Deeper than primes

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laca said:
Yes. So what? This might seem like some great idea to you but it is fairly self-evident to thinking people.
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No, it clearly seen that you do not understand the results of this abstract fact on your assertions.
 
doronshadmi said:
All we have is to combine your two "yes" answers in post http://www.internationalskeptics.com/forums/showpost.php?p=6584210&postcount=12476 .
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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?
 
doronshadmi said:
No, it clearly seen that you do not understand the results of this abstract fact on your assertions.
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Yes, doron, clearly my pathetically weak reasoning is incapable of comprehending countable and uncountable sets. Oh wait, that's you...
 
laca said:
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?
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(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).
 
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doronshadmi said:
(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.
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Why, doron? Show us how those two are equivalent. You do know what equivalent means, don't you?
 
laca said:
Why, doron? Show us how those two are equivalent. You do know what equivalent means, don't you?
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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).
 
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doronshadmi said:
You still do not get it.

Your two assertions do not have to be equivalent.
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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?

doronshadmi said:
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"
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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.

doronshadmi said:
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).
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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?
 
doronshadmi said:
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).
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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?
 
laca said:
Not my assertions, doron.
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It does not matter who expressed them, the important fact is that you agree with them, as follows:

Proposition A:
laca said:
doronshadmi said:
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?
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YES
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Laca, you agree with Proposition A.


Proposition B:
laca said:
doronshadmi said:
You actually assert that 1-dim space is actually a collection of 0-dim spaces.
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Well, yes I am.
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Laca, you agree with Proposition B.


laca said:
Now those two actually are equivalent.
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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.


laca said:
So you take two equivalent propositions
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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.

laca said:
and arrive at a third one with no apparent connection to them.
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This is the exact result that one gets, if one agrees with Proposition A AND Proposition B.

laca said:
Pray tell what kind of rule did you use when making the inference?
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The rule of the simplest common sense.
 
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zooterkin said:
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?
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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).
 
doronshadmi said:
By proposition A, it is not strictly clear that one defines 1-dimensional space in terms of a collection of distinct 0-dimesional spaces.
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Yes, it is. Not my problem that you don't get it.

doronshadmi said:
By proposition B, it is strictly clear that one defines 1-dimensional space in terms of a collection of distinct 0-dimensional spaces.
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Yes, it is equivalent to proposition A.

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

doronshadmi said:
This is the exact result that one gets, if one agrees with Proposition A AND Proposition B.
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Nope, sorry doron. You are at best deluded.

doronshadmi said:
The rule of the simplest common sense.
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Right...
 
doronshadmi said:
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).
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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.
 
doronshadmi said:
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).
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Why? Please explain.
 
laca said:
If you infer something from two equivalent propositions, it is redundant, because you can infer it from only one.
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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.
 
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doronshadmi said:
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).
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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.
 
zooterkin said:
Ah, we're back to your problems with needing to enumerate every member of an infinite set.
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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.
 
jsfisher said:
A line is completely covered by points.
No two points are adjacent.
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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.

jsfisher said:
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.
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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.
 
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doronshadmi said:
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.
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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.

doronshadmi said:
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.
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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.

doronshadmi said:
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.
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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.

doronshadmi said:
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.
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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).

doronshadmi said:
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.
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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.
 
doronshadmi said:
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.
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No, that is your delusion, not mine.

Perhaps you were hung up by that word, additional? True enough, there is no additional space between any two distinct points. It is the same space that was always there, and that space is completely covered by points. Those points aren't additional, either, at least not in the sense of needing to be added. They have always been there.

The collection of points that covered a line is complete because it includes all the points that cover the line. Nothing was omitted.

By about fourth grade, this concept becomes understandable.
 
jsfisher said:
Nothing was omitted.
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jsfisher, you can't get that the size of 1-dimensional space is strictly greater than the size of collections of 0-dimensional spaces, even if they are uncountable, because you can't think beyond the concept of Collection.

jsfisher said:
The collection of points that covered a line is complete because it includes all the points that cover the line. Nothing was omitted.
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So what? It is still a collection, and being a collection of 0-dimensional spaces is being strictly smaller that 1-dimensional space, which is a notion that you can't comprehend.
 
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laca said:
1 dimensional space (a line) is by definition a collection of points (you are calling those 0 dimensional).
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Only in your fantasy. Like jsfisher, you can't think beyond the concept of Collection.
 
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jsfisher said:
Those points aren't additional, either, at least not in the sense of needing to be added. They have always been there.
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There is no process of any kind here jsfisher.

