## Wednesday, September 13, 2006

### Heine-Cantor Theorem

In today's blog, I will talk about uniform continuity and the Heine-Cantor theorem. I use this theorem to establish the definition of the integral.

Definition 1: Uniform Continuity

Given any positive number ε, you can find a value δ such that for all values of u,v in the interval, if absolute(u-v) is less than δ, then absolute(f(u) - f(v)) is less than ε

NOTE: This is different from the definition of continuous at a point (see Definition 1, here for definition of continuous at a point). If a function f is continuous on [a,b], then for any point u in [a,b] and for any arbitrary positive number ε, there exists a value δ greater than 0 such that if v is a number in [a,b] such that abs(u - v) is less than δ, then abs(f(u) - f(v)) is less than ε.

In the case of uniform continuity, we can find a δ which works for all points u. If a function is continuous at a point but not uniformly continuous, then there is at least one δ for a given u that is different than the other δ for the other u's.

Definition 2: Topological Space

A topological space (X,T) is a set X paired with a T which is a set of subsets of X. (X,T) must meet the following conditions;

(a) The empty set ∅ ∈ T

(b) X ∈ T

(c) The intersection of any finite number of sets in T is also in T. If (ui) is a finite set of elements in T, then ∩ui ∈ T.

(d) The union of a elements in T is also in T. If (ui) is a set of elements in T, then ∪ ui ∈ T.

In the context of a topological space, any subset of X which is an element in T is said to be an open set in relation to this topological space. Any subset of X which is not an element in T is said to be a closed set in relation to this topological space. [See Definition 1, here for definition of open set and Definition 3, here for definition of closed set]

An element of X is said to be a point.

Theorem: Heine-Cantor Theorem

Every continuous function on a compact set is uniformly continuous there

Proof:

(1) Let X,Z be metric spaces. [See Definition 2, here for definition of a metric space]

(2) Let f be a function such that f: X → Z

(3) Let us assume that f is continuous on the compact subset K of X.

(4) Let ε be a real number greater than 0.

(5) Since f is continous on K (see Definition 1, here for Definition of continuous at a point), for each x ∈ K, there exists a real δ(x) greater than 0 such that for y ∈ K:

if d(x,y) is less than δ(x), then d(f(x),f(y)) is less than ε/2.

(6) We see that the set of open balls with center x of radius δ(x)/2 cover K. [See Definition 1, here for a definition of an open ball]

(7) Using the Heine-Borel Theorem (see here), we know that there exists a finite subcovering for K such that:

K = ∪ (i=1,p) Bi where each Bi is centered at xi.

(8) Let δ be the minimum(δ(xi/2)) where i is the set of values 1, 2, ..., p.

(9) Let's assume that x,y ∈ K and d(x,y) is less than δ.

(10) Since y ∈ K and since ∪ (i=1,p) Bi is a covering for K, there exists an i such that y ∈ Bi

(11) Since the radius of each ball Bi is δ(xi)/2, we have: d(y,xi) is less than δ(xi)/2.

(12) Since d(x,y) is less than δ and δ is the minimum of all δ(xi)/2, it follows that:

d(x,y) is less than δ(xi)/2.

(13) Further, d(x,xi) less than δ(xi) since d(x,xi) ≤ d(x,y) + d(y,xi) ≤ δ(xi)/2 + δ(xi)/2 = δ(xi) [ d(x,xi) ≤ d(x,y) + d(y,xi) is true by the "triangle inequality rule" of metric spaces, see Definition 1, here]

(14) Applying the triangle inequality rule, we have:

d(f(x),f(x1)) ≤ d(f(x),f(xi)) + d(f(xi),f(y))

(15) Since d(x,xi) is less than δ(xi), we have d(f(x),f(xi)) is less than ε/2. [See step #5]

(16) Since d(xi,y) is less than δ(xi), we have d(f(xi),f(y)) is less than ε/2. [See step #5]

(17) Putting #14, #15, and #16 together gives us:

d(f(x),f(x1)) ≤ d(f(x),f(xi)) + d(f(xi),f(y)) is less than ε/2 + ε/2 = ε

(18) Thus, we have shown for every ε greater than 0, there exists a δ such that for all x,y ∈ K,:

d(x,y) is less than δ → d(f(x),f(y)) is less than ε.

QED

References

#### 4 comments :

Keven said...

Nice proof. I recently had to do it similarly using Lebesgue number lemma.

Keven said...

I know you wrote this along time ago, I'm just seeing it now though. I gotta ask, why not use sequential compactness? it shortens the proof considerably.

I just discovered this blog, gotta tell you, I think it's great!

Larry Freeman said...

Hi Keven,

Thanks the for your thoughts on the blog!

Feel free to add your ideas for shortening the proof!

My proof is taken straight out of a textbook. If it can be improved, I think that others would like to know.

Cheers,

-Larry

Anonymous said...

This is an example of proof involving sequential compactness
http://en.wikipedia.org/wiki/Heine%E2%80%93Cantor_theorem