## 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

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!