Question of topology [closed]
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Is it true that the image of nowhere dense set under a continuous mapping is nowhere dense?
general-topology
asked Dec 19 '18 at 15:59
Issayeva ZereIssayeva Zere
42
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closed as off-topic by amWhy, Brahadeesh, Lord_Farin, ncmathsadist, Jyrki Lahtonen Dec 23 '18 at 6:40
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Brahadeesh, Lord_Farin, ncmathsadist, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Is it true that the image of nowhere dense set under a continuous mapping is nowhere dense?
general-topology
asked Dec 19 '18 at 15:59
Issayeva ZereIssayeva Zere
42
$endgroup$
closed as off-topic by amWhy, Brahadeesh, Lord_Farin, ncmathsadist, Jyrki Lahtonen Dec 23 '18 at 6:40
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Brahadeesh, Lord_Farin, ncmathsadist, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
3
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No take $mathbb{R}to{*}$ be the constant map (here ${*}$ is a one point topological space). Then the image of ${0}$ (which is nowhere dense) is the whole space.
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– Roland
Dec 19 '18 at 16:01
1
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Consider a map from the reals to a point. This sends every non-empty set to the whole space
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– ThorbenK
Dec 19 '18 at 16:01
2
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Here is a less obvious example where the source and the target have the same dimension : consider a space filling curve $c:mathbb{R}tomathbb{R}$ (there exist such a surjective continuous map). Consider now the map $mathbb{R}^2tomathbb{R}^2$ such that $(x,y)mapsto c(x)$. Then the image of $mathbb{R}times{0}$ is $mathbb{R}^2$.
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– Roland
Dec 19 '18 at 16:06
add a comment |
$begingroup$
Is it true that the image of nowhere dense set under a continuous mapping is nowhere dense?
general-topology
asked Dec 19 '18 at 15:59
Issayeva ZereIssayeva Zere
42
$endgroup$
Is it true that the image of nowhere dense set under a continuous mapping is nowhere dense?
general-topology
general-topology
asked Dec 19 '18 at 15:59
Issayeva ZereIssayeva Zere
42
asked Dec 19 '18 at 15:59
Issayeva ZereIssayeva Zere
42
asked Dec 19 '18 at 15:59
Issayeva ZereIssayeva Zere
42
asked Dec 19 '18 at 15:59
Issayeva ZereIssayeva Zere
42
asked Dec 19 '18 at 15:59
Issayeva ZereIssayeva Zere
42
42
closed as off-topic by amWhy, Brahadeesh, Lord_Farin, ncmathsadist, Jyrki Lahtonen Dec 23 '18 at 6:40
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Brahadeesh, Lord_Farin, ncmathsadist, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, Brahadeesh, Lord_Farin, ncmathsadist, Jyrki Lahtonen Dec 23 '18 at 6:40
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Brahadeesh, Lord_Farin, ncmathsadist, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
3
$begingroup$
No take $mathbb{R}to{*}$ be the constant map (here ${*}$ is a one point topological space). Then the image of ${0}$ (which is nowhere dense) is the whole space.
$endgroup$
– Roland
Dec 19 '18 at 16:01
1
$begingroup$
Consider a map from the reals to a point. This sends every non-empty set to the whole space
$endgroup$
– ThorbenK
Dec 19 '18 at 16:01
2
$begingroup$
Here is a less obvious example where the source and the target have the same dimension : consider a space filling curve $c:mathbb{R}tomathbb{R}$ (there exist such a surjective continuous map). Consider now the map $mathbb{R}^2tomathbb{R}^2$ such that $(x,y)mapsto c(x)$. Then the image of $mathbb{R}times{0}$ is $mathbb{R}^2$.
$endgroup$
– Roland
Dec 19 '18 at 16:06
add a comment |
3
$begingroup$
No take $mathbb{R}to{*}$ be the constant map (here ${*}$ is a one point topological space). Then the image of ${0}$ (which is nowhere dense) is the whole space.
