Results 1 to 2 of 2

Math Help - Continuity mk4

  1. #1
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782

    Continuity mk4

    I just want to make sure i'm doing these correctly.

    Prove that f(x)=\begin{cases}x \ \ \ \ \ x \ \ rational\\ x^2 \ \ \ \ x \ irrational \end{cases} is continuous at 0.
    I need to find |f(x)-f(x_0)|<\epsilon when 0<|x-x_0|<\delta.
    Let x_0=0 (since i'm interested in continuity at the origin).
    Therefore I need to find |f(x)|<\epsilon when 0<|x|<\delta.

    Let 0<\delta<1.
    |f(x)|<\epsilon \Leftrightarrow \ |x|<\epsilon for x \in \mathbb{Q}.
    |f(x)|<\epsilon \Leftrightarrow \ |x^2|<\epsilon \Leftrightarrow |x|< \epsilon (since |x^2| \leq |x|) for x \in \mathbb{R}\ \mathbb{Q}.

    Therefore pick \delta such that \delta=min\{\frac{1}{2}, \epsilon \}.

    That's the preliminary work done, so here's the proof.

    Let \epsilon>0 and define \delta:=min\{\frac{1}{2}, \epsilon \}. When 0<|x|< \delta:

    |f(x)|=|x|< \epsilon for x \in \mathbb{Q}
    |f(x)|=|x^2|<|x|<\epsilon for x \in \mathbb{R}\ \mathbb{Q}

    My one issue is that the method i've used is extremely similar to a method someone showed me how to use for another question. I also haven't used the properties of rational and irrational numbers as I would have liked.

    Is this right?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    You are simply over-thinking this problem.
    In any \varepsilon-region of zero the numbers are themselves near zero.
    So if \varepsilon  > 0 \text{ set }\delta  = \varepsilon .
    Doing that simple step solves this problem.

    \varepsilon  > 0\;\& \,\delta  = \varepsilon \, \Rightarrow \left[ {\,\left| {x - 0} \right| < \delta \, \Rightarrow \,\left| {x - 0} \right| < \varepsilon } \right]
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. continuity
    Posted in the Calculus Forum
    Replies: 5
    Last Post: April 20th 2011, 01:36 PM
  2. y=x^3 (continuity)
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: May 16th 2010, 10:50 PM
  3. continuity
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: April 9th 2010, 04:23 PM
  4. Continuity
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 7th 2009, 09:10 PM
  5. Continuity, Uniform Continuity
    Posted in the Calculus Forum
    Replies: 0
    Last Post: February 1st 2009, 08:36 PM

Search Tags


/mathhelpforum @mathhelpforum