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
- Charles R. MacCluer, Honors Calculus, Princeton University Press, 2006
- Topological Space, Wikipedia.Com
- "An Introduction to Metric Spaces", BBC Mathematics Page
4 comments :
Nice proof. I recently had to do it similarly using Lebesgue number lemma.
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!
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
This is an example of proof involving sequential compactness
http://en.wikipedia.org/wiki/Heine%E2%80%93Cantor_theorem
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