uniform continuity

Quote:

Originally Posted by particlejohn

Show that the function $f: (0,2) \to \bold{R}$ defined by $f(x) := 1/x$ is not uniformly continuous. I want to show this using the following corollary: Let $X$ be a subset of $\bold{R}$ , let $f: \bold{X} \to \bold{R}$ be a uniformly continuous function, and let $x_0$ be an adherent point of $X$ . Then $\lim_{x \to x_{0}: x \in \bold{X}} f(x)$ exists and is a real number.

So $0$ is an adherent point of $X = (0,2)$ , because $[0- \varepsilon, 0+ \varepsilon]$ contains a point of $X$ for every $\varepsilon > 0$ (its probably assumed that $0 < \varepsilon < 1$ ). And $\lim_{x \to 0; x \in \bold{X}} f(x)$ does not exist, which implies that $f(x)$ is not uniformly continuous. I used the contrapositive of the corollary.

Is this correct?

Yes, but you need more carefull wording, something like:

Assume $f(x)$ is uniformly continuous, then by the corrolarly

$\lim_{x \to 0: x \in \bold{X}} f(x)=c; \ \ c \in \mathbb{R},\ \ \ \ \ ...(1)$

that is the limit exists and is a real number as $0$ is an adherent point of $X = (0,2)$ .

But

$\lim_{x \to 0: x \in \bold{X}} f(x) =\infty$

which contradicts $(1)$ , hence the assumption that $f(x)$ is uniformly continuous is false and $f(x)$ is not uniformly continuous.

RonL