$endgroup$
– Roland
Dec 19 '18 at 16:01
1
$begingroup$
Consider a map from the reals to a point. This sends every non-empty set to the whole space
$endgroup$
– ThorbenK
Dec 19 '18 at 16:01
2
$begingroup$
Here is a less obvious example where the source and the target have the same dimension : consider a space filling curve $c:mathbb{R}tomathbb{R}$ (there exist such a surjective continuous map). Consider now the map $mathbb{R}^2tomathbb{R}^2$ such that $(x,y)mapsto c(x)$. Then the image of $mathbb{R}times{0}$ is $mathbb{R}^2$.
$endgroup$
– Roland
Dec 19 '18 at 16:06
3
3
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No take $mathbb{R}to{*}$ be the constant map (here ${*}$ is a one point topological space). Then the image of ${0}$ (which is nowhere dense) is the whole space.
$endgroup$
– Roland
Dec 19 '18 at 16:01
$begingroup$
No take $mathbb{R}to{*}$ be the constant map (here ${*}$ is a one point topological space). Then the image of ${0}$ (which is nowhere dense) is the whole space.
$endgroup$
– Roland
Dec 19 '18 at 16:01
1
1
$begingroup$
Consider a map from the reals to a point. This sends every non-empty set to the whole space
$endgroup$
– ThorbenK
Dec 19 '18 at 16:01
$begingroup$
Consider a map from the reals to a point. This sends every non-empty set to the whole space
$endgroup$
– ThorbenK
Dec 19 '18 at 16:01
2
2
$begingroup$
Here is a less obvious example where the source and the target have the same dimension : consider a space filling curve $c:mathbb{R}tomathbb{R}$ (there exist such a surjective continuous map). Consider now the map $mathbb{R}^2tomathbb{R}^2$ such that $(x,y)mapsto c(x)$. Then the image of $mathbb{R}times{0}$ is $mathbb{R}^2$.
$endgroup$
– Roland
Dec 19 '18 at 16:06
$begingroup$
Here is a less obvious example where the source and the target have the same dimension : consider a space filling curve $c:mathbb{R}tomathbb{R}$ (there exist such a surjective continuous map). Consider now the map $mathbb{R}^2tomathbb{R}^2$ such that $(x,y)mapsto c(x)$. Then the image of $mathbb{R}times{0}$ is $mathbb{R}^2$.
$endgroup$
– Roland
Dec 19 '18 at 16:06
add a comment |
2 Answers
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This is false. Consider $f: mathbb{R}to {0}$ by $f(x) = 0$, $forall xinmathbb{R}$. Note that $mathbb{Z}$ is nowhere dense in $mathbb{R}$ but you are mapping to the entire topological space.
answered Dec 19 '18 at 16:05
kminikmini
913
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add a comment |
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Very false. The (uniformly) continuous Cantor function from $[0,1]$ to $[0,1]$ maps the nowhere dense Cantor set onto $[0,1]$. In fact we can have any compact metric space $X$ as the image of such a continuous map from a nowhere dense set. The Cantor fucntion is just a convenient concrete example of this.
answered Dec 19 '18 at 22:07
Henno BrandsmaHenno Brandsma
110k347116
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add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is false. Consider $f: mathbb{R}to {0}$ by $f(x) = 0$, $forall xinmathbb{R}$. Note that $mathbb{Z}$ is nowhere dense in $mathbb{R}$ but you are mapping to the entire topological space.
answered Dec 19 '18 at 16:05
kminikmini
913
$endgroup$
add a comment |
$begingroup$
This is false. Consider $f: mathbb{R}to {0}$ by $f(x) = 0$, $forall xinmathbb{R}$. Note that $mathbb{Z}$ is nowhere dense in $mathbb{R}$ but you are mapping to the entire topological space.
