Theoretical part about limits and continuity




In this article all numbers are real numbers.

Dedekind's Axiom for real numbers

If L and H are two subsets of the set R of all real numbers and
  1. L and H are not empty
  2. L and H have no common element
  3. The union of L and H is R
  4. For each x in L and each y in H, we have x < y
Then, there is just one real number l such that
  1. no element of L exceeds l
  2. no element of H is smaller than l
The unique element l is called a dedekind cut.

Upper bound and lower bound of a set S

Say S is a set of real numbers.
 
                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

Bounded set

If a set S has an upper bound and a lower bound, we say that the set is bounded.

A least upper bound

Theorem
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

  1. no element of L exceeds l
  2. 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.

A greatest lower bound

Theorem
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.

About image sequences

Say f: R -> R : x -> f(x) is a real function and there is a strictly positive real number e such that ]b-e,b+e[ or ]b-e,b[ or ]b,b+e[ is part of the domain of the function f. Take the set of all sequences {x(n)} such that With each sequence {x(n)} corresponds an 'image sequence' {f(x(n))}. Suppose all image sequences converge.

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

Important property of the limit concept

 
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| < e
Proof:
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.

Theorem 1

If f(x) is continuous in b and if f(b) is positive,then there is a small environment of b such that f(x) stays positive. Proof:
 
        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) + e
Analogous we have

Theorem 2

If f(x) is continuous in b and if f(b) is negative,then there is a small environment of b such that f(x) stays negative.

Bolzano's theorem

If f(x) is continuous in [a,b] and f(a).f(b) < 0
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.

Theorem 3

If f(x) is continuous in b then f(x) is bounded in a suitable small environment of b. Proof:
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) + e
Hence, f(x) is bounded in a suitable small environment of b.

Theorem 4

If f(x) is continuous in [a,b], then f(x) is bounded in [a,b]. Proof:
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].

Theorem of Weierstrass.

If f is continuous in [a,b] , then f(x) attains a maximal and a minimal image in [a,b]. We'll prove the theorem for the maximal image.

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].


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