Epsilon = Invariant Proportion

[bloodhound]

First of all I am not familar with all the terms you use, can u define the following:-

 

"Accurate Definition"

"endless thing"

"accurate result"

"endless state"

 

Identity of S? I was talking about the Identity map of S. i dont even know what identiry of S is.

 

You claim the identity map is not a bijection...

 

I will prove it is for [math]S=\mathbb{R}[/math]

 

showing injectivity of the identity map

 

if [math]f(a)=f(b)[/math]

then, by definition of the mapping [math]a=b[/math]. Therefore the map is injective.

 

Showing surjectivity:-

For each element in the target say [math]x\in \mathbb{R}[/math] there is an element in the source [math]y\in \mathbb{R}[/math] which is mapped by the function to x. In this case y=x

 

Therefore the identity map is a bijection.

 

Can u find a flaw in that arguement?

 

[edit] Oh identity of S. u mean the additive or the multiplicative identity[/edit]

[Dave]

Here's an ultimatum for you Doron:

 

Either post some mathematics that uses formal definitions and makes sense or I will be forced to exclude you from the mathematics forum. I have warned you about posting your "ideas" on here and I'm not prepared to tolerate yet more talk of xor's and how everyone in mathematics is wrong about everything.

 

Posting bad mathematics is acceptable. Ignoring everyone else's criticism and posting the same tripe over and over again is not.

[lama]

Dear Bloodhound,

 

Thank you for you questions:

 

"Accurate Definition" is Well-defined.

 

"endless thing" is unbounded.

 

"accurate result" is one and only one result.

 

"endless state" is infinitely many ...

For each element in the target say [math]x\in \mathbb{R}[/math] there is an element in the source [math]y\in \mathbb{R}[/math] which is mapped by the function to x. In this case y=x

Since [math]\mathbb{R}[/math] is a collection of infinitely many elements' date=' then we need a universal quantification in order to prove the bijection in this case.

 

But then we see that is we use a universal quntification, then our set does not exist or our set is a finite set.

 

Since [math']\mathbb{R}[/math] has infinitely members, and since we need a universal quantification in order to prove the bijection, we cannot prove that there is a bijection.

[bloodhound]

ok u still havent commented on my verification that the identity map a bijection... in the proof take S=Z for example .. z is well defined and unbounded.

 

Also what is "finite or infinitely many products of definition"

In you last post can u explain the second paragraph properly. Include definitions of all unorthodox mathematical terms you use please.

[matt grime]

"Completed" means that its end can be found.

 

define end. do you mean max or sup or something? what if it weren't an ordered set? Is the set of all elliptic curves one with an end?

[lama]

Dear Bloodhound,

 

I'll do my best to explain again post #25 in a more formal way, but I need your help in order to do it right, so here it is again:

 

If [math]S[/math] has is well-defined, then the definition is [math]S[/math] and not any finite or infinitely many products of this definition.

 

So mapping between infinitely many elements cannot return the Identity map of [math]S[/math] (please look at post #23 in order to understand this).

 

If you want to find the Identity map of [math]S[/math], look at its definition (which does not depend on 'Quantity', 'Size', 'Magnitude', 'ALL' s in S, etc...).

 

But if we say that one of the properties of [math]S[/math] is to be infinite, then [math]s\mapsto s[/math] (identity map) is exactly this unbounded state, and not a bijection of it.

 

Please read again post #23 in order to understan my idea here, and also ask me about any non-convetional expressions of me, so we can together write it in its formal way.

 

Thank you.

 

Yours,

 

Doron

[matt grime]

f(s)->s isn't the correct notation for a map, Doron. Tut, and you think you know something about mathematics?

[lama]

Well, in this case I used bloodhound notations for identity map, so if they are not correct then please correct them, thank you.

 

As for your question about ends.

 

Take any non-empty finite collection and then you can find its finite cardinal and also (if this collection is well ordered) its first AND lest elements.

[bloodhound]

lol. i corrected mine.... u didn't even see the schoolboy error... tut tut... i am sure you should know how to write a mapping instead of using others notations. reply to your post coming soon. hopefullly

[matt grime]

But what about infinte sets that have maximal and minimal elements, Doron?

 

Oh, please define cardinal too.

 

Bloodhound is a second year undergraduate, I can forgive him his errors. People who claim to have earth shattering views on mathematics that are plain silly can't be forgiven for repeating mistakes. Not least since it is immaterial whether you think there are infinte sets or not. If you disallow them then it's your problem - we have them and they cause no problem to us, just interesting questions.

 

Oh, and yo'uve not defined a set either. Clearly it can't be a collection of things, since that allows for infinite sets and here you say:

 

"But then we see that is we use a universal quntification, then our set does not exist or our set is a finite set"

 

of course, it may just be that you don't understand universal quantification...