answered Dec 19 '18 at 16:05
kminikmini
913
$endgroup$
add a comment |
$begingroup$
This is false. Consider $f: mathbb{R}to {0}$ by $f(x) = 0$, $forall xinmathbb{R}$. Note that $mathbb{Z}$ is nowhere dense in $mathbb{R}$ but you are mapping to the entire topological space.
answered Dec 19 '18 at 16:05
kminikmini
913
$endgroup$
This is false. Consider $f: mathbb{R}to {0}$ by $f(x) = 0$, $forall xinmathbb{R}$. Note that $mathbb{Z}$ is nowhere dense in $mathbb{R}$ but you are mapping to the entire topological space.
answered Dec 19 '18 at 16:05
kminikmini
913
answered Dec 19 '18 at 16:05
kminikmini
913
answered Dec 19 '18 at 16:05
kminikmini
913
answered Dec 19 '18 at 16:05
kminikmini
913
913
add a comment |
add a comment |
$begingroup$
Very false. The (uniformly) continuous Cantor function from $[0,1]$ to $[0,1]$ maps the nowhere dense Cantor set onto $[0,1]$. In fact we can have any compact metric space $X$ as the image of such a continuous map from a nowhere dense set. The Cantor fucntion is just a convenient concrete example of this.
answered Dec 19 '18 at 22:07
Henno BrandsmaHenno Brandsma
110k347116
$endgroup$
add a comment |
$begingroup$
Very false. The (uniformly) continuous Cantor function from $[0,1]$ to $[0,1]$ maps the nowhere dense Cantor set onto $[0,1]$. In fact we can have any compact metric space $X$ as the image of such a continuous map from a nowhere dense set. The Cantor fucntion is just a convenient concrete example of this.
answered Dec 19 '18 at 22:07
Henno BrandsmaHenno Brandsma
110k347116
$endgroup$
add a comment |
$begingroup$
Very false. The (uniformly) continuous Cantor function from $[0,1]$ to $[0,1]$ maps the nowhere dense Cantor set onto $[0,1]$. In fact we can have any compact metric space $X$ as the image of such a continuous map from a nowhere dense set. The Cantor fucntion is just a convenient concrete example of this.
answered Dec 19 '18 at 22:07
Henno BrandsmaHenno Brandsma
110k347116
$endgroup$
Very false. The (uniformly) continuous Cantor function from $[0,1]$ to $[0,1]$ maps the nowhere dense Cantor set onto $[0,1]$. In fact we can have any compact metric space $X$ as the image of such a continuous map from a nowhere dense set. The Cantor fucntion is just a convenient concrete example of this.
answered Dec 19 '18 at 22:07
Henno BrandsmaHenno Brandsma
110k347116
edited Dec 20 '18 at 10:26
answered Dec 19 '18 at 22:07
Henno BrandsmaHenno Brandsma
110k347116
answered Dec 19 '18 at 22:07
Henno BrandsmaHenno Brandsma
110k347116
answered Dec 19 '18 at 22:07
Henno BrandsmaHenno Brandsma
110k347116
110k347116
add a comment |
add a comment |
3
$begingroup$
No take $mathbb{R}to{*}$ be the constant map (here ${*}$ is a one point topological space). Then the image of ${0}$ (which is nowhere dense) is the whole space.
$endgroup$
– Roland
Dec 19 '18 at 16:01
1
$begingroup$
Consider a map from the reals to a point. This sends every non-empty set to the whole space
$endgroup$
– ThorbenK
Dec 19 '18 at 16:01
2
$begingroup$
Here is a less obvious example where the source and the target have the same dimension : consider a space filling curve $c:mathbb{R}tomathbb{R}$ (there exist such a surjective continuous map). Consider now the map $mathbb{R}^2tomathbb{R}^2$ such that $(x,y)mapsto c(x)$. Then the image of $mathbb{R}times{0}$ is $mathbb{R}^2$.
$endgroup$
– Roland
Dec 19 '18 at 16:06