Deeper than primes

[doronshadmi]

Yes. So what? This might seem like some great idea to you but it is fairly self-evident to thinking people.

No, it clearly seen that you do not understand the results of this abstract fact on your assertions.

 

[laca]

All we have is to combine your two "yes" answers in post http://www.internationalskeptics.com/forums/showpost.php?p=6584210&postcount=12476 .

No, doron. Those two yes answers are both to the same question. Moreover, I even provided an equivalent formulation of my assertion. This is not woo-woo land. Just because you say something it does not make it so. You have to actually show how my assertion is equivalent to what you're thinking I'm saying. You have utterly failed to do so. Care to try again?

 

[laca]

No, it clearly seen that you do not understand the results of this abstract fact on your assertions.

Yes, doron, clearly my pathetically weak reasoning is incapable of comprehending countable and uncountable sets. Oh wait, that's you...

 

[doronshadmi]

No, doron. Those two yes answers are both to the same question. Moreover, I even provided an equivalent formulation of my assertion. This is not woo-woo land. Just because you say something it does not make it so. You have to actually show how my assertion is equivalent to what you're thinking I'm saying. You have utterly failed to do so. Care to try again?

(You assert that a 1-dimensional space is completely covered by distinct 0-dim spaces) AND (You assert that 1-dim space is actually a collection of 0-dim spaces).

The result of these two assertions is exactly this:

Given two 0-dimensional spaces along 1-dimensional space, (they are both distinct) AND (have 0 distance between them).

 

[laca]

(You assert that a 1-dimensional space is completely covered by distinct 0-dim spaces) AND (You assert that 1-dim space is actually a collection of 0-dim spaces).

The result of these two assertions is exactly this:

Given two 0-dimensional spaces along 1-dimensional space, they are both distinct AND have 0 distance between them.

Why, doron? Show us how those two are equivalent. You do know what equivalent means, don't you?

 

[doronshadmi]

Why, doron? Show us how those two are equivalent. You do know what equivalent means, don't you?

You still do not get it.

Your two assertions do not have to be equivalent.

All we need is to get the result of these assertions, which you claim that they are true propositions.

Proposition A = "1-dimensional space is completely covered by distinct 0-dim spaces"

Proposition B = "1-dimensional space is actually a collection of distinct 0-dim spaces"

By (Proposition A) AND (Proposition B) we get this result:

Given two 0-dimensional spaces along 1-dimensional space, (they are both distinct) AND (have 0 distance between them).

 

[laca]

You still do not get it.

Your two assertions do not have to be equivalent.

Not my assertions, doron. I know it is hard, but do try to keep up with the conversation. The assertion you claim that I'm making has to be
actually equivalent with my assertion. Otherwise you're putting words in my mouth. You wouldn't want to be called a liar, do you?

All we need is to get the result of these assertions, which you claim that they are true propositions.

Proposition A = "1-dimensional space is completely covered by distinct 0-dim spaces"

Proposition B = "1-dimensional space is actually a collection of distinct 0-dim spaces"

Now those two actually are equivalent. You may reasonably demand that if one agrees with proposition A, then one has to agree with proposition B.

By (Proposition A) AND (Proposition B) we get this result:

Given two 0-dimensional spaces along 1-dimensional space, (they are both distinct) AND (have 0 distance between them).

So you take two equivalent propositions, combine them (redundancy, anyone?) and arrive at a third one with no apparent connection to them. Pray tell what kind of rule did you use when making the inference?

 

[zooterkin]

You still do not get it.

Your two assertions do not have to be equivalent.

All we need is to get the result of these assertions, which you claim that they are true propositions.

Proposition A = "1-dimensional space is completely covered by distinct 0-dim spaces"

Proposition B = "1-dimensional space is actually a collection of distinct 0-dim spaces"

By (Proposition A) AND (Proposition B) we get this result:

Given two 0-dimensional spaces along 1-dimensional space, (they are both distinct) AND (have 0 distance between them).

Doron, what you are saying is nonsense.

Are you actually trying to say that if the line is completely covered in points, then any particular point must have another point adjacent to it, touching it, in effect?

 

[doronshadmi]

Not my assertions, doron.

