Results 1 to 2 of 2

Thread: uniform continuity

  1. #1
    Member
    Joined
    Jun 2008
    Posts
    170

    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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    5
    Quote Originally Posted by particlejohn View Post
    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. uniform continuity of x^2
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: Apr 11th 2011, 04:17 PM
  2. uniform differentiable => uniform continuity
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: Nov 30th 2009, 03:19 PM
  3. Uniform Continuity
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: Nov 24th 2009, 05:43 AM
  4. uniform continuity
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: Oct 14th 2009, 05:51 AM
  5. uniform continuity
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Aug 11th 2009, 07:48 PM

Search Tags


/mathhelpforum @mathhelpforum