y))

you wrote:

IF (for all x and for all y (such that x is a member of B AND y is a member of B) implies (x < y)) THEN (A < B)
 
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you wrote:

IF (for all x and for all y (such that x is a member of B AND y is a member of B) implies (x < y)) THEN (A < B)

No, I didn't. I wrote in plain English, as requested.


I also don't think that 'iff' means the same as 'equals'.

ETA:
Regardless, you are quibbling about a trivial point.

What is the successor to [3, 5)?
 
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No, I didn't. I wrote in plain English, as requested.


I also don't think that 'iff' means the same as 'equals'.

You are right, I mean =, and = in plain English is "if and only if".

Let us fix it in the previous posts.
ETA:
Regardless, you are quibbling about a trivial point.

What is the successor to [3, 5)?

Any number that is greater than eny number of [3,... interval.
 
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