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  1. MHB Master
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    #1
    Hey!!

    Let $D\subseteq \mathbb{R}$ be a non-empty set. I want to show that $D$ ist compact if and only if each continuous function is bounded on $D$.

    I have done the following:

    We suppose that $D$ is compact. Since $f$ is continuous, we have that $f(D)$ is also compact, right?
    We have that a set is compact iff it is bounded and closed.
    Therefore, we have that $f(D)$ is bounded, and so $f$ is bounded on $D$.

    Let $f$ be a continuous function that is bounded on $D$.
    Since $f$ is bounded on $D$, we have that $f(D)$ is bounded.
    To show that $D$ is compact we have to show that $D$ is bounded and closed.
    Could you give me a hint how we could show that?

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    #2
    Quote Originally Posted by mathmari View Post
    Hey!!

    Let $D\subseteq \mathbb{R}$ be a non-empty set. I want to show that $D$ ist compact if and only if each continuous function is bounded on $D$.

    I have done the following:

    We suppose that $D$ is compact. Since $f$ is continuous, we have that $f(D)$ is also compact, right?
    We have that a set is compact iff it is bounded and closed.
    Therefore, we have that $f(D)$ is bounded, and so $f$ is bounded on $D$.

    Let $f$ be a continuous function that is bounded on $D$.
    Since $f$ is bounded on $D$, we have that $f(D)$ is bounded.
    To show that $D$ is compact we have to show that $D$ is bounded and closed.
    Could you give me a hint how we could show that?
    Suppose that each continuous function on $D$ is bounded.

    We argue that $D$ is closed. Suppose not. Then there is a point $b\notin D$ such that $b$ is in the boundary of $D$. Define $f: D\to \mathbf R$ as $f(x)=1/(x-b)$. Then $f$ is an unbounded continuous function on $D$. This is a contradiction.

    Now since the map $g: D\to \mathbf R$ defined as $g(x)=x$ for all $x\in D$ is bounded, we infer that $D$ is bounded.

    Therefore $D$ is closed and bounded, and thus compact.

  3. MHB Master
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    #3 Thread Author
    Quote Originally Posted by caffeinemachine View Post
    Then there is a point $b\notin D$ such that $b$ is in the boundary of $D$. Define $f: D\to \mathbf R$ as $f(x)=1/(x-b)$. Then $f$ is an unbounded continuous function on $D$. This is a contradiction.
    Why is $f$ unbounded?

    Do we apply $f$ at $x_n$, and since $x_n\rightarrow b$, $x_n$ is arbitrarily close to $b$, therefore $x_n-b$ is arbitrarily close to $0$, and so $f(x)$ goes to infinity?
    Last edited by mathmari; January 4th, 2017 at 05:10.

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    #4
    Quote Originally Posted by mathmari View Post
    Why is $f$ unbounded?

    Do we apply $f$ at $x_n$, and since $x_n\rightarrow b$, $x_n$ is arbitrarily close to $b$, therefore $x_n-b$ is arbitrarily close to $0$, and so $f(x)$ goes to infinity?
    That is correct.

  5. MHB Master
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    #5 Thread Author
    Quote Originally Posted by caffeinemachine View Post
    That is correct.
    Thank you!!

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