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anonymous

  • one year ago

please proving continuity help

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  1. imqwerty
    • one year ago
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    ???

  2. anonymous
    • one year ago
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    let x and y be subspace of R given by x=[0,1]U[2,4] defined by f: x--->R by f(x)=1 : x E [0,1] and 2 :x E [2,4]

  3. anonymous
    • one year ago
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    prove that F is continues

  4. anonymous
    • one year ago
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    @zzr0ck3r

  5. anonymous
    • one year ago
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    @Loser66

  6. Loser66
    • one year ago
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    let fix \(x_0\in [0,1]\), By definition, we must find \(\delta >0\) for given \(\epsilon >0\) such that |\(||x-x_0||<0\) implies \(||f(x)-f(x_0)||<\epsilon \). Choose \(\delta=\epsilon\), the definition becomes \(||x-x_0||<\delta\) implies \(||x-x_0||<\epsilon\). Hence f is continuous. Do the same with \(x_0\in [2,4]\)

  7. Loser66
    • one year ago
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    Other proof is valid also, since [0,1] and [2,4] are compact sets, hence f is continuous but I am not sure whether you study compact sets or not.

  8. anonymous
    • one year ago
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    thank you sir

  9. zzr0ck3r
    • one year ago
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    @Loser66 you proved for \(f(x) = x\) We need to prove if for \[f(x) = \begin{cases} 1 & 0\le x\leq 1 \\ 2 & 2\leq x\leq 4 \end{cases}\]

  10. Loser66
    • one year ago
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    Please, fix it. :) Much appreciate.

  11. zzr0ck3r
    • one year ago
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    I need to know what form he wants, I think, because of the subspace, this is a topology question.

  12. zzr0ck3r
    • one year ago
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    Suppose \(U\) is opoen in \(\mathbb{R}\). Then \(U= \cup_{i\in I}B_i\) for some index \(I\) where the \(B_i\) are intervals of the form \((a,b)\). Now \(f^{-1}(\cup_{i\in I}B_i)=U_{i\in I}f^{-1}(B_i)\). Consider \(i\in I\). If \(1\in B_i\) and \(2\notin B_i\) then \(f^{-1}(B_i)=[0,1]\) which is open in the subspace \([0,1]\cup[2,4]\) because \([0,1]=(-1, 1.5)\cap [0,1]\cup [2,4]\) If \(2\in B_i\) and \(1\notin B_i\) then \(f^{-1}(B_i)=[2,4]\) which is open in the subspace \([0,1]\cup[2,4]\) because \([2,4]=(1.5, 5)\cap [0,1]\cup [2,4]\). If both \(1, 2\in f^{-1}(B_i)\) then \(f^{-1}(B_i)=[0,1]\cup[2,4]\) which is open in \([0,1]\cup[2,4]\) because it is the entire subspace. Finally, if \(1,2\notin f^{-1}(B_i)\), then \(F^{-1}(B_i)=\emptyset\) which is of course open. So for each \(i\in I\) we have \(f^{-1}(B_i)\) is open and thus so is \(\cup_{i\in I}f^{-1}(B_i)=f^{-1}(\cup_{i\in I}B_i)=f^{-1}(U)\) showing that \(f\) is indeed continuous.

  13. zzr0ck3r
    • one year ago
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    Make sense?

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