Deeper than primes

[doronshadmi]

No, it doesn't w.r.t. what you're trying to demonstrate. I just wanted to bring to your attention the fact that your statements about properties of various objects may not be always true as you think they are.

Show some case that there are different AND smallest elements, without an ever smaller element between them, which is located at once at both smallest elements.

Show some smallest element which is located at once in more than one location.

 

[doronshadmi]

I think that your attempt to prove that the definition

A line segment is a part of a line that is bounded by two end points, and contains every point on the line between its end points.

implies nothing but that the points in question are adjacent to each other amounts to saying that the mathematicians who formulated the definition were simply insane.

Wrong.

I simply show the co-existence of ever smaller elements AND smallest elements along a given line segment.

 

[doronshadmi]

In other words, as you keep dividing infinitely, points Log[t] or Sin[t] must and will show up sooner or later.

There is no process here of any kind.

There is always an ever smaller element between the smallest element [t] and some ever closer smallest element (which is defiantly not [t]).

 

[epix]

Wrong.

I simply show the co-existence of ever smaller elements AND smallest elements along a given line segment.

Do you care to disclose the reason why? I don't think that dividing on smaller pieces is something that you can run into the patent office with.

The beginning of the Stone Age marks the era when hominins first began manufacturing stone tools, and evidence of these tools dates back at least 2.6 million years in Ethiopia. One of the earliest distinguishable stone tool forms is the hand axe.

Tough luck. Someone beat you to it.

 

[doronshadmi]

Do you care to disclose the reason why?

Do you care to read http://www.internationalskeptics.com/forums/showpost.php?p=7079152&postcount=15021 ?

 

[doronshadmi]

I don't think that dividing on smaller pieces is something that you can run into the patent office with.

You are right, there are probably other persons along the human history, which said that a line-segment (no matter what length it has) can't be an 0 length element and still be considered as a line segment.

EDIT:

For example:

http://setis.library.usyd.edu.au/stanford/entries/settheory-alternative/#WhySetThe

Even an early modern thinker like Spinoza could comment that it is obvious that a line is not a collection of points (whereas for us it may hard to see what else it could be) (Ethics, I.15, scholium IV, 96).

In other words, Spinoza thought that an ever smaller element is different than the collection of all the smallest elements along it.

 

[zooterkin]

You are right, there are probably other persons along the human history, which said that a line-segment (no matter what length it has) can't be an 0 length element and still be considered as a line segment.

EDIT:

For example:

http://setis.library.usyd.edu.au/stanford/entries/settheory-alternative/#WhySetThe

In other words, Spinoza thought that an ever smaller element is different than the the collection of all the smallest elements along it.

You do understand that the page you've linked to is saying that Spinoza was wrong, don't you?


ETA: I also wouldn't vouch for your interpretation of what Spinoza said as being correct; it wasn't there when I hit the "Reply" button.

 

[doronshadmi]

You do understand that the page you've linked to is saying that Spinoza was wrong, don't you?

Do you understand that Spinoza is right about this subject, and the scholars that say otherwise about this subject, are wrong?

EDIT:

Do you care to read http://www.internationalskeptics.com/forums/showpost.php?p=7079152&postcount=15021 ?

Do you get http://www.internationalskeptics.com/forums/showpost.php?p=7069331&postcount=14958 ?

 

[doronshadmi]

ETA: I also wouldn't vouch for your interpretation of what Spinoza said as being correct; it wasn't there when I hit the "Reply" button.

What do you mean by "it wasn't there"?

 

[doronshadmi]

Let us demonstrate Cantor's construction method, by using |P(S)| proper subsets of P(S) members, where each P(S) proper subset has |S| different P(S) members that are mapped with all S members.

