Deeper than primes

[doronshadmi]

Sung to the tune of

"AT SEVENTEEN" By Janis Ian





"Of Cardinality"

I learned the truth of cardinality
That mathematicians just can’t see
Among existing things with some magnitude
Of values to which I’ll just allude
The concepts that I never knew
The musing of kindergarten youth
Were taken as demonstrable
Of cardinality I learned the truth...

And those of us with 1-D spaces
Just can’t cover all the places
With just points all alone
Imagining math of my own
Replete with inconsistencies
And repeated vague obscurities
It isn't all it seems, of cardinality...

It’s from Fullness that it all abounds
For any non-empty set around
By some vague non-local property
I’ll claim with all sincerity
The set exists beyond itself
and add to it my imagination’s wealth
With even more obscurity
A linkage called complexity...

So remember those who use a set
And haven’t got that linage yet
From claims lacking quality and dubious integrity
I’ll ascribe to them claims of my own design
And unflinchingly claim a line
Exceeds the points it has, in cardinality...

From Emptiness we still must gain
The explanation that never came
Of the linkage from which it all ensues
You don’t get it? Well you lose
That’s too bad, it just a shame
If I can’t explain, it’s just you I’ll blame
Your direct perception just can’t see
This linkage of complexity…

It’s a fantasy that’s just repeated
An infinite set can’t be completed
With it’s members all alone
Imagining math of my own
Replete with inconsistencies
And repeated vague obscurities
It just isn't anything, of cardinality...

Maybe I am wrong, but it is reasonable that after this poetry, you are going to compose some symphony or even some choreography of a modern dance about this subject, instead of simply reply in details to http://www.internationalskeptics.com/forums/showpost.php?p=6577818&postcount=12392 about this subject, isn't it The Man?

 

[doronshadmi]

Perhaps you can find a simple counter example?

Perhaps you can reply in details to http://www.internationalskeptics.com/forums/showpost.php?p=6577818&postcount=12392 ?

 

[doronshadmi]

Doron,
Here, let's give you a leg-up on that "someday." All you need do is disprove the following statement. Then that nasty old summation won't work any more.

[latex]$$$\displaystyle
\forall {\epsilon > 0} ,\: \exists N \; n > N \Rightarrow \epsilon > \left| 1 - \sum_{i=1}^{n} {1 \over {2^i}} \right|
$$$[/latex]​

Perhaps you can find a simple counter example?

Your reasoning fails right at the beginning because there is no such a thing like "for all epsilon > 0" if "epsilon" represents infinitely many distinct values.

You still do not get that the cardinality of any given infinite collection < ∞, which is the cardinality of "that has no successor" that its minimal form is the non-local property of a 1-dimensional space, which exists simultaneously at and beyond the location of any given 0-dimensional space along it.

 

[The Man]

Maybe I am wrong, but it is reasonable that after this poetry, you are going to compose some symphony or even some choreography of a modern dance about this subject, instead of simply reply in details to http://www.internationalskeptics.com/forums/showpost.php?p=6577818&postcount=12392 about this subject, isn't it The Man?

Yep, as usual you are wrong; too bad you never considered that option before wasting these past decades. Though your latter proposals might actually be more entertaining, since even though you do seem to go out of your way to jazz things up with new ways of showing you’re wrong it still comes down to your same worn out old shtick of you just being wrong because of some, possibly deliberate, misinterpretation or misrepresentation on your part.

Doron rewording your argument as colored segments and some imaginary dialogue doesn’t change its invalidity that even you assert in that very dialoged. You fail right from the start by asserting a claim you agree to be invalid.

“If 1/2+1/4+1/8+1/16+1/32+1/64+… = 1, as you assert, then the most right (Red or Blue) segment must be reduced into a single 0-dimensional space.”

Really? Well keep drawing Doron, when you actually find “the most right (Red or Blue) segment” you be sure to let us know.

