1. ## uniform continuity

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

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

Is this correct?

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

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

Is this correct?
Yes, but you need more carefull wording, something like:

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

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

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

But

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

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

RonL