It does not matter who expressed them, the important fact is that you agree with them, as follows:

Proposition A:

Laca, please answer only by yes or no to the following question:

Do you claim that a 1-dimensional space is completely covered by distinct 0-dim spaces?

YES

Laca, you agree with Proposition A.


Proposition B:

You actually assert that 1-dim space is actually a collection of 0-dim spaces.

Well, yes I am.

Laca, you agree with Proposition B.

Now those two actually are equivalent.

By proposition A, it is not strictly clear that one defines 1-dimensional space in terms of a collection of distinct 0-dimesional spaces.

By proposition B, it is strictly clear that one defines 1-dimensional space in terms of a collection of distinct 0-dimensional spaces.

So you take two equivalent propositions

They are not equivalent propositions, and also if they were equivalent, as you claim, then still "equivalence" is not the same as "equality", so there is no redundancy here.

and arrive at a third one with no apparent connection to them.

This is the exact result that one gets, if one agrees with Proposition A AND Proposition B.

Pray tell what kind of rule did you use when making the inference?

The rule of the simplest common sense.

 

[doronshadmi]

Are you actually trying to say that if the line is completely covered in points, then any particular point must have another point adjacent to it, touching it, in effect?

No, I say that if one agree with the notion that a collection of distinct 0-dimensional spaces completely cover 1-dimensional space, than one can't the avoid the following contradiction:

There are two 0-dimensional spaces along 1-dimensional space, (which are distinct) AND (there is 0 distance between them).

 

[laca]

By proposition A, it is not strictly clear that one defines 1-dimensional space in terms of a collection of distinct 0-dimesional spaces.

Yes, it is. Not my problem that you don't get it.

By proposition B, it is strictly clear that one defines 1-dimensional space in terms of a collection of distinct 0-dimensional spaces.

Yes, it is equivalent to proposition A.

They are not equivalent propositions, and also if they were equivalent, as you claim, then still "equivalence" is not the same as "equality", so there is no redundancy here.

If you infer something from two equivalent propositions, it is redundant, because you can infer it from only one.

This is the exact result that one gets, if one agrees with Proposition A AND Proposition B.

Nope, sorry doron. You are at best deluded.

The rule of the simplest common sense.

Right...

 

[laca]

No, I say that if one agree with the notion that a collection of distinct 0-dimensional spaces completely cover 1-dimensional space, than one can't the avoid the following contradiction:

There are two 0-dimensional spaces along 1-dimensional space, (which are distinct) AND (there is 0 distance between them).

Oh, poor doron... You can't infer from that proposition the self-contradictory gibberish you continue to spout despite several attempts to explain it to you. I'm beginning to believe you don't have the necessary mental capabilities. Which kind of makes all attempts to educate you futile. But it sure is fun.

 

[zooterkin]

No, I say that if one agree with the notion that a collection of distinct 0-dimensional spaces completely cover 1-dimensional space, than one can't the avoid the following contradiction:

There are two 0-dimensional spaces along 1-dimensional space, (which are distinct) AND (there is 0 distance between them).

Why? Please explain.

 

[doronshadmi]

If you infer something from two equivalent propositions, it is redundant, because you can infer it from only one.

laca it really does not matter, the fact is that you actually claim that 1-dimensional space is no more than a collection of distinct 0-dimensional spaces.

In that case this collection is complete only if given two distinct 0-dimensional spaces there is no additional place for more 0-dimensional spaces between them, because if there is an additional place for more 0-dimensional spaces between them, then the completeness of this collection is not satisfied.

In that case you have to decide to take one of these two options:

Option 1: The collection of 0-dimensional spaces is complete, but then we get the following contradiction: Two distinct 0-dimensional spaces have 0 distance between them.

So option 1 is invalid.

Option 2: The collection of 0-dimensional spaces is incomplete, because given arbitrary distinct 0-dimensional spaces upon infinitely many scale levels, there is always more space for more 0-dimensional spaces.

Option 2 is indeed the valid one, but this more space is strictly greater than any amount of distinct 0-dimensional spaces, which suppose to completely fill (or cover) it.

This strictly greater space is, surprise surprise, not less then 1-dimensional space, and we realize that 1-dimensional space is not defined as a collection of 0-dimensional spaces.

