Uniform Continuity

Uniformly Continuous Function

We say that

f : S \subseteq ℝ^{n} \to ℝ^{m}

is uniformly continuous if for every

ϵ \in ℝ^{+}

there exists some

r \in ℝ^{+}

such that for every

x, a \in S

we have

‖ x - a ‖ < r ⟹ ‖ f (x) - f (a) ‖ < ϵ

Every Lipschitz function is Uniformly Continuous

As per title.

Going to Infinity Implies not Uniformly Continuous

Suppose that

f

is continuous on

(0, 1)

and that

\lim_{x \to 0^{+}} f (x) = \infty

show that

f

is not uniformly continuous

Suppose for the sake of contradiction that $f$ is uniformly continuous and take $ϵ = 1$ therefore we obtain some $r \in ℝ^{+}$ such that for any $x, a \in S$ if $| x - a | < r$ implies that $| f (x) - f (a) | < ϵ = 1$ .

Note that this implies that for any $a, b \in (0, r)$ we have $| f (a) - f (b) | < 1$ clearly this will lead to nonsense as it blows up around $0$ and so we should be able to find two points whose vertical distance is greater or equal to $1$ .

To do this let $x \in (0, r)$ , since we know that $\lim_{x \to 0^{+}} f (x) = \infty$ therefore by taking $δ = r, M = f (x) + 1$ we obtain some $a \in (0, r)$ such that $f (a) > M = f (x) + 1$ therefore we have that $f (a) - f (x) > 1$ which implies that $1 > | f (a) - f (x) | = | f (x) - f (a) |$ which is a contradiction, so that $f$ is uniformly continuous.

The above prove never used the fact that $f$ was continuous, is there a problem with it or can that assumption be removed?