Existence of a continuous function with zeros in a closed set
If $E subset mathbb{R}$ is a closed subset show that there is a continuous real-valued function on $mathbb{R}$ such that $f(x) = 0$ if and only if $x in E$.
I've considered trying to show that the preimage of any open or closed set in $mathbb{R}$ must be open or closed, but this didn't seem to get me anywhere.
real-analysis continuity
asked Nov 29 '18 at 0:46
A. SmithA. Smith
385
add a comment |
If $E subset mathbb{R}$ is a closed subset show that there is a continuous real-valued function on $mathbb{R}$ such that $f(x) = 0$ if and only if $x in E$.
I've considered trying to show that the preimage of any open or closed set in $mathbb{R}$ must be open or closed, but this didn't seem to get me anywhere.
real-analysis continuity
asked Nov 29 '18 at 0:46
A. SmithA. Smith
385
The preimage of what?
– Andrés E. Caicedo
Nov 29 '18 at 0:59
add a comment |
If $E subset mathbb{R}$ is a closed subset show that there is a continuous real-valued function on $mathbb{R}$ such that $f(x) = 0$ if and only if $x in E$.
I've considered trying to show that the preimage of any open or closed set in $mathbb{R}$ must be open or closed, but this didn't seem to get me anywhere.
real-analysis continuity
asked Nov 29 '18 at 0:46
A. SmithA. Smith
385
If $E subset mathbb{R}$ is a closed subset show that there is a continuous real-valued function on $mathbb{R}$ such that $f(x) = 0$ if and only if $x in E$.
I've considered trying to show that the preimage of any open or closed set in $mathbb{R}$ must be open or closed, but this didn't seem to get me anywhere.
real-analysis continuity
real-analysis continuity
asked Nov 29 '18 at 0:46
A. SmithA. Smith
385
asked Nov 29 '18 at 0:46
A. SmithA. Smith
385
asked Nov 29 '18 at 0:46
A. SmithA. Smith
385
asked Nov 29 '18 at 0:46
A. SmithA. Smith
385
asked Nov 29 '18 at 0:46
A. SmithA. Smith
385
385
The preimage of what?
– Andrés E. Caicedo
Nov 29 '18 at 0:59
add a comment |
The preimage of what?
– Andrés E. Caicedo
Nov 29 '18 at 0:59
The preimage of what?
– Andrés E. Caicedo
Nov 29 '18 at 0:59
The preimage of what?
– Andrés E. Caicedo
Nov 29 '18 at 0:59
add a comment |
1 Answer
1
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oldest
votes
$f(x) = min[text{dist}(x,E),1]$, where $text{dist}(x,E)=inf{|x-y|:yin E}$
Based on comments, we could also more simply use
$g(x) =inf{|x-y|:yin E}$
answered Nov 29 '18 at 0:51
DunhamDunham
2,084613
2
+1 Why not just the distance?
– Ethan Bolker
Nov 29 '18 at 0:52
+1: but I suspect the OP won't know what $mbox{dist}(x, E)$ means.
– Rob Arthan
Nov 29 '18 at 0:53
I think I unconsciously added bounded as a condition when I read the question
– Dunham
Nov 29 '18 at 0:54
1
More interestingly, you can arrange for $f$ to be $C^infty$.
– Andrés E. Caicedo
Nov 29 '18 at 1:00
Ah, thank you. I actually had the graph of this function in mind when trying to intuitively think about this problem, but couldn't think of a way to mathematically describe it.
– A. Smith
Nov 29 '18 at 1:12
|
show 2 more comments
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1 Answer
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active
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1 Answer
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active
oldest
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oldest
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oldest
votes
$f(x) = min[text{dist}(x,E),1]$, where $text{dist}(x,E)=inf{|x-y|:yin E}$
Based on comments, we could also more simply use
$g(x) =inf{|x-y|:yin E}$
answered Nov 29 '18 at 0:51
DunhamDunham
2,084613
2
+1 Why not just the distance?
– Ethan Bolker
Nov 29 '18 at 0:52
+1: but I suspect the OP won't know what $mbox{dist}(x, E)$ means.
– Rob Arthan
Nov 29 '18 at 0:53
I think I unconsciously added bounded as a condition when I read the question
– Dunham
Nov 29 '18 at 0:54
1
More interestingly, you can arrange for $f$ to be $C^infty$.
– Andrés E. Caicedo
Nov 29 '18 at 1:00
Ah, thank you. I actually had the graph of this function in mind when trying to intuitively think about this problem, but couldn't think of a way to mathematically describe it.