My heart with you when I see how your notion is closed under the concept of Collection, such that you can't get anything beyond it, but your notion's problems do not change the abstract notion of 1-dimensional space that is strictly greater than any collection of 0-dimesional spaces along it.

Furthermore 1-dimensional space exists even if there are no 0-dimesinal spaces along it and it has the magnitude of the continuum, but this notion is really beyond your head.
 
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doronshadmi said:
Welcome HatRack,

Do you assert that a 1-dimensional space is completely covered by a collection of 0-dimensional spaces?

Please answer only by yes or no.
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Doron, I will respond to your question when you respond to my original post. 1/2 + 1/4 + 1/8 + ... is equal to the limit at infinity of 1 - 1/2^n, according to the definition of the sum of an infinite series as found in any calculus text. From the standard definition of limits and all of the standard theorems which follow from them, which all in turn follow from the axioms of ZFC set theory given the appropriate set of definitions, we have 1/2 + 1/4 + 1/8 + ... = 1.

If you assert anything different, then you are calling one of the ZFC axioms or one of the traditional definitions found in analysis into question. I asked you to clearly and precisely state which axiom or definition you are changing to derive the proposition 1/2 + 1/4 + 1/8 + ... < 1.
 
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jsfisher said:
there is no additional space between any two distinct points. It is the same space that was always there, and that space is completely covered by points.
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If 1-dimensional space is completely covered by 0-dimensional spaces, then 1-dimensional space does not exist.

In that case we are left with collection of distinct 0-dimensional spaces.

jsfisher said:
No two points are adjacent.
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Again you are using top to bottom (macro) reasoning, which according to it no two points are adjacent.

But this is not the only possible reasoning, because by bottom to top (micro) reasoning, we are still deal with collection of distinct 0-dimesional spaces, and it this micro level a collection is complete only of there is exactly 0 distance between distinct 0-dimesional spaces, which is a contradiction, because 0-dimensional spaces can't be both distinct AND have 0 distances between them.

Since this is a contradiction, then given uncountable scale levels, it is always true that no two points are adjacent, such that there is always uncovered space between them, because being distinct is an essential property of any collection of 0-dimensional spaces.

This uncovered space is 1-dimensional space, which its magnitude is greater than the magnitude of the distinct 0-dimensional spaces along it.
 
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HatRack said:
Doron, I will respond to your question when you respond to my original post. 1/2 + 1/4 + 1/8 + ... is equal to the limit at infinity of 1 - 1/2^n, according to the definition of the sum of an infinite series as found in any calculus text. From the standard definition of limits and all of the standard theorems which follow from them, which all in turn follow from the axioms of ZFC set theory given the appropriate set of definitions, we have 1/2 + 1/4 + 1/8 + ... = 1.

If you assert anything different, then you are calling one of the ZFC axioms or one of the traditional definitions found in analysis into question. I asked you to clearly and precisely state which axiom or definition you are changing to derive the proposition 1/2 + 1/4 + 1/8 + ... < 1.
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HatRack,

ZFC is based on the concept Set, where a set is based on the concept of collection of distinct objects.

My question is essential to the concept of collection of distinct objects.

Please answer by yes or no to this essential question:

Do you claim that a collection of distinct 0-dimenssional spaces completely cover 1-dimensional space?
 
doronshadmi said:
HatRack,

ZFC is based on the concept Set, where a set is based on the concept of collection of distinct objects.

My question is essential to the concept of collection of distinct objects.
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Doron, your line of reasoning is useless if it leads you to the proposition 1/2 + 1/4 + 1/8 + ... < 1. This is because it is ultimately derivable from the ZFC axioms, which give us the basic properties of sets, that 1/2 + 1/4 + 1/8 + ... = 1. By the trichotomy law, which is also derivable from ZFC, a number cannot simultaneously equal 1 and be less than 1. This means that, in the ZFC system alone, any proof you give that 1/2 + 1/4 + 1/8 + ... < 1 is flawed and needs no further examination.

Thus, the only way to reconcile your line of reasoning with standard mathematics is to change one of the ZFC axioms or change a definition so that the proposition 1/2 + 1/4 + 1/8 + ... = 1 is no longer derivable.

My question is simple and direct, and you still haven't answered it. Which ZFC axiom or definition are you modifying/abolishing to make way for your line of reasoning?
 
jsfisher said:
It is the same space that was always there, and that space is completely covered by points.
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Utter nonsense.

