Theorem: If f(x) is continuous on [a,b], it is uniformly continuous.
Proof: Given ε, assume there is an x from [a,b] such that no matter how small I make δ,
│f(x) - f(p)│not < ε if │x - p │ < δ.
Then f(x) is not continuous at x → there is a δ independent of x and f(x) is uniformly continuous. Same wording applies for a compact set.
Why bother with a Covering Theorem?
the trouble here, is that you don't want to be forced into picking a minimum delta that works for all x in [a,b] of 0.
for example, f(x) = 1/x is continuous on (0,1), but it's not uniformly continuous, because as x gets really close to 0, we keep having to pick smaller and smaller delta's.
so just forcing delta "small enough" isn't "good enough". we need for delta to stop shy of 0 (given epsilon).
that is: we need inf({δ > 0: |x-p| < δ implies |f(x) - f(p)| < ε for all p in [a,b]}) > 0.
you are missing something essential, here: [a,b] is a closed interval. prove the image of f is contained within another closed interval. break THAT closed interval into subintervals of length < ε. that will give you a FINITE set of deltas to choose from.
If f(x) is continuous on [a,b], given ε, for every x there is a δ >0 st │f(x) - f(p)│ < ε if │x - p│ < δ. Choose the smallest δ. f(x) is uniformly continuous on [a,b].
EDIT: I really didn't want to go to this level, but for every x I can choose delta rational above so then the delta I am looking for is the minimum of all the deltas. (if delta irrational, there is a rational number between 0 and delta). -> f(x) is uniformly continuous.
Here is a difficulty that you cannot get around.
There are infinitely many in
With each there is a from continuity.
There may be no smallest . Such a number may not exist.
There is no smallest positive number.
"Choose the smallest δ" cannot be done.
If there were finitely many, then yes it can be done. That is where the compactness of comes in.
The Weak Link in Post 1:
Given f(x) continuous on [a,b], given ε, for each x you can find a δ > 0 st │f(x) - f(p)│ < ε if │x - p│ < δ.
Superficially, the set S of δ’s has a minimum value which proves the theorem.
Weak Link:
1) ε → 0.
2) S is infinite.
1) No matter what ε is, δ > 0.
As for 2): Houston, we have a problem. If I have an infinite set of numbers all of which are > 0, the set may not have a minimum greater than 0. For example, the set of all rational numbers whose square is > 2 does not have a minimum, it has a glb. In the case of S, a glb of 0 doesn’t work.
Assume S doesn’t have a minimum δ.
Divide [a,b] into 2 closed intervals. At least one of them doesn’t have a minimum δ.
Keep dividing [a,b] into closed intervals which don’t have a minimum δ.
The nested intervals contain a limit point.
We now have a point on [a,b] which doesn’t have a minimum δ. Contradiction because f(x) is continuous at that point. Therefore S has a minimum δ proving that f(x) is uniformly continuous.