Continuous but not Uniformly Continuous: An Example

A uniformly continuous function is always continuous. But the converse is not true. For example, f(x)= 1/x on (0, 2). In this post, we will provide an example which is continuous but not uniformly continuous. Before we do that let us recall their definitions.

Continuous Function: A function f(x) is said to be continuous at x=a if for every ε>0 there exists a δ>0 (depending on ε and the point x=a, so we write δ:=δ(ε, a)) such that

|f(x)-f(a)| < ε for all |x-a| < δ.

Uniform Continuous Function: A function f(x) is said to be uniformly continuous at x=a if for every ε>0 there exists a δ>0 (not depend on ε and the point x=a) such that for any two points x₁ and x₂ with |x₁ – x₂| < δ we have

|f(x₁)-f(x₂)| < ε.

To show continuity does not imply uniform continuity, let us consider the following function.

An Example

f(x) = 1/x is continuous on (0,2) but not uniformly continuous

Let

$f(x)=\dfrac{1}{x}$ on (0, 2).

See that f(x) is continuous on (0,2) as the only point of discontinuity is x=0. We now show that f(x) is not uniformly continuous.

Note that if f(x) = 1/x was uniformly continuous on (0, 2) then we would have that ∀ ε>0 ∃ a positive δ such that

|f(x)-f(y)| < ε whenever |x-y| < δ.

That is,

$|\dfrac{1}{x}-\dfrac{1}{y}|$ < ε whenever |x-y| < δ …(∗)

Take ε = 1 and x:= min{δ, ε} = min{δ, 1} where δ>0 is arbitrary. Also, consider $y=\dfrac{x}{2}$. Then

$|x-y|=|x-\dfrac{x}{2}| =\dfrac{x}{2}<\delta$.

But,

$|\dfrac{1}{x}-\dfrac{1}{y}|$ = $|\dfrac{1}{x}-\dfrac{2}{x}|$ = $|\dfrac{1}{x}|$ ≥ 1 = ε.

This contradicts the fact (∗) above, showing that f(x) is not uniformly continuous on(0, 2). So f(x) = 1/x is continuous on (0, 2) but not uniformly continuous.

Another Example

Consider the function

$f(x)=\dfrac{1}{x}$ on (0, 1].

It is a continuous function. To show it is a uniformly continuous, for every pair of sequences {a_n} and {b_n} in (0, 1] with lim |a_n – b_n| = 0 the following holds:

lim |f(a_n) – f(b_n)| = 0.

Take a_n = 1/n and b_n = 1/2n. Then

lim |a_n – b_n| = lim | 1/n – 1/2n | = 0.

But, lim |f(a_n) – f(b_n)| = lim n ≠ 0.

So f(x) = 1/x on (0, 1] is not uniformly continuous but it is a continuous function.

You Can Read:

• Intermediate Value Theorem
• Fixed Point Theorem
• Completeness Axiom of Real Numbers
• Supremum & Infimum of a Set
• Archimedean Property

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