Teacher: “Nope, we have been over this before, in order for the current segment being summed to be zero length (a point) the previous sum must already have reached the limit.”

Student: “I agree with you, this reduction is impossible exactly because 1/2+1/4+1/8+1/16+1/32+1/64+… are the most right points of the left segment of any arbitrary convergent pair of Blue\Red Red\blue equal segments, upon infinitely many scale levels”.

So you must not actually agree as you said you did, so why the lie?

Have you yet to learn that insisting on replies never bodes well for you?

 

[jsfisher]

Perhaps you can reply in details to http://www.internationalskeptics.com/forums/showpost.php?p=6577818&postcount=12392 ?

I did. It was back a bit, though, at an earlier repetition by you.

 

[jsfisher]

Your reasoning fails right at the beginning because there is no such a thing like "for all epsilon > 0" if "epsilon" represents infinitely many distinct values.

You continue to try to disprove definitions. Not a good ploy. Despite your resistance, the for-all concept continues to flourish.

You still do not get that the cardinality of any given infinite collection....

No, the failing is completely yours. Cardinality has a definition; you refuse to comprehend it. Don't blame the rest of us just because you have a blind spot in your understanding, especially since it is deliberate.

 

[epix]

[latex]$$$\displaystyle
\forall {\epsilon > 0} ,\: \exists N \; n > N \Rightarrow \epsilon > \left| 1 - \sum_{i=1}^{n} {1 \over {2^i}} \right|
$$$[/latex]​

For all epsilon that are greater than zero and belong to no particular set, there exists N that belongs to no particular set as well, but is greater than n and therefore epsilon is greater than the absolute value of 1 minus the given sum.


Can you post a link to that implication?

 

[epix]

“If 1/2+1/4+1/8+1/16+1/32+1/64+… = 1, as you assert, then the most right (Red or Blue) segment must be reduced into a single 0-dimensional space.”

That's not true. The sum can be expressed as

f = 1 - 1/2ⁿ
where

[lim n → ∞] 1 - 1/2ⁿ = 1

The convergence is basically a case of infinite deceleration and so the sum can reach 1, as some believe, without some nth term equaling zero. What happens after the sum reaches 1 is never mentioned though (that's what you refer to with the 0-dimensional segment argument) -- it is probably absorbed in the formula that proves the sum identical to 1. That's the formula with all intermediate steps missing.

You need to show the imposibility of the most right segment being reduced into a single 0-dimensional space through an equation and not through another verbal assertion, coz math mostly speaks that way.

 

[doronshadmi]

I did. It was back a bit, though, at an earlier repetition by you.

jsfisher http://www.internationalskeptics.com/forums/showpost.php?p=6577818&postcount=12392 is different.

If you get it as a repetition of some previous post, it simply shows that you did not bother to carefully read it, and then reply to each part of it in details.

If you can't do that, we have a concrete proof of your inability to deal with this fine subject.

 

[doronshadmi]

Really? Well keep drawing Doron, when you actually find “the most right (Red or Blue) segment” you be sure to let us know.

So you must not actually agree as you said you did, so why the lie?

Have you yet to learn that insisting on replies never bodes well for you?

It does not work that way The Man.

This time please read all of http://www.internationalskeptics.com/forums/showpost.php?p=6577818&postcount=12392 any try to reply to each part of it, in details.

If you can't do that, we have a concrete proof of your inability to deal with this fine subject.

 

[Astrodude]

As I said, your reasoning is limited to Quantity, by avoiding any reasoning which deals with the foundation that enables Quantity, in the first place.

I think confronting that reasoning put Cantor in a mental institution, though I'm by no means an authority.

Infinitesimals are perfectly legitimate based on Robinson's hyperreal set as long as you use Zermelo-Fraenkel set theory which pretty much everyone does on the blackboard I think.

