- Dedekind's Axiom for real numbers

- Upper bound and lower bound of a set S

- Bounded set

- A least upper bound

- A greatest lower bound

- About image sequences

- Important property of the limit concept

- Theorem 1

- Theorem 2

- Bolzano's theorem

- Theorem 3

- Theorem 4

- Theorem of Weierstrass.

In this article all numbers are real numbers.

- L and H are not empty
- L and H have no common element
- The union of L and H is
**R** - For each x in L and each y in H, we have x < y

- no element of L exceeds l
- no element of H is smaller than l

A number y is an upper bound of S <=> no element of S exceeds y A number x is a lower bound of S <=> no element of S is smaller than x

If S is a (not empty) set of real numbers and S has an upper bound y,

then there is a least upper bound of the set S.

That least upper bound is also called the supremum of S.

proof:

Let H be the set of all upper bounds of S.

Let L be the set **R** \ H

By virtue of Dedekind's Axiom, there is just one real number l such that

- no element of L exceeds l
- no element of H is smaller than l (*)

1) We'll show that l is an upper bound of S

Suppose, l isn't an upper bound of S, then there is an element (l + h) in S with h > 0 .

Then, the number (l+h/2) is not an upper bound of S and so it is in L.

But, the number (l+h/2) is greater than l and so it is in H.

This is impossible because L an H have no element in common.

2)From (*) it follows that no upper bound of S is smaller than l.

If S is a (not empty) set of real numbers and S has a lower bound y,

then there is a greatest lower bound of the set S.

That greatest lower bound is called the infimum of S.

- lim x(n) = b ;
- for each n, x(n) is different from b ;
- for each n, x(n) is in the domain of f.

We'll show that all these image sequences have the same limit.

Choose such sequence x(1),x(2), ... and say the corresponding image sequence converges to a limit A. Choose a second sequence x'(1),x'(2), ... and say the corresponding image sequence converges to a limit B.

With these sequences, we construct a third sequence x(1),x'(1),x(2),x'(2), ... . Since all image sequences converge, the image sequence of the last sequence converges to C.

The first image sequence is a subsequence of the third one. So they have the same limit. A = C.

The second image sequence is a subsequence of the third one. So they have the same limit. B = C.

Therefore A = B and so, all the image sequences have the same limit c.

Then, appealing on this property we say that

lim f(x) = c b

IF lim f(x) = c x->b THEN with each strictly positive number e, corresponds a suitable strictly positive number d, such that |x - b| < d => |f(x) - c| < eProof:

If it is not so. Then there is at least one exception value e, such that it is impossible to find a strictly positive number d,such that |x - b| < d => |f(x) - c| < e .

In other words, for each arbitrary small value d, there is at least one x1,

such that | x1 - b | < d => | f(x1) - c | >= e

Now we take a sequence of descending d values with limit 0.

d1, d2, d3 ....

For all these d values we have :

For d1 there is at least an x1 such that |x1 - b| < d1 => |f(x1) - c| >= e For d2 there is at least an x2 such that |x2 - b| < d2 => |f(x2) - c| >= e For d3 there is at least an x3 such that |x3 - b| < d3 => |f(x3) - c| >= e ... ... ... .... ... ...Since the sequence {dn} converges to 0, the sequence {xn} converges to b and the image sequence {f(xn)} converges to c.

This is in contradiction with

|f(x1) - c| >= e |f(x2) - c| >= e |f(x3) - c| >= e ...QED.

lim f(x) = f(b) b <=> with each strictly positive number e, corresponds a suitable strictly positive number d, such that |x - b| < d => |f(x) - f(b)| < e Choose 0 < e < f(b) , then b - d < x < b + d => 0 < f(b) - e < f(x) < f(b) + eAnalogous we have

Then there is a real number c in ]a,b[ such that f(c) = 0.

Proof:

We'll prove the theorem for f(a) < 0 and f(b) > 0.

Take the set D = {x in [a,b] ; f(x) < 0}

D is not empty and has an upper bound b, so there is a supremum m.

We'll show that f(m) = 0.

If f(m) < 0 then there is a small environment of m such that f(x) stays negative. Then, there are elements in D greater than m and this is impossible.

If f(m) > 0 then there is a small environment of m such that f(x) stays positive. Then, there is a number smaller than m that are not exceeded by an element of D. This is impossible.

Choose a strictly positive number e. Since f(x) is continuous in b, there is a strictly positive number d such that

b - d < x < b + d => f(b) - e < f(x) < f(b) + eHence, f(x) is bounded in a suitable small environment of b.

Take the set D = { x in [a,b] ; f(x) is bounded in [a,x] }

The set D is not empty and there is an upper bound b. Hence the set D has a supremum m.

If m < b then, from previous theorem, f(x) is bounded in a suitable small environment ]m-d, m+d[ . But then, f(x) is bounded in [a,m+d[ and m is not the supremum of D.

Therefore m = b, and since f(b) is a real number, f(x) is bounded in in [a,b].

From previous theorem we know that f(x) has the least upper bound M in [a,b].

Suppose there is no x in [a,b], such that f(x)=M.
Then M - f(x) > 0 for all x in [a,b] .
Now, we construct the function g(x) = 1/(M - f(x)). This function is
continuous and strictly positive in [a,b]. From previous theorem it is bounded in [a,b]. Say s > 0 is the least upper bound. Then 1/(M - f(x)) does not exceeds s in [a,b]. Hence M - 1/s >= f(x) . This is impossible since M is the least upper bound M in [a,b].

The tutorial address is http://home.scarlet.be/math/

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