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Thread: I need some examples

  1. #1
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    I need some examples

    Is there a function which is defined on $\displaystyle \mathbb{R}$ but which is NOT continuous on ANY point of $\displaystyle \mathbb{R}$?

    What about a function which is defined on $\displaystyle \mathbb{R}$ and is continuous at exactly TWO points of $\displaystyle \mathbb{R}$?
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by binkypoo View Post
    Is there a function which is defined on $\displaystyle \mathbb{R}$ but which is NOT continuous on ANY point of $\displaystyle \mathbb{R}$?

    What about a function which is defined on $\displaystyle \mathbb{R}$ and is continuous at exactly TWO points of $\displaystyle \mathbb{R}$?
    $\displaystyle f(x)=\left\{\begin{array}{lr}1:&x\in\mathbb{Q}\\0: &x\notin\mathbb{Q}\end{array}\right\}$

    $\displaystyle g(x)=\left\{\begin{array}{lr}x:&x\in\mathbb{Q}\cap (-\infty,1]\\x-2:&x\in\mathbb{Q}\cap(1,\infty)\\0:&x\notin\mathbb {Q}\end{array}\right\}$

    $\displaystyle f(x)$ is continuous nowhere, and $\displaystyle g(x)$ is continuous only at $\displaystyle x=0$ and $\displaystyle x=2$.
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  3. #3
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    what about a function f defined on an interval [a,b] which is continuous at 2 points of [a,b]?
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  4. #4
    Super Member redsoxfan325's Avatar
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    Use the exact same function; just restrict the domain.
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  5. #5
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    $\displaystyle g(x)=\left\{\begin{array}{lr}x:&x\in\mathbb{Q}\cap (-\infty,1]\\x-b:&x\in\mathbb{Q}\cap(1,\infty)\\a:&x\notin\mathbb {Q}\end{array}\right\}
    $
    would this work>?
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  6. #6
    Super Member redsoxfan325's Avatar
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    It needs to be defined on $\displaystyle [a,b]$; that function there is still defined on $\displaystyle (-\infty,\infty)$.
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  7. #7
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    $\displaystyle g(x)=\left\{\begin{array}{lr}x:&x\in\mathbb{Q}\cap[a,a+\delta]\\x-b:&x\in\mathbb{Q}\cap[a+\delta,b]\\a:&x\notin\mathbb{Q}\end{array}\right\}
    $

    I'm shooting in the dark here, not too sure
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  8. #8
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by binkypoo View Post
    $\displaystyle g(x)=\left\{\begin{array}{lr}x:&x\in\mathbb{Q}\cap[a,a+\delta]\\x-b:&x\in\mathbb{Q}\cap[a+\delta,b]\\a:&x\notin\mathbb{Q}\end{array}\right\}
    $

    I'm shooting in the dark here, not too sure
    Take any three constants $\displaystyle c,d,e\in[a,b]$ such that $\displaystyle a<c<d<e<b$. Define a function:

    $\displaystyle g(x)=\left\{\begin{array}{lr}x-c:&x\in\mathbb{Q}\cap[a,d]\\x-e:&x\in\mathbb{Q}\cap(d,b]\\0:&x\notin\mathbb{Q}\end{array}\right\}$

    $\displaystyle g$ will be continuous at $\displaystyle x=c$ and $\displaystyle x=e$.
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