Sequential Criterion for Absence of Uniform Continuity

A function $f : A \to R$ fails to be uniformly continuous on $A$ iff $\exists ε_{0} > 0$ and two sequences $(x_{n}), (y_{n})$ in $A$ satisfying

| x_{n} - y_{n} | \to 0, | f (x_{n}) - f (y_{n}) | \geq ε_{0}

Proof

Using the negation of Uniform Continuity, then $f$ is not uniformly continuous iff $\exists ε_{0} \geq 0$ such that $\forall δ > 0$ we can find two points $x, y$ satisfying $| x - y | < δ$ but with $| f (x) - f (y) | \geq ε_{0}$ . Thus, if we set $δ_{1} = 1$ then $\exists x_{1}, y_{1}$ where $| x_{1} - y_{1} | < 1$ but $| f (x_{1}) - f (y_{1}) | \geq ε_{0}$ .

In a similar way if we set $δ_{n} = \frac{1}{n}$ for any $n \in N$ then $\exists x_{n}, y_{n}$ where $| x_{n} - y_{n} | < \frac{1}{n}$ but $| f (x_{n}) - f (y_{n}) | \geq ε_{0}$ . The resulting $(x_{n}), (y_{n})$ sequences satisfy the requirements described in the theorem.

Conversely, if $ε_{0}$ and $(x_{n}), (y_{n})$ are as described, it's clear that no $δ > 0$ is a suitable response for $ε_{0}$ .

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Example

Consider the function $h (x) = \sin (\frac{1}{x})$ .

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f(x) = sin(1/x)

It is continuous at every point in the open interval $(0, 1)$ but not uniformly continuous on the interval. Near 0, where the rapid oscillations take place, the domain values are close together but the range values approach being a distance of 2 apart. For example, take $ε_{0} = 2$ and set:

x_{n} = \frac{1}{\frac{π}{2} + 2 n π}, y_{n} = \frac{1}{\frac{3 π}{2} + 2 n π}

Because each of these sequences approach 0, we have $| x_{n} - y_{n} | \to 0$ and a short calculation gives $| h (x_{n}) - h (y_{n}) | = 2 \geq ε_{0}$ for all $n \in N$ . Thus, this shows that it is not uniformly continuous at this point.