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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 $f: [0,1]to mathbb R$

$$f(x) =begincases 1/q & textif x= p/qin mathbb Q, \ 0 & textotherwise.endcases$$

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

asked 6 hours ago

Arctic CharArctic Char

157112

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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 $f: [0,1]to mathbb R$

$$f(x) =begincases 1/q & textif x= p/qin mathbb Q, \ 0 & textotherwise.endcases$$

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

asked 6 hours ago

Arctic CharArctic Char

157112

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

$begingroup$

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 $f: [0,1]to mathbb R$

$$f(x) =begincases 1/q & textif x= p/qin mathbb Q, \ 0 & textotherwise.endcases$$

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

asked 6 hours ago

Arctic CharArctic Char

157112

$endgroup$

share|cite|improve this question

asked 6 hours ago

Arctic CharArctic Char

157112

share|cite|improve this question

asked 6 hours ago

Arctic CharArctic Char

157112

asked 6 hours ago

Arctic CharArctic Char

157112

asked 6 hours ago

Arctic CharArctic Char

157112

1 Answer
1

active

oldest

votes

13

$begingroup$

The number of points $xin[alpha,beta]$ at which $|f(x)|>1/n$, for a fixed $ninmathbbN$, 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

answered 6 hours ago

user647486user647486

34615

$endgroup$

add a comment | 

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13

$begingroup$

The number of points $xin[alpha,beta]$ at which $|f(x)|>1/n$, for a fixed $ninmathbbN$, 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

answered 6 hours ago

user647486user647486

34615

$endgroup$

add a comment | 

13

$begingroup$

The number of points $xin[alpha,beta]$ at which $|f(x)|>1/n$, for a fixed $ninmathbbN$, 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

answered 6 hours ago

user647486user647486

34615

$endgroup$

add a comment | 

$begingroup$

The number of points $xin[alpha,beta]$ at which $|f(x)|>1/n$, for a fixed $ninmathbbN$, 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

answered 6 hours ago

user647486user647486

34615

$endgroup$

share|cite|improve this answer

answered 6 hours ago

user647486user647486

34615

answered 6 hours ago

user647486user647486

34615

answered 6 hours ago

user647486user647486

34615