A limit with limit zero everywhere must be zero somewhere












8












$begingroup$


I wish to know if the following is true:




Let $f : [alpha, beta]to mathbb R$ be a function so that
$$ lim_{xto x_0} f(x) = 0$$
for all $x_0 in [alpha, beta]$. Then $f(x) = 0$ for some $xin [alpha, beta]$.




The Thomae's function $f: [0,1]to mathbb R$



$$f(x) =begin{cases} 1/q & text{if }x= p/qin mathbb Q, \ 0 & text{otherwise.}end{cases}$$



leads me to the above question. The Thomae function has limit zero everywhere, althought it is nonzero in $mathbb Q$. I think I can take any countable dense subset $Dsubset [alpha, beta]$ and construct a function which is nonzero in $D$ but limit equals zero everywhere. But I can't think of a function that is nonzero everywhere but has zero limit everywhere.






real-analysis limits







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$endgroup$












  • $begingroup$
    See math.stackexchange.com/q/1801935/72031 and math.stackexchange.com/q/980022/72031
    $endgroup$
    – Paramanand Singh
    48 mins ago
















8












$begingroup$


I wish to know if the following is true:




Let $f : [alpha, beta]to mathbb R$ be a function so that
$$ lim_{xto x_0} f(x) = 0$$
for all $x_0 in [alpha, beta]$. Then $f(x) = 0$ for some $xin [alpha, beta]$.




The Thomae's function $f: [0,1]to mathbb R$



$$f(x) =begin{cases} 1/q & text{if }x= p/qin mathbb Q, \ 0 & text{otherwise.}end{cases}$$



leads me to the above question. The Thomae function has limit zero everywhere, althought it is nonzero in $mathbb Q$. I think I can take any countable dense subset $Dsubset [alpha, beta]$ and construct a function which is nonzero in $D$ but limit equals zero everywhere. But I can't think of a function that is nonzero everywhere but has zero limit everywhere.






real-analysis limits







share|cite|improve this question









$endgroup$












  • $begingroup$
    See math.stackexchange.com/q/1801935/72031 and math.stackexchange.com/q/980022/72031
    $endgroup$
    – Paramanand Singh
    48 mins ago














8












8








8


1



$begingroup$


I wish to know if the following is true:




Let $f : [alpha, beta]to mathbb R$ be a function so that
$$ lim_{xto x_0} f(x) = 0$$
for all $x_0 in [alpha, beta]$. Then $f(x) = 0$ for some $xin [alpha, beta]$.




The Thomae's function $f: [0,1]to mathbb R$



$$f(x) =begin{cases} 1/q & text{if }x= p/qin mathbb Q, \ 0 & text{otherwise.}end{cases}$$



leads me to the above question. The Thomae function has limit zero everywhere, althought it is nonzero in $mathbb Q$. I think I can take any countable dense subset $Dsubset [alpha, beta]$ and construct a function which is nonzero in $D$ but limit equals zero everywhere. But I can't think of a function that is nonzero everywhere but has zero limit everywhere.






real-analysis limits







share|cite|improve this question









$endgroup$




I wish to know if the following is true:




Let $f : [alpha, beta]to mathbb R$ be a function so that
$$ lim_{xto x_0} f(x) = 0$$
for all $x_0 in [alpha, beta]$. Then $f(x) = 0$ for some $xin [alpha, beta]$.




The Thomae's function $f: [0,1]to mathbb R$



$$f(x) =begin{cases} 1/q & text{if }x= p/qin mathbb Q, \ 0 & text{otherwise.}end{cases}$$



leads me to the above question. The Thomae function has limit zero everywhere, althought it is nonzero in $mathbb Q$. I think I can take any countable dense subset $Dsubset [alpha, beta]$ and construct a function which is nonzero in $D$ but limit equals zero everywhere. But I can't think of a function that is nonzero everywhere but has zero limit everywhere.






real-analysis limits




real-analysis limits






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asked 6 hours ago









Arctic CharArctic Char

157112




157112












  • $begingroup$
    See math.stackexchange.com/q/1801935/72031 and math.stackexchange.com/q/980022/72031
    $endgroup$
    – Paramanand Singh
    48 mins ago


















  • $begingroup$
    See math.stackexchange.com/q/1801935/72031 and math.stackexchange.com/q/980022/72031
    $endgroup$
    – Paramanand Singh
    48 mins ago
















$begingroup$
See math.stackexchange.com/q/1801935/72031 and math.stackexchange.com/q/980022/72031
$endgroup$
– Paramanand Singh
48 mins ago




$begingroup$
See math.stackexchange.com/q/1801935/72031 and math.stackexchange.com/q/980022/72031
$endgroup$
– Paramanand Singh
48 mins ago










1 Answer
1






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13












$begingroup$

The number of points $xin[alpha,beta]$ at which $|f(x)|>1/n$, for a fixed $ninmathbb{N}$, has to be finite. Otherwise, since the interval is compact, they accumulate somewhere and at that point the limit wouldn't be zero.



Therefore, the points $xin[alpha,beta]$ at which $|f(x)|neq0$ is countable, but $[alpha,beta]$ isn't (unless $alpha=beta$ but that case follows directly).






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    1 Answer
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    1 Answer
    1






    active

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    active

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    13












    $begingroup$

    The number of points $xin[alpha,beta]$ at which $|f(x)|>1/n$, for a fixed $ninmathbb{N}$, has to be finite. Otherwise, since the interval is compact, they accumulate somewhere and at that point the limit wouldn't be zero.



    Therefore, the points $xin[alpha,beta]$ at which $|f(x)|neq0$ is countable, but $[alpha,beta]$ isn't (unless $alpha=beta$ but that case follows directly).






    share|cite|improve this answer









    $endgroup$


















      13












      $begingroup$

      The number of points $xin[alpha,beta]$ at which $|f(x)|>1/n$, for a fixed $ninmathbb{N}$, has to be finite. Otherwise, since the interval is compact, they accumulate somewhere and at that point the limit wouldn't be zero.



      Therefore, the points $xin[alpha,beta]$ at which $|f(x)|neq0$ is countable, but $[alpha,beta]$ isn't (unless $alpha=beta$ but that case follows directly).






      share|cite|improve this answer









      $endgroup$
















        13












        13








        13





        $begingroup$

        The number of points $xin[alpha,beta]$ at which $|f(x)|>1/n$, for a fixed $ninmathbb{N}$, has to be finite. Otherwise, since the interval is compact, they accumulate somewhere and at that point the limit wouldn't be zero.



        Therefore, the points $xin[alpha,beta]$ at which $|f(x)|neq0$ is countable, but $[alpha,beta]$ isn't (unless $alpha=beta$ but that case follows directly).






        share|cite|improve this answer









        $endgroup$



        The number of points $xin[alpha,beta]$ at which $|f(x)|>1/n$, for a fixed $ninmathbb{N}$, has to be finite. Otherwise, since the interval is compact, they accumulate somewhere and at that point the limit wouldn't be zero.



        Therefore, the points $xin[alpha,beta]$ at which $|f(x)|neq0$ is countable, but $[alpha,beta]$ isn't (unless $alpha=beta$ but that case follows directly).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 6 hours ago









        user647486user647486

        34615




        34615






























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