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Math Help - 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: R \rightarrow R be a continuous function and let
    C=\{x \in R \mid f(x)=0\}
    Prove that C is a closed subset of 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 f(x_0)\ne 0 then  \left( {\exists \delta  > 0} \right) such that 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 f(x_0)\ne 0 then  \left( {\exists \delta  > 0} \right) such that 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 \epsilon neighborhood which makes V_\epsilon \in set?
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  8. #8
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    Exactly. So pick a point in \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 \left( {x_0  - \delta ,x_0  + \delta } \right) is an open set containing 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 \mathbb{R} :
    A \subseteq \mathbb{R} is closed iff the limit of every convergent sequence \{x_{n}\}_{n\in \mathbb{N}}, with x_{n} \in A, stays in A.

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

    Btw. more generally for any continous function f, it is true that 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 A=\{0\}, which is closed and so 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 f(x_0)\ne 0 then  \left( {\exists \delta > 0} \right) such that 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 \delta such that this would be true. Perhaps \delta = min \{x_0 - c, c - x_0\}, where c \in \mathbb{R} is the closest element of C to 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 \{x_{n}\}_{n\in \mathbb{N}} be any sequence in C, which has a limit x.
    For all n x_{n} \in C => f(x_{n})=0
    => \lim_{n\to \infty} f(x_{n})=0=f(x) , because f is continous => x \in C.
    qed

    Hi,thank you for your time. But I just wonder that what is the purpose to prove this \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 \delta such that this would be true.
    If f is continuous then so is |f| and |f|(x_0)>0.
    So in the definition take \varepsilon  = |f|(x_0 ) then find the corresponding \delta.
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