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Thread: Combinations of Continuous Functions

  1. #1
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    Combinations of Continuous Functions

    (1) Let $\displaystyle g(x) = \sqrt[3]{x}$.
    (a) Prove that $\displaystyle g$ is continuous at $\displaystyle c=0$.
    (b) Prove that $\displaystyle g$ is continuous at a point $\displaystyle c \not=0$. (The identity $\displaystyle a^3-b^3=(a-b)(a^2+ab+b^2$) will be helpful)

    (2) Using the $\displaystyle \epsilon-\delta$ characterization of continuity (and thus using no previous results about sequences), show that the linear function $\displaystyle f(x)= ax+b$ is continuous at every point of $\displaystyle \mathbb{R}$.

    (3)
    (a) Using the definition: ( A function $\displaystyle f: A \rightarrow \mathbb{R}$ is continuous at a point $\displaystyle c \in A$ if, for all $\displaystyle \epsilon > 0$, there exists a $\displaystyle \delta > 0$ such that whenever $\displaystyle |x-c| < \delta$ (and $\displaystyle x \in A$) it follows that $\displaystyle |f(x)-f(c)| < \epsilon$. If $\displaystyle f$ is continuous at every point in the domain $\displaystyle A$, then we say that $\displaystyle f$ is continuous on $\displaystyle A$.) show that any function $\displaystyle f$ with domain $\displaystyle \mathbb{Z}$ will necessarily be continuous at every point in its domain.

    (b) Show in general that if $\displaystyle c$ is an isolated point of $\displaystyle A$ $\displaystyle \subseteq \mathbb{R}$, then $\displaystyle f: A\rightarrow R$ is continuous at $\displaystyle c$.


    (4) Assume $\displaystyle h: \mathbb{R} \rightarrow \mathbb{R} $ is continuous on $\displaystyle \mathbb{R}$ and let $\displaystyle K = {x: h(x) = 0}$. Show that $\displaystyle K$ is a closed set.

    If anyone could figure out any of these, I would greatly appreciate it! Thanks!
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  2. #2
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    $\displaystyle \forall \epsilon > 0, \exists \delta > 0 $ such that $\displaystyle |x - c| < \delta \: \: \Rightarrow \: \: |f(x) - f(c)| < \epsilon$

    1.(a) c = 0:
    $\displaystyle |f(x) - f(c)| = |\sqrt[3]{x} - \sqrt[3]{0}| < \epsilon$
    $\displaystyle |\sqrt[3]{x}| < \epsilon$
    $\displaystyle |x| < \epsilon^{3}$

    We know that |x - c| = |x - 0| = |x| < delta and we know that |x| < epsilon cubed. What should we choose delta to be?

    (b) c not 0:
    $\displaystyle |f(x) - f(c)| = |\sqrt[3]{x} - \sqrt[3]{c}| < \epsilon$
    $\displaystyle \iff |x^{\frac{1}{3}}-c^{\frac{1}{3}}| \cdot \left|\frac{ x^{\frac{1}{9}} + x^{\frac{1}{9}}c^{\frac{1}{9}} + c^{\frac{1}{9}} }{x^{\frac{1}{9}} + x^{\frac{1}{9}}c^{\frac{1}{9}} + c^{\frac{1}{9}} }\right| < \epsilon$
    (Think of x - c as a difference of cubes. So, the $\displaystyle x^{\frac{1}{3}} - c^{\frac{1}{3}}$ corresponds to the $\displaystyle (a-b)$ part of $\displaystyle a^{3} - b^{3} = (a-b)(a^2 + ab +b^2)$)

    $\displaystyle \iff \frac{|x-c|}{|x^{\frac{1}{9}} + x^{\frac{1}{9}}c^{\frac{1}{9}} + c^{\frac{1}{9}}|} < \epsilon$

    etc.

    2. $\displaystyle |f(x) - f(c)| < \epsilon$
    $\displaystyle |ax + b - (ac + b)| < \epsilon$
    $\displaystyle |ax - ac| < \epsilon$
    $\displaystyle |a| \: |x - c| < \epsilon$
    $\displaystyle |x-c| < \frac{\epsilon}{|a|}$

    and the conclusion follows.
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    Quote Originally Posted by shadow_2145 View Post
    (4) Assume $\displaystyle h: \mathbb{R} \rightarrow \mathbb{R} $ is continuous on $\displaystyle \mathbb{R}$ and let $\displaystyle K = \{x: h(x) = 0\}$. Show that $\displaystyle K$ is a closed set.
    Hint: If $\displaystyle A$ is a closed set then $\displaystyle h^{-1}(A)$ would be a closed set.
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