In this demonstration we are using the all members of S={a,b,c} and |P(S)| proper subsets of different members of P(S)={{},{a},{b},{c},{a,b},{a,c},{b,c},{a,b,c}}, where each used P(S) proper subset has |S| different P(S) members, as follows:

.
.
.
|
|
|-----------------------
|a ↔ {a} |
|b ↔ {b} | Provides {}
|c ↔ {c} |
|-----------------------
|
|
|-----------------------
|a ↔ { } |
|b ↔ {b} | Provides {a}
|c ↔ {c} |
|-----------------------
|
|
|-----------------------
|a ↔ {a} |
|b ↔ { } | Provides {b}
|c ↔ {c} |
|-----------------------
|
|
|-----------------------
|a ↔ {a} |
|b ↔ {b} | Provides {c}
|c ↔ { } |
|-----------------------
|
|
|-----------------------
|a ↔ { } |
|b ↔ {c} | Provides {a,b}
|c ↔ {a,c} |
|-----------------------
|
|
|-----------------------
|a ↔ { } |
|b ↔ {b} | Provides {a,c}
|c ↔ {a,b} |
|-----------------------
|
|
|-----------------------
|a ↔ {a} |
|b ↔ {c} | Provides {b,c}
|c ↔ {a,b} |
|-----------------------
|
|
|-----------------------
|a ↔ { } |
|b ↔ {a} | Provides {a,b,c}
|c ↔ {b} |
|-----------------------
|
|
.
.
.

As can be seen, we have constructed the all P(S) members without exceptional, by using Cantor's construction method, and it does not matter if |P(S)| and |S| are finite or infinite.

In the case where |P(S)| and |S| are infinite, Dedekind's infinite holds (where S is a proper subset of P(S)).

 

[doronshadmi]

[SIZE=+1]Again I ask: Can we expect to see your bijections between the elements of {} and its power set ... ?[/SIZE]

First we are using Cantor's construction method in order to define the all members of {{}}.

↔ {} Provides {} simply because {} is always the result if the member of S has an image in the mapped member of P(S) (and in this case the image is Emptiness).

Now you can define a function between {} and some set (a, for example):

a ↔ {}

 

[epix]

You are right, there are probably other persons along the human history, which said that a line-segment (no matter what length it has) can't be an 0 length element and still be considered as a line segment.

EDIT:

For example:

http://setis.library.usyd.edu.au/stanford/entries/settheory-alternative/#WhySetThe

In other words, Spinoza thought that an ever smaller element is different than the collection of all the smallest elements along it.

http://www.englishexercises.org/makeagame/viewgame.asp?id=1876
Can you rephrase your conclusion now.

Things are simple: You misinterpreted the definition of line segment and you've been trying to show that the definition doesn't reflect upon the intuitive perception of a line segment. I have to ask The Man about that type of fallacy -- I can't place it.

 

[epix]

First we are using Cantor's construction method in order to define the all members of {{}}.

↔ {} Provides {} simply because {} is always the result if the member of S has an image in the mapped member of P(S) (and in this case the image is Emptiness).

Now you can define a function between {} and some set (a, for example):

a ↔ {}

I guess this is the most ingenius way to convert an unsuspecting atheist into a theist through a bijection in two easy mathematical steps:

1. atheist - a = theist

2. a ↔ {}

 

[epix]

Let us demonstrate Cantor's construction method, by using |P(S)| proper subsets of P(S) members, where each P(S) proper subset has |S| different P(S) members that are mapped with all S members.

In this demonstration we are using the all members of S={a,b,c} and |P(S)| proper subsets of different members of P(S)={{},{a},{b},{c},{a,b},{a,c},{b,c},{a,b,c}}, where each used P(S) proper subset has |S| different P(S) members, as follows:

.
.
.
|
|
|-----------------------
|a ↔ {a} |
|b ↔ {b} | Provides {}
|c ↔ {c} |
|-----------------------
|
|
|-----------------------
|a ↔ { } |
|b ↔ {b} | Provides {a}
|c ↔ {c} |
|-----------------------
|
|
|-----------------------
|a ↔ {a} |
|b ↔ { } | Provides {b}
|c ↔ {c} |
|-----------------------
|
|
|-----------------------
|a ↔ {a} |
|b ↔ {b} | Provides {c}
|c ↔ { } |
|-----------------------
|
|
|-----------------------
|a ↔ { } |
|b ↔ {c} | Provides {a,b}
|c ↔ {a,c} |
|-----------------------
|
|
|-----------------------
|a ↔ { } |
|b ↔ {b} | Provides {a,c}
|c ↔ {a,b} |
|-----------------------
|
|
|-----------------------
|a ↔ {a} |
|b ↔ {c} | Provides {b,c}
|c ↔ {a,b} |
|-----------------------
|
|
|-----------------------
|a ↔ { } |
|b ↔ {a} | Provides {a,b,c}
|c ↔ {b} |
|-----------------------
|
|
.
.
.

As can be seen, we have constructed the all P(S) members without exceptional, by using Cantor's construction method, and it does not matter if |P(S)| and |S| are finite or infinite.

In the case where |P(S)| and |S| are infinite, Dedekind's infinite holds (where S is a proper subset of P(S)).

Your schematics is too confusing. Can you draw a pic how is it done? See, I have two electrical outlets A and B in the kitchen and five appliances C, D, E, F, and G. Can you draw a sketch that connects all appliances to both electrical outlets without using a splitter?

 

[epix]

You are right, there are probably other persons along the human history, which said that a line-segment (no matter what length it has) can't be an 0 length element and still be considered as a line segment.

EDIT:

For example:

http://setis.library.usyd.edu.au/stanford/entries/settheory-alternative/#WhySetThe

Your example speaks of unimaginable ignorance on your part. You don't even bother to copy/paste an excerpt from the link that explicitly address the raised issue concerning line segments. I guess that's because the reference isn't there.

Look at those straight lines bellow.

Can't you comprehend that a vector with 0 magnitude (length) does exists and you can locate it on a plane as point [x,y]?

 

[jsfisher]

Let us demonstrate Cantor's construction method, by using |P(S)| proper subsets of P(S) members, where each P(S) proper subset has |S| different P(S) members that are mapped with all S members.

In this demonstration we are using the all members of S={a,b,c} and |P(S)| proper subsets of different members of P(S)={{},{a},{b},{c},{a,b},{a,c},{b,c},{a,b,c}}, where each used P(S) proper subset has |S| different P(S) members, as follows:

.
.
.
|
|
|-----------------------
|a ↔ {a} |
|b ↔ {b} | Provides {}
|c ↔ {c} |
|-----------------------

The standard proof for Cantor's Theorem posits a bijection between the elements of S and P(S). Yours is not a bijection, just a random injection from S to P(S).

...more arbitrarily selected injections snipped...

As can be seen, we have constructed the all P(S) members without exceptional

A totally trivial feat, a totally unnecessary feat, since we started with all the members of P(S).

...by using Cantor's construction method

Not without bijections, you didn't.

Still, let's reapply Doron's abortion of the proof to S={A,B}. Doron didn't mention it this time, but we need 2^|S| "rounds" of his nonsense.

Round 1:
A ↔ {A}, B ↔ {B} which yields {} (which we already knew was a member of P(S), but whatever).

Round 2:
A ↔ {A}, B ↔ {A,B} which yields {} again.

Round 3:
A ↔ {A,B}, B ↔ {B} which yields {} a third time.

Round 4:
A ↔ {B}, B ↔ {A} which yields {A,B}.


So, according to Doron if S = {A,B} then P(S) = {{},{A,B}}

 

[epix]

Still, let's reapply Doron's abortion of the proof to S={A,B}. Doron didn't mention it this time, but we need 2^|S| "rounds" of his nonsense.

Round 1:
A ↔ {A}, B ↔ {B} which yields {} (which we already knew was a member of P(S), but whatever).

Round 2:
A ↔ {A}, B ↔ {A,B} which yields {} again.

