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Thread: Continuity mk4

  1. #1
    Super Member Showcase_22's Avatar
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    Continuity mk4

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

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

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

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

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

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

    $\displaystyle |f(x)|=|x|< \epsilon$ for $\displaystyle x \in \mathbb{Q}$
    $\displaystyle |f(x)|=|x^2|<|x|<\epsilon$ for $\displaystyle x \in \mathbb{R}$\$\displaystyle \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?
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  2. #2
    MHF Contributor

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    You are simply over-thinking this problem.
    In any $\displaystyle \varepsilon-$region of zero the numbers are themselves near zero.
    So if $\displaystyle \varepsilon > 0 \text{ set }\delta = \varepsilon $.
    Doing that simple step solves this problem.

    $\displaystyle \varepsilon > 0\;\& \,\delta = \varepsilon \, \Rightarrow \left[ {\,\left| {x - 0} \right| < \delta \, \Rightarrow \,\left| {x - 0} \right| < \varepsilon } \right]$
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