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Wow ! What fun !
CracnkFan, you missed the main point of my argument.CrankFan said:it was a major mistake of yours to think that f:A->B alone, implies that f is a bijection.
If you do not understand my paper, then you cannot show us any meaningful conclusion about it.Your paper is most understandable at the beginning ...
What are you talking about?CrankFan said:Matt already pointed out if your "reals" aren't equivalent to our reals...
ex-xian said:Thanks for the kind words...(this thread is becoming a mush fest).
Lama said:Please give your detailed epsilon-delta proof, which is not based on a proof by contradiction, and please add an informal explanation to each formal part of it, thank you.
ex-xian said:The definition of a limit is as follows:
lim_x->c f(x) = L (the limit, as x tends to c, of f(x) is L) means that given an ε>0, there exists a δ>0 such that 0<|x-c|<δ implies |f(x) - L|<ε.
The point is that given any epsilon, you can always find a delta that corresponds to that epsilon so that the above implication holds. There is no magic delta that holds for all epsilon.
Here's a very simple example of a proof that the lim_x->2 2x+3 = 7, written in line form rather than paragraph form to make it easier to refer to.
1) Let ε>0
2) We seek a δ such that if 0<|x-2|<δ then |(2x+3)-7|<ε.
3) Let δ=ε/2.
4) Let |x-2|<δ, that is |x-2|<ε/2.
5) Then 2|x-2|<ε,
|2x-4|<ε, and
|(2x+3)-7|<ε.
6)That is |f(x)-L|<ε. So, by definition, lim_x->2 2x+3 = 7. qed
1) An arbitrary epsilon is "chosen." The epsilon is abitraray so no generality is lost, but in no way does it stand for every epsilon greater than zero.
2) Just a statement of what we're trying to do. For a given epsilon, there is one of more corresponding deltas that fulfill the requirments. For example, if ε=1, then a delta of 1/2 would be sufficient.
3) The delta is chosen. This is accomplished by manipulating |f(x)-L|<ε to get it in the form of |x-c|<(an expression that involves ε). In this example:
|(2x+3)-7|<ε
|2x-4|<ε
2|x-2|<ε
|x-2|<ε/2.
This gives the connection between the chosen epsilon and the delta.
4)Statement of definition.
5)Showing definition holds.
6), 7) Closing statements of proof.
Thank you,Lama said:Hi ex-xian,
Thank you for the very clear post about epsilon-delta proof.
It is based on a bijection (edit: insted of bijection please read injection) between infinitely many arbitrary ε to their unique δ, where each connection is a unique 1-1 case (there is no magic δ for all ε, as you said).
Well my friend, in my paper I am talking about the logical (general) meaning behind any epsilon-delta proof, which in this case uses ε and δ connection to show how any given interval |x-x0| implies δ=0.
Let us write this logical proof by contradiction:
If |a-b| = δ < all ε > 0 then δ = 0.
Proof:
Let us say that δ > 0
1) δ < all ε > 0
2) δ > 0
Since δ < all ε > 0 and d > 0 then δ<δ that cannot be true, so (1) and (2) cannot both be true.
Therefore, it is true that If (1), then not (2) --> δ = 0, QED (a proof by contradiction).
In my paper I use this proof to show that it is limited to an excluded-middle logical reasoning (exactly as there is no magic δ for all ε).
This is some example of re-examination of fundamental mathematical concepts, and in this particular paper I re-examine:
1) Logical reasoning.
2) Limit.
3) universal quantification.
Please read again http://www.geocities.com/complementarytheory/ed.pdf , thank you.
Lama said:Hi ex-xian,
Thank you for the very clear post about epsilon-delta proof.
It is based on a bijection between infinitely many arbitrary ε to their unique δ
where each connection is a unique 1-1 case (there is no magic δ for all ε, as you said).
ex-xian said:No such bijection exists. Although no delta works for every epsilon, for a given epsilon there is an infinite number of possible deltas.
Lama said:Well my friend, in my paper I am talking about the logical (general) meaning behind any epsilon-delta proof, which in this case uses ε and δ connection to show how any given interval |x-x0| implies δ=0.
ex-xian said:That's not what a delta-epsilon proof is supposed to show. I gave the formal definition in my previous post.
Lama said:Let us write this logical proof by contradiction:
If |a-b| = δ < all ε > 0 then δ = 0.
ex-xian said:Well, again, this isn't what a delta-epsilon proof is for. Also, this is trivially true. The only non-negative number that is less than every postitive number is 0.
A delta-epsilon is used to show the existence of a limit. What limit and function are you working with?
ex-xian said:for a given epsilon there is an infinite number of possible deltas.
Lama said:...in this case we use ε and δ connection to show how any given interval |x-x0| implies δ=0.
Lama:ex-xian said:this is trivially true
Yes.kaiser soze said:Did you mean "let us say that there exists delta such that delta >0" ?
Lama said:Dear kaiser soze,
Math has many faces, and if you choose to ingore it, then we have no dialog between us.
Dear Russell E. Rierson, thank you, but I want to add that I have no problem with dogmatic approach of others, if I can use it to develop my work.Don't let the dogmatic zealots dampen your spirits with their flatulant barkings.
It depends on the framework that we choose to work with.So we see that:
[paradox] = not-[paradox]
is a paradox of course!
Lama said:Dear Russell E. Rierson, thank you, but I want to add that I have no problem with dogmatic approach of others, if I can use it to develop my work.
Take for example persons like Matt Grime, which in my opinion make here a very good job as the bodyguard of Math.
It took me some time (almost 2 years) to understand that I am talking to a full time job bodyguard, so now I take what I take and I do not care anymore that full time job bodyguards do not want to or can’t understand my work.
It depends on the framework that we choose to work with.
In an excluded-middle reasoning a = not_a is nothing but a false statement.
Lama said:There is no question here.
a is not_a is nothing but a false statement in boolean logic, because no identity can be in more than one unique state in boolean logic.
In logic we can say that our true result is a false statemant.If the statement contradiction = not-contradiction is a contradiction is false, the statement contradiction = not-contradiction is not a contradiction is true?
'=' is used here for the tautology of a = a.Matt Grime said:note the correct use of iff, sometimes denoted <=>, and not =, since 'equals' is not an operator in boolean logic
As usual, you miss the point.Matt Grime said:A and not_A
If we look at this propositoin, we can say that within an excluded-middle reasoning, if a self reference of a proposition changes the propositon, then and only then it cannot be referred to itsef, because in an excluded-middle reasoning, each element has exactly one and only one uniqe identity.No proposition can make a statement about itself...