Laca, your reasoning is an invalid dead end street, exactly because you claim that a 1-dimensional space is no more than a collection of 0-dimensional spaces.

 

[doronshadmi]

Why? Please explain.

http://www.internationalskeptics.com/forums/showpost.php?p=6585117&postcount=12494

 

[zooterkin]

Ah, we're back to your problems with needing to enumerate every member of an infinite set.

 

[jsfisher]

No, I say that if one agree with the notion that a collection of distinct 0-dimensional spaces completely cover 1-dimensional space, than one can't the avoid the following contradiction:

There are two 0-dimensional spaces along 1-dimensional space, (which are distinct) AND (there is 0 distance between them).

Well, at least you have switched from the other ridiculous statement to this. I bet, though, you cannot tell the two are different.

Still, it all boils down to you, Doron, not understanding how the continuum behaves. You want it to behave like a finite set, or perhaps even a countably infinite set. Too bad for you; it doesn't.

A line is completely covered by points.
No two points are adjacent.

Sorry if you don't understand that. It wouldn't be grasped by most kindergarteners, either, if that is any consolation. Their cognitive abilities haven't developed enough.

 

[doronshadmi]

Ah, we're back to your problems with needing to enumerate every member of an infinite set.

No, we do not need to anumetate anything.

An infinite collection of 0-dimensioal spaces is strictly smaller the the space that enables them to be placed upon infinitely many scale levels, and this space is called 1-dimensional space.

 

[doronshadmi]

A line is completely covered by points.
No two points are adjacent.

Poor jsfisher, he can't understand that ("completely covered by points") AND ("No two points are adjacent") is Option 1 of http://www.internationalskeptics.com/forums/showpost.php?p=6585117&postcount=12494.

Still, it all boils down to you, Doron, not understanding how the continuum behaves. You want it to behave like a finite set, or perhaps even a countably infinite set. Too bad for you; it doesn't.

Wrong again poor jsfisher, a person that can't grasp that the uncountable size of distinct 0-dimensional spaces is strictly smaller than the size of 1-dimesional space.

He will never get the valid Option 2 of http://www.internationalskeptics.com/forums/showpost.php?p=6585117&postcount=12494.

 

[laca]

laca it really does not matter, the fact is that you actually claim that 1-dimensional space is no more than a collection of distinct 0-dimensional spaces.

That I do. And I agree that it is irrelevant to the point in question, it is just serving to show the depth of your incompetence.

In that case this collection is complete only if given two distinct 0-dimensional spaces there is no additional place for more 0-dimensional spaces between them, because if there is an additional place for more 0-dimensional spaces between them, then the completeness of this collection is not satisfied.

Wrong. What you're missing is that the "additional space" is covered with more of those pesky 0-dim points. Repeat ad infinitum. Problem solved. No contradiction. No "additional space". Just points. Lots of them.

In that case you have to decide to take one of these two options:

Option 1: The collection of 0-dimensional spaces is complete, but then we get the following contradiction: Two distinct 0-dimensional spaces have 0 distance between them.

So option 1 is invalid.

Option 2: The collection of 0-dimensional spaces is incomplete, because given arbitrary distinct 0-dimensional spaces upon infinitely many scale levels, there is always more space for more 0-dimensional spaces.

Option 2 is indeed the valid one, but this more space is strictly greater than any amount of distinct 0-dimensional spaces, which suppose to completely fill (or cover) it.

I choose secret option 3! You have not even attempted to show that those two are the only valid options. They are actually not even internally consistent. Sorry, you fail again.

This strictly greater space is, surprise surprise, not less then 1-dimensional space, and we realize that 1-dimensional space is not defined as a collection of 0-dimensional spaces.

Well, at this point it's not surprising at all that you seem to be incapable to grasp basic mathematical concepts. 1 dimensional space (a line) is by definition a collection of points (you are calling those 0 dimensional).

Laca, your reasoning is an invalid dead end street, exactly because you claim that a 1-dimensional space is no more than a collection of 0-dimensional spaces.

No, my reasoning is sound. You have failed to show any discrepancies in it. Not to mention that it is not actually "my" reasoning. I'm not the one who came up with it. I just came to the conclusion that it is sound based on some very basic studying of the concepts involved.