However, in real life, we can't deal with actual infinity so we just adopt Kroneker's intuitionism and Gauss' views usually. Would you actually ever contemplate slicing a three dimensional sandwich into an infinite number of two dimensional cross sections?

You seem smart, but you're fatigued and bewildered by a Cantorian view. Avoid his fate and adopt Constructivism.

 

[epix]

EDIT:

Student: “I have a proof against the assertion that 1/2+1/4+1/8+1/16+1/32+1/64+… = 1, as follows:”

“First, we express 1/2+1/4+1/8+1/16+1/32+1/64+… etc ... by the following diagram:”

“If 1/2+1/4+1/8+1/16+1/32+1/64+… = 1, as you assert, then the most right (Red or Blue) segment must be reduced into a single 0-dimensional space.”

You have nothing, Doron. Your argument is based on an assumption that there is an instance where the sum equals 1 and since the series is infinite, each term after the identity takes place must equal zero, which is impossible. But your assumption is wrong.

If you convert the whole term into workable format 1/2+1/4+1/8+1/16+1/32+1/64+… = 1 - 1/2ⁿ where n goes to infinity, you may try to find out when the series becomes equal to 1. In other words, there should be n that belongs to N for which

1 - 1/2ⁿ = 1

Such n doesn't exist, coz

IF
1 - 1/2ⁿ = 1
THEN
1/2ⁿ = 0
THEREFORE
1 = 0

and that's false.

To make sure that the contradiction holds, you can actually go ahead and isolate n in the initial equation with this result:

n= log(0)/log(-2)

where log(0) is undefined and therefore n is indeterminable.

The conclusion is simple: there is no n that belongs to N, not even to R for which 1 - 1/2ⁿ = 1.

However, there is this term n → ∞ that cannot substitute any real number in any equation and where n doesn't have a unique numerical equivalent.

 

[laca]

For all epsilon that are greater than zero and belong to no particular set, there exists N that belongs to no particular set as well, but is greater than n and therefore epsilon is greater than the absolute value of 1 minus the given sum.


Can you post a link to that implication?

Well, I'm no mathematician, but here's my take.

[latex]A sequence $a_n$, $a_n\in\Re$ is converging to $A$ if and only if $\forall \epsilon \in \Re, \epsilon>0\ \exists N \in \aleph,\ \forall n>N \Rightarrow |a_n-A| < \epsilon$[/latex]
See here^WP.

[latex]
This holds for $a_n=\sum_{k=1}^{k=n}\frac{1}{2^k}$, since $a_n=1-\frac{1}{2^{n+1}}$ (you yourself pointed this out too), and if we take $N=-\log_2\epsilon$, we have $\forall n > N, 1-a_n=\frac{1}{2^{n+1}}<\epsilon$. This shows that the sum converges to 1.[/latex]
(sorry for the overly long latex text, but mixed html and latex looked even more weird)

My guess is that jsfisher was alluding to doron's incapability to disprove this, which he must do to show that the series does not converge to 1.

 

[doronshadmi]

Well, I'm no mathematician, but here's my take.

[latex]A sequence $a_n$, $a_n\in\Re$ is converging to $A$ if and only if $\forall \epsilon \in \Re, \epsilon>0\ \exists N \in \aleph,\ \forall n>N \Rightarrow |a_n-A| < \epsilon$[/latex]
See here^WP.

[latex]
This holds for $a_n=\sum_{k=1}^{k=n}\frac{1}{2^k}$, since $a_n=1-\frac{1}{2^{n+1}}$ (you yourself pointed this out too), and if we take $N=-\log_2\epsilon$, we have $\forall n > N, 1-a_n=\frac{1}{2^{n+1}}<\epsilon$. This shows that the sum converges to 1.[/latex]
(sorry for the overly long latex text, but mixed html and latex looked even more weird)

My guess is that jsfisher was alluding to doron's incapability to disprove this, which he must do to show that the series does not converge to 1.

