~UNIFORM CONTINUITY~

~Being connected is a nice way of thinking of continuity. (see link, differentiability & continuity in the review section).

~Question: "Is there only one type of continuity?"
  Answer: "no"

~A function can be connected (continuous) in a "stronger way". To understand this, first review basic continuity at my review link.

~You will see that being continuous at a point depends on a special limit holding:   

 If  lim  f(x)  =  f(a),  then f is continuous at x = a.
    x→a

~Using Cauchy's definition for this limit (see link in review section), for any positive e, there exist a positive d, such that if x is in the d-neighborhood of a, then the f(x) is within e of f(a). In this case, the d depends on the choice of e and the specific point a.

~In essence, uniform continuity states that the choice of d will depend on e alone and not any specific point in the interval.

~Informally, this means that for any e>0 and any two x values in the interval, we can find a d >0, such that, whenever these x values are within d, their functional values are within e.

~Saying it another way, this means that x values close to each other, give functional values close to each other. This is a stronger type of continuity and a more rigid requirement than regular continuity.

~Of course, if a function meets this requirement, it certainly will meet the lesser one (being uniformly continuous implies regular continuity).

~It's important to note that the same d must work throughout the interval for x values in the interval. Hence, d depends on e alone not where we are in the interval.

~To illustrate this with an example, let's take the classic example often given in advanced calculus & real analysis courses. That is , f(x) = 1/x over the open interval (0;1).

~First, let me prove that it is continuous over (0;1). The easy way to do this is to establish the existence of a derivative at each point in (0;1).
Since f '(x)=- 1/x², we can see that f is differentiable there, so, f is continuous there (see link, differentiability & continuity in the review section).

~Now, to show f(x) = 1/x is not uniformly continuous in (0;1). This will take a bit more work.

~I will be using an indirect proof (very important type & used widely in advanced courses). It's a proof of contradiction. That is, we will assume f is uniformly continuous over (0;1) and show that this is impossible (this is the contradiction). So, our assumption is false & f will not be uniformly continuous there.

~Note: In Latin it's called "Reductio Ad Absurdum" or RAA for short. Basically meaning, "reduction to the absurd". The absurd is the resulting contradiction achieved in this "indirect proof".

~Here we go.  Assume f(x) = 1/x is uniformly continuous in (0;1). That means for any e>0 we should be able to find a d>0, such that , for x₁, x₂ in (0;1), whenever these values are within d of each other, then f(x₁) and f(x₂) are within e of each other.

~Using more mathematical formulation,  let e>0 be given, and since we are assuming f uniformly continuous, there is a d>0, such that, |x₁ - x₂| < d
gives |f(x₁) - f(x₂)| <e, by definition.

~But |f(x₁) - f(x₂)| = | (1/x₁) - (1/x₂)| = | (x₂ - x₁)/x₁x₂ | = |x₁ - x₂ |/|x₁x₂| < d/x₁x₂ < e.  (remember e has been chosen already and the d must work for that choice).

~This is impossible, since x₁ and x₂ taken sufficiently close to zero will make that fraction as large as we want (denominator can be made as small as we like by taking very small x values)(the product of two numbers near zero can be made arbitrarily small). So, it can be made to be greater than e, a contradiction of the uniform continuity property. So, we reject our assumption that f is uniformly continuous on (0;1) and conclude that is isn't.

~In essence, the functional values can be made to separate as far apart as we like by taking x values very close to zero.

~Generally speaking, any function which is unbounded (see link on LUB & GLB in enrichment section) on an open interval, cannot be uniformly continuous on that interval.

~Also, if a function is continuous on a closed interval, it will automatically be uniformly continuous on that interval.

~A very important place where this is used is in the proof of the fundamental tie between the limit of a Riemann Sum and the area under a curve (see link, Area under a curve in the review section).

~This is an advanced concept and is useful in proving theorems (as noted) and other important properties in real analysis.