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Thread: proof of closed subset

  1. #1
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    proof of closed subset

    Hi,can anyone please help me with this question? I cannot even start it.

    Let f: $\displaystyle R \rightarrow R$ be a continuous function and let
    $\displaystyle C=\{x \in R \mid f(x)=0\}$
    Prove that C is a closed subset of $\displaystyle R$.
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  2. #2
    A Plied Mathematician
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    What ideas have you had so far?
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  3. #3
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    Quote Originally Posted by Ackbeet View Post
    What ideas have you had so far?
    All I can think about is go back to definition, so I should try to show that R\C is open, which means C is closed. Am I right? But I really don't know how to even start. So C is a constant function, and R\C should be R\{0}, am I right so far?
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  4. #4
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    That is a good approach.
    Recall that for continuous functions if $\displaystyle f(x_0)\ne 0$ then $\displaystyle \left( {\exists \delta > 0} \right)$ such that $\displaystyle x \in \left( {x_0 - \delta ,x_0 + \delta } \right)\, \Rightarrow \,f(x) \ne 0.$
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  5. #5
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    Quote Originally Posted by Plato View Post
    That is a good approach.
    Recall that for continuous functions if $\displaystyle f(x_0)\ne 0$ then $\displaystyle \left( {\exists \delta > 0} \right)$ such that $\displaystyle x \in \left( {x_0 - \delta ,x_0 + \delta } \right)\, \Rightarrow \,f(x) \ne 0.$

    Hi, thanks for your reply. However, I actually still don't know how to continue, could you pleas;e give me a bit more details? I just started learning Real Analysis and I'm still trying to improve it at this moment. I will really appreciate for your time. Thanks a lot.
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  6. #6
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    How do you normally show that a set is open?
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  7. #7
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    Quote Originally Posted by Ackbeet View Post
    How do you normally show that a set is open?
    To show that for any point in the set, can always find $\displaystyle \epsilon$ neighborhood which makes $\displaystyle V_\epsilon \in$ set?
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  8. #8
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    Exactly. So pick a point in $\displaystyle \mathbb{R}\setminus C$ and try to get your neighborhood. How would you proceed?
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  9. #9
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    @stang, Do you realize that $\displaystyle \left( {x_0 - \delta ,x_0 + \delta } \right)$ is an open set containing $\displaystyle x_0?$
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  10. #10
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    Alternatively you could also prove it with the sequences definition of closed sets in $\displaystyle \mathbb{R}$ :
    $\displaystyle A \subseteq \mathbb{R}$ is closed iff the limit of every convergent sequence $\displaystyle \{x_{n}\}_{n\in \mathbb{N}}$, with $\displaystyle x_{n} \in A$, stays in A.

    Let $\displaystyle \{x_{n}\}_{n\in \mathbb{N}}$ be any sequence in C, which has a limit x.
    For all n $\displaystyle x_{n} \in C => f(x_{n})=0$
    => $\displaystyle \lim_{n\to \infty} f(x_{n})=0=f(x)$ , because f is continous => $\displaystyle x \in C$.
    qed

    Btw. more generally for any continous function f, it is true that $\displaystyle f^{-1}(A)$ is always a closed set, when A is a closed set and is always an open set, when A is an open set.
    In your exercise, we just have the special case $\displaystyle A=\{0\}$, which is closed and so $\displaystyle C=f^{-1}(\{0\})$ must be closed too.
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  11. #11
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    Quote Originally Posted by Plato View Post
    That is a good approach.
    Recall that for continuous functions if $\displaystyle f(x_0)\ne 0$ then $\displaystyle \left( {\exists \delta > 0} \right)$ such that $\displaystyle x \in \left( {x_0 - \delta ,x_0 + \delta } \right)\, \Rightarrow \,f(x) \ne 0.$
    I'm just trying to think of a reasonable condition on $\displaystyle \delta$ such that this would be true. Perhaps $\displaystyle \delta = min \{x_0 - c, c - x_0\}$, where $\displaystyle c \in \mathbb{R}$ is the closest element of $\displaystyle C$ to $\displaystyle x_0$? Just a guess; I'm not too sure.
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  12. #12
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    Quote Originally Posted by Iondor View Post
    Let $\displaystyle \{x_{n}\}_{n\in \mathbb{N}}$ be any sequence in C, which has a limit x.
    For all n $\displaystyle x_{n} \in C => f(x_{n})=0$
    => $\displaystyle \lim_{n\to \infty} f(x_{n})=0=f(x)$ , because f is continous => $\displaystyle x \in C$.
    qed

    Hi,thank you for your time. But I just wonder that what is the purpose to prove this $\displaystyle \lim_{n\to \infty} f(x_{n})=0=f(x)$? Could you please just give me a bit more details that why you are doing this? Thank you.
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  13. #13
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    He proved that the set C is closed using the definition of sequences for a closed set, which he said at the start of his post.
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  14. #14
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    Quote Originally Posted by Ares_D1 View Post
    I'm just trying to think of a reasonable condition on $\displaystyle \delta$ such that this would be true.
    If $\displaystyle f$ is continuous then so is $\displaystyle |f|$ and $\displaystyle |f|(x_0)>0$.
    So in the definition take $\displaystyle \varepsilon = |f|(x_0 )$ then find the corresponding $\displaystyle \delta.$
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