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Sometime I read that Dedekind's axiom is a continuity axiom, and sometimes I read that it's a completeness axiom. Besides Dedekind's axiom is equivalent to other properties as I read here in The Main Theorems of Calculus. Are all theese called continuity properties or completness properties ? if there is a difference what is it ?

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    Dedekind himself called it continuity, he meant it in the sense that we now call completeness. – Sort of Damocles Jan 21 '18 at 14:49
  • Well those different versions (mentioned in my answer which you linked) are different ways/perspectives to answer the following question: how is the system of real numbers different from that of the rational numbers? Also these versions may not be equivalent when we go to more general systems like metric spaces/topological spaces. The relationship between these notions in general spaces is one of the interesting topics in abstract analysis. – Paramanand Singh Jan 21 '18 at 15:09

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As far as continuity

There are certain gaps on the ordered set of rational numbers. Consider the quantity $\tau$ characterized by the property that its square is $2$. There does not exist such a rational number. But there are rational numbers which are less than $\tau$ and there are rational numbers which are greater than the same. The two sets partition the set the rational numbers. This is a Dedekind cut with $$A=\{r\mid r<\tau\}\ \text{ and } \ B=\{r\mid r>\tau \}.$$ We can identify all rational numbers with a Dedekind cut. But some (most) Dedekind cuts cannot be identified with rational numbers. There are holes on the rational line. In order to fill these wholes one can consider all the Dedekind cuts as numbers -- an extension to the set of rational numbers. This extension makes the rational line continuous : in some sense the holes of the rational numbers have been patched.

So, the rational numbers are not Dedekind complete because the set of Dedekind cuts, the set of reals, is larger than the set of rational numbers.

As far as completeness

But, is the set of real numbers Dedekind complete? Let's not call the set of Dedekind cuts of rational numbers reals, call them ratcuts and ask the question again: Is the set of ratcuts Dedekind complete? What does characterize the ratcuts? Ratcuts is an ordered field. Can we add further elements, by the Dedekind method, to the field of ratcuts in such a way that the result remains an ordered field? Consider, as an example, the ordered field called surreal numbers. This is an extension of the ratcuts!

Then what makes the real numbers Dedekind complete? Nothing but the axiom claiming that there are no more Dedekind cuts but those which are considered to be real numbers. Better said: we don't consider more Dedekind cuts but those given by the rational cuts.

Hey, then what is an axiom? Is an axiom not an obvious statement that we don't want to prove. As the example of the real numbers shows, axioms are not related to some "reality" in the mirror of which a statement is obvious. Axioms form the (Platonic) reality... Axioms lead, not follow.

zoli
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  • It would have been better to avoid the symbol $\sqrt{2}$ in defining $A, B$. Also the rationals don't have any gaps inherently. It is rather that we want some properties in our number system (motivated by the idea of an abstract geometric notion of a straight line, at least that was the case for Dedekind) which are simply not possessed by rationals. – Paramanand Singh Jan 21 '18 at 15:13
  • @ParamanandSingh: Than you. I will try to improve my answer. – zoli Jan 21 '18 at 15:15
  • By the way rest of your answer is great and +1 for "axioms lead not follow". – Paramanand Singh Jan 21 '18 at 15:16
  • @ParamanandSingh: Thank you again. I've edited the first part. – zoli Jan 21 '18 at 15:23
  • And I forgot to mention about invention of the word ratcuts!! Funny term indeed. – Paramanand Singh Jan 21 '18 at 15:26
  • You can write $$A={x\mid x\in\mathbb{Q}, x\leq 0}\cup {x\mid x\in\mathbb{Q}, x>0,x^2<2} ,B=\mathbb{Q}-A$$ This avoids any facts related to the existence of some number whose square is $2$. – Paramanand Singh Jan 21 '18 at 15:30