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?