Round 3:
A ↔ {A,B}, B ↔ {B} which yields {} a third time.

Round 4:
A ↔ {B}, B ↔ {A} which yields {A,B}.


So, according to Doron if S = {A,B} then P(S) = {{},{A,B}}

I don't understand any of it. I thought that the bijection between S and P(S) can be compared to plugging kitchen appliances to the electrical outlets. So I asked Doron to draw a pic showing how to plug in 4 appliances to 2 outlets without using a splitter. His initial idea using outlets a, b, c and 8 appliances calls for plugging in the lines so that only three appliances can be on simultaneously:

|-----------------------
|a ↔ {a} |
|b ↔ {b} | Provides {}
|c ↔ {c} |
|-----------------------
|
|
|-----------------------
|a ↔ { } |
|b ↔ {b} | Provides {a}
|c ↔ {c} |
|-----------------------

But I want all the appliances connected at the same time, coz he claimed that |S| can equal |P(S)|

According to your conclusion if S={A,B} then P(S) = {{},{A,B}, not only that Doron can't hook it up the way that I could serve dinner to Cantor on time, but Doron also seems to be specific about which two appliances out of those eight can be connected to both outlets. That sucks.

 

[doronshadmi]

The standard proof for Cantor's Theorem posits a bijection between the elements of S and P(S). Yours is not a bijection, just a random injection from S to P(S).



A totally trivial feat, a totally unnecessary feat, since we started with all the members of P(S).


Not without bijections, you didn't.

Still, let's reapply Doron's abortion of the proof to S={A,B}. Doron didn't mention it this time, but we need 2^|S| "rounds" of his nonsense.

Round 1:
A ↔ {A}, B ↔ {B} which yields {} (which we already knew was a member of P(S), but whatever).

Round 2:
A ↔ {A}, B ↔ {A,B} which yields {} again.

Round 3:
A ↔ {A,B}, B ↔ {B} which yields {} a third time.

Round 4:
A ↔ {B}, B ↔ {A} which yields {A,B}.


So, according to Doron if S = {A,B} then P(S) = {{},{A,B}}

So according to jsfisher if S = {A,B} then P(S) = {{},{A,B}}.

But it is just jsfisher's box-reasoning.

The right reasoning is this:

S = {A,B}

P(S) = {{},{A},{B},{A.B}}

Cantor's constriction method of all the above P(S) members, works like this (without using any rounds, because it is constructed at once in parallel):

|
|
|-----------------------
|a ↔ {a} |
|b ↔ {b} | Provides {}
|-----------------------
|
|
|-----------------------
|a ↔ { } |
|b ↔ {b} | Provides {a}
|-----------------------
|
|
|-----------------------
|a ↔ {a} |
|b ↔ { } | Provides {b}
|-----------------------
|
|
|-----------------------
|a ↔ { } |
|b ↔ {a} | Provides {a,b}
|-----------------------

Then it is possible to define the following bijection, which is based on the fact that no P(S) member is missing:

{} ↔ a
{a} ↔ b
{b} ↔ c
{a,b} ↔ d

In other words, jsfisher's reasoning simply does not get the considered fine subject.

 

[doronshadmi]

I don't understand any of it.

So do the needed work in your mind, in order to understand it, before you air your view about the considered fine subject.

First you really have to get:

http://www.internationalskeptics.com/forums/showpost.php?p=7066349&postcount=14908

http://www.internationalskeptics.com/forums/showpost.php?p=7066490&postcount=14911

If you do not get them, you are not able to air a meaningful view about the considered fine subject.

 

[doronshadmi]

Things are simple: You misinterpreted the definition of line segment and you've been trying to show that the definition doesn't reflect upon the intuitive perception of a line segment. I have to ask The Man about that type of fallacy -- I can't place it.

You still don't get it epix.

There is only one person that really can help you to understand this considered fine subject.

The name of this person is called epix, and the needed work to get this subject has to be done in his own mind.