[lama]

ok, corrected to [math]s\mapsto s[/math] , thank you.

[lama]

Matt,

 

We are talking here about ideas and not about accurate formal nonations, so please do not use your energy on less important things like corrected formal notations, and open your mind to the ideas.

 

The Ideas are not any formal notations of them.

 

Oh' date=' and yo'uve not defined a set either. Clearly it can't be a collection of things, since that allows for infinite sets and here you say:

 

"But then we see that if we use a universal quntification on a collection of infinitely many elements, then our set does not exist or our set is a finite set"

[/quote']

Please read Post #23, and also please explain us what is a universal quantification, thank you.

 

But what about infinte sets that have maximal and minimal elements' date=' Doron?

[/quote']

 

If you mean to the set which is based on [0,1] then if 0 and 1 are included in the set then this set has finitely many elements, because in any case that the maximum and the minimum are in the set, then there is a discontinuous function between any arbitrary member, which is not 0 or 1 (in this case) and 0 or 1, and the rusult is a set with finitely many elements.

[bloodhound]

If you mean to the set which is based on [0' date=1] then if 0 and 1 are included in the set then this set has finitely many elements, because in any case that the maximum an the minimum are in the set, then there is a diccontinuous function between any arbitrary member, which is not 0 or 1 (in this case) and 0 or 1, and the rusult is a set with finitely many elements.

I got no idea what this paragraph says..

 

"A discontinuous function between any arbitary member"? what is that? once you explained what it is, i would like to see the proof of existance.

[lama]

Matt wrote:

 

"But what about infinte sets that have maximal and minimal elements, Doron?"

 

So dear bloodhound, please show us such a well-ordered set where both its minimanl and maximal members are included in it, and it is also a set with infinitely many members.

 

If you read again posts #1 , #19 and #23 you will find out why this set cannot be found.

[bloodhound]

u still havent replied to my last question. I havent officially studied "well ordered set" but have only a vague notion of it. so i can't give u an example.

[bloodhound]

offtopic: i just found out that doron and matt have a long history at physicsforum... just wondering what happened? did he get banned from there?

[lama]

"A discontinuous function between any arbitary member"? what is that? once you explained what it is' date=' i would like to see the proof of existance.

[/quote']

 

Between 0 and 1 there are infinitely many [math]\mathbb{R}[/math] members as long as there is a continuous function between them.

 

But if we want to include 0 and 1 in [math]\mathbb{R}[/math], then it can be done only by a discontinuous function between any [math]\mathbb{R}[/math] member (which is not 0 or 1) and between 0 or 1.

 

If both discontinuous functions between [math]\mathbb{R}[/math] members (which are not 0 or 1) and 0 and 1 are defined, then the collection of [math]\mathbb{R}[/math] members between 0 and 1 is a finite collection.

 

The reason is very simple:

 

Any function between epsilon=0 and epsilon>0 can be only a discontinuous function.

[matt grime]

I'll give you an example of a well ordered set with infintely many elements and a maximal element:

 

the ordinals

 

w+1, w+2,.. w+r..., pw+r,..

 

are all well ordered sets with a maximal element.

 

A model of one of them in the reals would be

 

{1/2,2/3,4/5.5/6,...} u{3/2}

 

interestingly you can use infinite ordinals such as these and transfinite induction to show that a continuous function on a closed boudned interval is bounded.

 

you might want to look up the definition of continuous Doron (or function for that matter) since you appear not to know what that means either.

[lama]

Dear Matt,

 

but in order to define that w is a limit ordinal (has no predecessor) you first have to show that the transfinite universe exists.

 

And it cannot exist, because no set of infinitely many elements is a complete collection (include also its maximal AND minimal elements).

 

So, your example is circular because you are using a transfinite element in order to show the existence of a transfinite element, but you do not show how this transfinite element (w in this case) can exist, by using a universal quantification on a collection of infinitely many elements and also to save it as a magnitude of infinitely many elements.

 

I clearly and simply show why it can be done, but you did not show how it can be done.

 

All you show is that some how out of the blue, there exists w.

 

If you cannot explain how w exists, then you cannot use it as a model.

 

you might want to look up the definition of continuous Doron (or function for that matter) since you appear not to know what that means either.

In a continuous function a small input gives a small output, and this is not the case if we define a function between one of the memebers, which is not maximal or minimal, and one of the members which is maximal or minimal (as I clearly show in post #42).

[bloodhound]

 

If both discontinuous functions between [math]\mathbb{R}[/math] members (which are not 0 or 1) and 0 and 1 are defined' date=' then the collection of [math']\mathbb{R}[/math] members between 0 and 1 is a finite collection.

it seems that with each new post you introduce terms that highly unorthodox.

