Questions about completeness axiom I'm now going through the completeness axiom. What is it used for? What can you do with it? Why is it called completeness?
And beside that, how can you proof this theorem?
Suppose that S is a nonempty subset of R and k is an upper bound of S. Then k is the least upper bound of S if and only if for each $\epsilon > 0$ there exists $s \in S$ such that $k - \epsilon < s$.
I tried picking a random $\epsilon$, but then I come to the point $k-s<-\epsilon$..
 A: The completeness axiom is probably the most important concept in real analysis. Every theorem in real analysis follows from it; for instance, every convergent sequence of real numbers has a real limit (which is not the case for, say, rational numbers, which are not a complete field). The fact that real numbers are a continuum (which is implied by completeness) allows you to derive most results in calculus, etc.
Also, as its name implies, the completeness axiom is an axiom, not a theorem, therefore, there's no proof for it (at least if you're using an axiomatic, that is, non-constructive, definition of real numbers).
A: "Completeness" refers to a wide variety of results that describe how the real line is, well, 'complete' -- that there are no holes, that it's not missing anything that it ought to have.
For example, consider the "Elementary Continuity Principle" from Euclidean geometry:

Let C be a circle. Let P be a point inside of C. Let Q be a point outside of C.
  Then, the line segment PQ intersects C.

This type of geometric fact is an example of something that depends on completeness. An algebraic example (which is effectively the same as the above) is

Every positive real number has a square root.

With just rational numbers, for example, both of the above fail. For example, 2 doesn't have a square root -- the rational numbers are "missing" some numbers. A counterexample to the geometric version is to consider the rational plane (every point has rational coordinates), and let


*

*C be the circle centered at (0,0) with radius 1

*P be the origin (0,0)

*Q be the point (1,1)


Then the line segment PQ starts inside the circle, ends outside the circle, but doesn't ever intersect the circle -- the circle has "holes"!
Whatever particular statement you have been introduced to as the "completeness axiom" was chosen for the sake of pedagogy and/or tradition. You shouldn't get the impression that that axiom is what completeness 'means' -- instead, that axiom is just one of a wide variety of facts related to completeness.
A: As for the theorem you're trying to prove: Suppose $k$ is the least upper bound of $S$ and that there's an $\epsilon$ such that $k-\epsilon>s$ for any $s\in S$; but then $k-\epsilon$ is an upper bound of $S$ and $k-\epsilon<k$ so $k$ is not the least upper bound of $S$. The proof for the converse of this statement is quite similar.
This fact can be used to prove that, given a bounded, non-empty set $A$, there's a sequence $(x_n)$ of elements of $A$ that converge to $a$, where $a$ is $A$'s least upper bound. You can construct such a sequence choosing $x_n\in A$ such that $a−\frac 1n<x\leq a$. Then $|x_n−a|<\frac 1n$ for all $n$, so, given $\epsilon>0$, $|x_n−a|<\epsilon$ for a big enough $n$. This uses the fact we have proven to choose an appropriate $x_n$. As a sidenote, we can choose the smallest $x_n$ possible at each step to avoid using the axiom of choice.
A: completeness axiom is ''every non empty subset of R which is bounded above has the least upper bound''
or,
''every non empty subset of R which is bounded below has the greatest lower bound''
The above two are axioms so there is no need to prove it, if any one wants to prove 1st of axioms then he has to use 2nd or for trying to prove 2nd then must aply 1st.
