Theoretical part about limits and continuity

• 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.
  

Dedekind's Axiom for real numbers

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

1. no element of L exceeds l
2. no element of H is smaller than l

Upper bound and lower bound of a set S

 
                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

A least upper bound

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

About image sequences

• 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

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

 
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
... ...                 ...             ....            ...     ...

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

Theorem 1

 
        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

Theorem 2

Bolzano's theorem

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

 
        b - d < x < b + d  =>      f(b) - e < f(x) < f(b) + e

Theorem 4

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.

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].