 

you can find many discontinuous functions in the interval [0,1]

 

the most famous being

[math]\chi \colon [0,1] \to \{0,1\}[/math] where

[math]\chi(x)=0[/math] if x is rational, and

[math]\chi(x)=1[/math] if x is irrattional.

 

Do u still contend that [0,1] is a finite collection?

 

[altough i do not understand what the existance of the discontinuous function has to do with anything at all]

 

Recommendation: Look up Cantor's Diagonal Arguement.

[lama]

Dear Bloodhound,

 

Let us take for example the real line.

 

If we say that our collection is the closed interval [0,1], then 0 AND 1 are members in our collection.

 

If we use a function between [0 and x, than this function cannot be but a discontinuous function, and there are no members in the domain of the discontinuous function.

 

The same holds between y and 1] so we get:

 

Let cf be a continuous function.

 

Let df be a discontinuous function.

 

[0 <-df-> x1 <-df-> x2 ,... , Epsilon>0 ... <-cf-> ... Epsilon>0 , ..., y2 <-df-> y1 <-df-> 1]

 

As you can see, we always has two separated collections:

 

1) A collection which is based on a continuous function that has infinitely many elements.

 

2) A collection of infinitely many elements which is based on infinitely many discontinuous functions (where each member in this set is the result of a quantum leap between the continuous function to a discontinuous function.

 

Since we get two disjoined sets (one is based on cf and the other is based on df), we cannot use a universal quantification on both of them, and also not on each one of them, because in the case of infinitely many elements in both of them, there is always an Epsilon>0 between them.

 

In other words, because we have a permanent Epsilon>0, no Universal Quantification can be used on a collection of infinitely many elements.

 

By the way there is also the case that the collection of discontinuous functions is a finite collection (as I wrote in my previous posts), but then it is understood that this set has a universal quantification, where the set that is based on a continuous function, has no universal quantification (because Epsilon > 0 ).

 

This is, by the way, the reason why Dedekind's Cut is problematic, because also in this case we have Epsilon>0, for example:

 

L Epsilon>0 c Epsilon>0 R and we cannot find all of Q members in L and R sets.

 

Also we have to understand that if Epsilon=0 then L member = c = R member.

[matt grime]

Dear Matt' date='

 

but in order to define that w is a limit ordinal (has no predecessor) you first have to show that the transfinite universe exists.

 

And it cannot exist, because no set of infinitely many elements is a complete collection (include also its maximal AND minimal elements).

 

So, your example is circular because you are using a transfinite element in order to show the existence of a transfinite element, but you do not show how this transfinite element (w in this case) can exist, by using a universal quantification on a collection of infinitely many elements and also to save it as a magnitude of infinitely many elements.

 

I clearly and simply show why it can be done, but you did not show how it can be done.

 

All you show is that some how out of the blue, there exists w.

 

If you cannot explain how w exists, then you cannot use it as a model.

 

 

In a continuous function a small input gives a small output, and this is not the case if we define a function between one of the memebers, which is not maximal or minimal, and one of the members which is maximal or minimal (as I clearly show in post #42).[/quote']

 

 

So even after I explicitly show you an infinite well ordered set (as a collection of rational s yuo deny such a thing is possible? Wow, you are blind. Incidentally, what on earth do you mean by "exist"?

 

w is the set of natural numbers, something no one vut you seems to have a problem with. It exists since we declare it to (or have you got a "2" hanging around -an you deny being a constructivist?)

[lama]

w is the set of natural numbers' date='

[/quote']

w is a limit ordinal and it is a transfinite element (by Cantor).

 

N is the set of Natural numbers, and the Natural numbers are defined by their axioms and not by N.

 

For example, in Peano's axiomatic system the Set concept does not exist.

you deny being a constructivist?)

Yes' date=' because there are infinitely many [i']n[/i] in N but because what a clearly show in post #1 |N| cannot be defined and all we can get is |N| - epsilon.

[bloodhound]

So even after I explicitly show you an infinite well ordered set (as a collection of rational s yuo deny such a thing is possible? Wow, you are blind.

Did u read his last post? he just have you an example of an infinite well ordered set

 

"A model of one of them in the reals would be

 

[math]\{1/2,2/3,4/5,5/6,...\} \cup \{3/2\}[/math]"

[matt grime]

You mean apart from the fact that peano's axioms are usually cosntructed by sets? N, the natural numbers is a model of w, the first infinite ordinal. You do understand that? w is a (an equivalence class of) set(s): a well ordered countable set with no maximal element, it is naturally isomorphic to N with its usual ordering.