If that space (1-dimensional space) is completely covered by 0-dimensioanl spaces, then that space does not exist, so there is no "the same space that was always there".

jsfisher said:
No two points are adjacent
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You are right, no two distinct 0-dimensional spaces are adjacent, and this abstraction is invariant upon any given infinitely many scale levels, which is exactly the 1-dimesional space, which its magnitude is strictly greater than any given collection of distinct 0-dimensioanal spaces along it.
 
doronshadmi said:
jsfisher, you can't get that the size of 1-dimensional space is strictly greater than the size of collections of 0-dimensional spaces, even if they are uncountable, because you can't think beyond the concept of Collection.
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As always, the limitation is yours, not mine.

doronshadmi said:
EDIT:If 1-dimensional space is completely covered by 0-dimensional spaces, then 1-dimensional space does not exist.
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What a bizarre comment. Baseless, untrue, and bizarre.

In that case we are left with collection of distinct 0-dimensional spaces.
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Come on, you can say: points.

...it is always true that no two points are adjacent
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Yes, I have said that before.

...such that there is always uncovered space between them
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Maybe that's true in bizarro Doronetics world; in the land of real Mathematics, not so.

...because being distinct is an essential property of any collection of 0-dimensional spaces.
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Great! You have thrown in yet another unrelated issue. Yes, all the points are distinct.

This uncovered space is 1-dimensional space, which its magnitude is greater than the magnitude of the distinct 0-dimensional spaces along it.
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Go ahead. Show us any place along a line that is not covered by points. Any place at all. You keep claiming they are there. Heck, you keep claiming they are everywhere, yet you can't show us one. Why is that?
 
doronshadmi said:
Utter nonsense.
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Is it? Then why can't you show us any contradiction from within Mathematics? Instead, you resort to misrepresentations, gibberish, undefined terms, and leaps of reasoning with no basis in logic.

Come on, Doron. Stop heckling. Show us something real. How about showing us any place along a line where there isn't a point? Surely such a place must exist, because you told us so. Show it to us.
 
HatRack said:
The notation 1/2 + 1/4 + 1/8 + 1/16 + ... is, loosely speaking, alternative notation for the limit as n approaches infinity of the finite sum of the series 1/2 + 1/4 + ... + 1/2^n. This finite sum is of course 1 - 1/2^n. Hence, 1/2 + 1/4 + 1/8 + ... = lim n->inf (1 - 1/2^n) = 1. Doronshadmi's assertion that 1/2 + 1/4 + 1/8 + ... < 1 is complete nonsense for that reason. For any natural number n, it is true that 1/2 + 1/4 + 1/8 + ... + 1/2^n < 1. But, having a "+ ..." at the end indicates that the limit is being taken.
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No. The notation 1/2 + 1/4 + 1/8 + 1/16 + ... is a descriptive and non-algebraic expression that merely establishes an infinite series. Since Doron is not familiar with the established symbolic language, he would use the description to indicate inequality 1/2 + 1/4 + 1/8 + 1/16 + ... < 1. This symbolism is normally avoided, coz it lacks necessary elements to prove or disprove the assertion. When this convergent series is converted into its algebraic form including the inequality

1 - 1/2n < 1 where n → ∞

it can be shown in three easy steps that the inequality holds.

1 - 1/2n < 1
-1/2n < 0
-1 < 0

However, the solution of the inequality is also good for any sum S ≥ 1, which allows the sum be greater than 1. That's why the limit needs to be computed. During the process

[lim n → ∞] 1 - 1/2n
the whole term 1/2n is held sufficiently close to zero to be removed under the rule that derives the limits. Obviously, the term never equals zero as n goes to infinity, and therefore the series cannot reach its limit.


No. The partial sums of the series are always less than 1 for any given natural number n. The infinite sum is defined as the limit of the sequence of partial sums, and the value of that limit is 1 in this particular case.
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The conversion of the series into function f(x) = 1 - 1/2x where x → ∞ leaves the issue of partial sums irrevelant, coz f(x) is a function and not a series comprising addends. But the "partiality" can be inherited and resurface in an assertion that cannot be easily defined. Let

f(x) = 1 - 1/2x where x → ∞

and

g(x) = 1 - 1/2x where x ≥ 1

In that case, f(x) ≠ g(x) as it is often asserted.

The concept of "partiality" must be properly translated into the function that substitutes the series and that's the point where it squeaks.
 
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doronshadmi said:
Exactly!

But since Traditional Math asserts that a 1-dim space is completely covered by a collection of distinct 0-dim spaces, then (according to Traditional Math) two points along a 1-dim space are both distinct AND there is no gap between them.

In that case Traditional Math has to provide the formal proof, which is resulted by the ability of two 0-dim spaces along a 1-dim space, to be both distinct AND gap-less.

Let's see how Traditional Math formally proves it :popcorn1
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There are two distinct points in "TRADITIONAL MATH" and that's R and N.

Isn't it enough?

