A limit with limit zero everywhere must be zero somewhere
$begingroup$
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
asked 4 hours ago
Arctic CharArctic Char
152112
$endgroup$
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 $f: [0,1]to mathbb R$
$$f(x) =begin{cases} 1/q & text{if }x= p/qin mathbb Q, \ 0 & text{otherwise.}end{cases}$$
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
asked 4 hours ago
Arctic CharArctic Char
152112
$endgroup$
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 $f: [0,1]to mathbb R$
$$f(x) =begin{cases} 1/q & text{if }x= p/qin mathbb Q, \ 0 & text{otherwise.}end{cases}$$
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
asked 4 hours ago
Arctic CharArctic Char
152112
$endgroup$
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
asked 4 hours ago
Arctic CharArctic Char
152112
asked 4 hours ago
Arctic CharArctic Char
152112
asked 4 hours ago
Arctic CharArctic Char
152112
asked 4 hours ago
Arctic CharArctic Char
152112
asked 4 hours ago
Arctic CharArctic Char
152112
152112
add a comment |
add a comment |
1 Answer
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$begingroup$
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).
answered 4 hours ago
user647486user647486
33615
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add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
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).
answered 4 hours ago
user647486user647486
33615
$endgroup$
add a comment |
$begingroup$
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).
answered 4 hours ago
user647486user647486
33615
$endgroup$
add a comment |
$begingroup$
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).
answered 4 hours ago
user647486user647486
33615
$endgroup$
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).
answered 4 hours ago
user647486user647486
33615
answered 4 hours ago
user647486user647486
33615
answered 4 hours ago
user647486user647486
33615
answered 4 hours ago
user647486user647486
33615
33615
add a comment |
add a comment |
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