No matter how many epsilon > delta are considered, delta > 0, because delta can't be a single point, exactly because 1-dimensional space (expressed as delta, in this case) is irreducible into 0-dimensional space.

As a result 1/2+1/4+1/8+1/16+1/32+1/64+ ... < 1.

Furthermore, persons like The Man and jsfisher have to explicitly demonstrate (by using bottom to top techniques) that two distinct points define a line segment, such that there is no gap between these two points.

According to their reasoning, such a thing is possible, because they claim that a line segment is completely covered by points, which means that there is no gap between these two points along a given line segment.

Let's see how they are formally prove it

 

[laca]

No matter how many epsilon > delta are considered, delta > 0, because delta can't be a single point, exactly because 1-dimensional space (expressed as delta, in this case) is irreducible into 0-dimensional space.

Delta? What delta? Are you now seeing things that aren't there too, doron? You really should have yourself examined.

As a result 1/2+1/4+1/8+1/16+1/32+1/64+ ... < 1

For finite members, yes. For infinite, it's exactly 1. I'm sorry you can't grasp infinity doron, but that's not really our problem, now is it?

 

[doronshadmi]

Delta? What delta?

[latex]$Delta=|a_n-A|[/latex]

For finite members, yes. For infinite, it's exactly 1. I'm sorry you can't grasp infinity doron, but that's not really our problem, now is it?

I am sorry that you can't grasp actual-infinity, which is "that has no successor" and its cardinality = ∞

Actual infinity can be expressed as 1-dimensional space, and any amount of objects (some collection) along
1-dimensional space has cardinality < ∞

In order to show otherwise, http://www.internationalskeptics.com/forums/showpost.php?p=6580213&postcount=12414 has to be proven.

 

[laca]

[latex]$Delta=|a_n-A|[/latex]

OK...

No matter how many epsilon > delta are considered, delta > 0, because delta can't be a single point, exactly because 1-dimensional space (expressed as delta, in this case) is irreducible into 0-dimensional space.

Well, still gibberish. Do you actually know the definition of a limit? This is one definition, since it is equivalent with another one. Still arguing definitions?

As a result 1/2+1/4+1/8+1/16+1/32+1/64+ ... < 1.

Furthermore, persons like The Man and jsfisher have to explicitly demonstrate (by using bottom to top techniques) that two distinct points define a line segment, such that there is no gap between these two points.

According to their reasoning, such a thing is possible, because they claim that a line segment is completely covered by points, which means that there is no gap between these two points along a given line segment.

Let's see how they are formally prove it

No, doron, you fail. Again. A line is defined by the points on it. It has an equation. A linear one, hence, line. It can be easily shown by reductio ad absurdum that for each point on the x axis there is a point on the y axis such that (x,y) is on the line. Q.E.D. Otherwise, you are still arguing definitions. It makes you look silly. Please stop it.

 

[doronshadmi]

It can be easily shown by reductio ad absurdum that for each point on the x axis there is a point on the y axis such that (x,y) is on the line.

No Laca, you are using two points which exist along different lines of a given plan.

I am talking about different points along the same line.

The Man and jsfisher claim that this single line is completely covered by points.

In that case they have no choice but to formally prove that there exist two distinct points along the same line, whiteout any gap between them.

Let's see how they are formally prove it

 

[sympathic]

No Laca, you are using two points which exist along different lines of a given plan.

I am talking about different points along the same line.

The Man and jsfisher claim that this single line is completely covered by points.

In that case they have no choice but to formally prove that there exist two distinct points along the same line, whiteout any gap between them.

Let's see how they are formally prove it

They do not need to. Maybe they are using direct perception.

 

[doronshadmi]

They do not need to. Maybe they are using direct perception.

Exactly the opposite.

Since they do not use direct perception they must formally prove that there exist two distinct points along the same line, whiteout any gap between them, exactly because they assert that this line is completely covered by points.

Let's see how they are formally prove it