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Thread: more counterexamples needed

  1. #1
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    more counterexamples needed

    I know that the following are false, I just cant come up with counterexamples:
    [f(x)]^2 continuous on (a,b) implies taht f is continuous on (a,b)

    f and g are not continuous on (a,b) implies that fg is not continuous on (a,b)

    f and g are not continuous on (a,b) implies that $\displaystyle f\circ g$ is not continuous on (a,b)
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by dannyboycurtis View Post
    I know that the following are false, I just cant come up with counterexamples:
    [f(x)]^2 continuous on (a,b) implies taht f is continuous on (a,b)

    f and g are not continuous on (a,b) implies that fg is not continuous on (a,b)

    f and g are not continuous on (a,b) implies that $\displaystyle f\circ g$ is not continuous on (a,b)
    This example works for the first two. Take $\displaystyle f(x)=g(x)=\left\{\begin{array}{lr}1:&x\in\mathbb{Q }\\-1:&x\notin\mathbb{Q}\end{array}\right\}$

    This is obviously discontinuous, but $\displaystyle f^2(x)=f(x)g(x)=1$, which is continuous.

    For the third one, let $\displaystyle f(x)=\left\{\begin{array}{lr}x:&x\in\mathbb{Q}\\0: &x\notin\mathbb{Q}\end{array}\right\}$ and $\displaystyle g(x)=\left\{\begin{array}{lr}\pi:&x\in\mathbb{Q}\\ 0:&x\notin\mathbb{Q}\end{array}\right\}$

    Clearly both $\displaystyle f(x)$ and $\displaystyle g(x)$ are discontinuous, but $\displaystyle f(g(x))=0$.
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  3. #3
    MHF Contributor chisigma's Avatar
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    I think taht the following function pair is a countexample in all the cases You have proposed...

    $\displaystyle f(x)= \left\{\begin{array}{cc}-1, &\mbox {if } -1\le x \le 0\\1, & \mbox {if } 0< x \le 1\end{array}\right. $

    $\displaystyle g(x)= -f(x)$

    $\displaystyle x \in [-1,1]$

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  4. #4
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    Where you put a pi, any number, including x would work, right?
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  5. #5
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by binkypoo View Post
    Where you put a pi, any number, including x would work, right?
    Any irrational number will work in place of $\displaystyle \pi$.
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  6. #6
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    how about a counterexample to this:
    if f is continuous at x=a, then f(a+)=f(a-)
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  7. #7
    Super Member redsoxfan325's Avatar
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    That's true. If a function $\displaystyle f$ is continuous at $\displaystyle x=a$, then $\displaystyle \lim_{x\to a^-}f(x)=f(a)=\lim_{x\to a^+}f(x)$
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  8. #8
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    wow, that one has been boggling me for a while, because I was sure it was true, but my book says its false! Thanks
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  9. #9
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    last question on this thread, Ive been trying to find a counterexample for this one for a while, Im sure its simple but I cant seem to see it:
    if |f| is cont. on (a,b) then f is cont. on (a,b).
    Again my book claims its false...
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  10. #10
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by binkypoo View Post
    last question on this thread, Ive been trying to find a counterexample for this one for a while, Im sure its simple but I cant seem to see it:
    if |f| is cont. on (a,b) then f is cont. on (a,b).
    Again my book claims its false...
    Consider $\displaystyle f(x)=\left\{\begin{array}{lr}1:&x\in\mathbb{Q}\\-1:&x\notin\mathbb{Q}\end{array}\right\}$
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