No? Okay . . .

When N opens its door, these guys get out: 1, 2, 3, 4, ...
They are distinct and there is no gap between them.
 
epix said:
No. The notation 1/2 + 1/4 + 1/8 + 1/16 + ... is a descriptive and non-algebraic expression that merely establishes an infinite series.
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I'm looking at a calculus textbook right now that agrees with me. It has this written almost exactly:

[latex]a + ar + ar^{2} + ... + ar^{n-1} + ... = \displaystyle\sum\limits_{n=1}^{\infty}ar^{n-1}=\lim_{n \to \infty}\displaystyle\sum\limits_{k=1}^{n}ar^{k-1}=\frac{a}{1-r}[/latex]

Substitute in 1 for a and 1/2 for r to get:

[latex]1 + 1/2 + 1/4 + ... + (1/2)^{n-1} + ... = \displaystyle\sum\limits_{n=1}^{\infty}(1/2)^{n-1}=\lim_{n \to \infty}\displaystyle\sum\limits_{k=1}^{n}(1/2)^{k-1}=\frac{1}{1-1/2}=2[/latex]

Subtract 1 from both sides and you have:

[latex]1/2 + 1/4 + 1/8 + ... = 1[/latex]

An analysis text I have from 1980 also uses the + ... to indicate the limit of the sequence of partial sums. It's not as precise as using the Sigma notation or using a limit symbol, but it's clear that it indicates an infinite summation, which is defined as the limit of the sequence of partial sums, which is 1 in this case. Hence, 1/2 + 1/4 + 1/8 + 1/16 + ... < 1 is nonsense.

If Doron defines 1/2 + 1/4 + 1/8 + 1/16 + ... in a nonstandard way, then he needs to specify that, which is what I've been asking him.
 
jsfisher said:
Yes, all the points are distinct.
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You are right jsfisher.

Being distinct point along 1-dimensional space is possible only if given any arbitrary pair of points, they are not adjacent.

You see jsfisher? No need to show any missing point, all we need is to understand how a given point is distinct along a given line, and it is distinct only if the distance to another point is > 0.

This is a simple abstract fact that can't be avoided, if all the points are distinct, exactly as you and me claim.

I understand my claim.

You do not undertand your claim.
 
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doronshadmi said:
You are right jsfisher.

Being distinct point along 1-dimensional space is possible only if given any arbitrary pair of points, they are not adjacent.

You see jsfisher? No need to show any missing point, all we need is to understand how a given point is distinct along a given line, and it is distinct only if the distance to another point is > 0.

This is a simple abstract fact that can't be avoided, if all the points are distinct, exactly as you and me claim.

I understand my claim.

You do not undertand your claim.
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No Dorn you actually and simply don’t understand either claim. Yes the points are “distinct only if the distance to another point is > 0”, but guess what, (and this has been explained to you multiple times before in multiple ways including this one) any distance >0 can be divided by 2 giving at least one other distinct point between those two. In fact the whole infinite series can be mapped between those two points or any two points. Once again that is precisely what makes it a continuous space as opposed to a discrete space that you are claiming.
 
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doronshadmi said:
You are right jsfisher.
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Tell my wife that.

Being distinct point along 1-dimensional space is possible only if given any arbitrary pair of points, they are not adjacent.
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Very not true. Consider the pair of points on the real number line, 2 and 2. (No, you didn't specify the points needed to be distinct.)

You see jsfisher? No need to show any missing point, all we need is to understand how a given point is distinct along a given line, and it is distinct only if the distance to another point is > 0.

This is a simple abstract fact that can't be avoided, if all the points are distinct, exactly as you and me claim.
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No two distinct points along a line are adjacent. Is that what you are trying to say? If so, then we all agree.

I understand my claim.
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The evidence you have provided makes this doubtful. You dwell on the trivial then jump to the irrational without the constraint of logic. Heck, you haven't even figured out, yet, the difference between "two distinct points" versus "two adjacent points."

You do not undertand your claim.
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Yep.
 
doronshadmi said:
jsfisher said:
Yes, all the points are distinct.
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You are right jsfisher.
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By the way, the quote of mine was stripped of its context, and even in context, its true meaning was probably lost. At any rate, given A as a point on line L and B as a point on line L, there is no implicit requirement that A and B be two different points.

I believe I have been consistent, always adding the qualifier, distinct, where appropriate.

Given two arbitrary points (along a line or in a plane, or in 3-space, or ...):
(1) The points are not adjacent.
(2) If distinct, then there are uncountably many points along the line segment with the two points as open end-points.

Standby for a classic Doron misinterpretation in 5...4...3...
 
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