– A. Smith
Nov 29 '18 at 1:12
|
show 2 more comments
$f(x) = min[text{dist}(x,E),1]$, where $text{dist}(x,E)=inf{|x-y|:yin E}$
Based on comments, we could also more simply use
$g(x) =inf{|x-y|:yin E}$
answered Nov 29 '18 at 0:51
DunhamDunham
2,084613
2
+1 Why not just the distance?
– Ethan Bolker
Nov 29 '18 at 0:52
+1: but I suspect the OP won't know what $mbox{dist}(x, E)$ means.
– Rob Arthan
Nov 29 '18 at 0:53
I think I unconsciously added bounded as a condition when I read the question
– Dunham
Nov 29 '18 at 0:54
1
More interestingly, you can arrange for $f$ to be $C^infty$.
– Andrés E. Caicedo
Nov 29 '18 at 1:00
Ah, thank you. I actually had the graph of this function in mind when trying to intuitively think about this problem, but couldn't think of a way to mathematically describe it.
– A. Smith
Nov 29 '18 at 1:12
|
show 2 more comments
$f(x) = min[text{dist}(x,E),1]$, where $text{dist}(x,E)=inf{|x-y|:yin E}$
Based on comments, we could also more simply use
$g(x) =inf{|x-y|:yin E}$
answered Nov 29 '18 at 0:51
DunhamDunham
2,084613
$f(x) = min[text{dist}(x,E),1]$, where $text{dist}(x,E)=inf{|x-y|:yin E}$
Based on comments, we could also more simply use
$g(x) =inf{|x-y|:yin E}$
answered Nov 29 '18 at 0:51
DunhamDunham
2,084613
edited Nov 29 '18 at 0:56
answered Nov 29 '18 at 0:51
DunhamDunham
2,084613
answered Nov 29 '18 at 0:51
DunhamDunham
2,084613
answered Nov 29 '18 at 0:51
DunhamDunham
2,084613
2,084613
2
+1 Why not just the distance?
– Ethan Bolker
Nov 29 '18 at 0:52
+1: but I suspect the OP won't know what $mbox{dist}(x, E)$ means.
– Rob Arthan
Nov 29 '18 at 0:53
I think I unconsciously added bounded as a condition when I read the question
– Dunham
Nov 29 '18 at 0:54
1
More interestingly, you can arrange for $f$ to be $C^infty$.
– Andrés E. Caicedo
Nov 29 '18 at 1:00
Ah, thank you. I actually had the graph of this function in mind when trying to intuitively think about this problem, but couldn't think of a way to mathematically describe it.
– A. Smith
Nov 29 '18 at 1:12
|
show 2 more comments
2
+1 Why not just the distance?
– Ethan Bolker
Nov 29 '18 at 0:52
+1: but I suspect the OP won't know what $mbox{dist}(x, E)$ means.
– Rob Arthan
Nov 29 '18 at 0:53
I think I unconsciously added bounded as a condition when I read the question
– Dunham
Nov 29 '18 at 0:54
1
More interestingly, you can arrange for $f$ to be $C^infty$.
– Andrés E. Caicedo
Nov 29 '18 at 1:00
Ah, thank you. I actually had the graph of this function in mind when trying to intuitively think about this problem, but couldn't think of a way to mathematically describe it.
– A. Smith
Nov 29 '18 at 1:12
2
2
+1 Why not just the distance?
– Ethan Bolker
Nov 29 '18 at 0:52
+1 Why not just the distance?
– Ethan Bolker
Nov 29 '18 at 0:52
+1: but I suspect the OP won't know what $mbox{dist}(x, E)$ means.
– Rob Arthan
Nov 29 '18 at 0:53
+1: but I suspect the OP won't know what $mbox{dist}(x, E)$ means.
– Rob Arthan
Nov 29 '18 at 0:53
I think I unconsciously added bounded as a condition when I read the question
– Dunham
Nov 29 '18 at 0:54
I think I unconsciously added bounded as a condition when I read the question
– Dunham
Nov 29 '18 at 0:54
1
1
More interestingly, you can arrange for $f$ to be $C^infty$.
– Andrés E. Caicedo
Nov 29 '18 at 1:00
More interestingly, you can arrange for $f$ to be $C^infty$.
– Andrés E. Caicedo
Nov 29 '18 at 1:00
Ah, thank you. I actually had the graph of this function in mind when trying to intuitively think about this problem, but couldn't think of a way to mathematically describe it.
– A. Smith
Nov 29 '18 at 1:12
Ah, thank you. I actually had the graph of this function in mind when trying to intuitively think about this problem, but couldn't think of a way to mathematically describe it.
– A. Smith
Nov 29 '18 at 1:12
|
show 2 more comments
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The preimage of what?
– Andrés E. Caicedo
Nov 29 '18 at 0:59