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Is there a continuous onto function $f:BbbD rightarrow [-1,1]$?
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Is there a continuous onto function $f:BbbD rightarrow [-1,1]$ ,
where $BbbD$ is a closed unit disk in $BbbR^2$ ?
I think the answer is no, otherwise $f(BbbD)=[-1,1]$, so removing one point in the domain is still connected but the image is not. Am I right? Any hint?
real-analysis continuity
asked Jul 27 at 5:36
Learning Mathematics
469213
add a comment |Â
up vote
0
down vote
favorite
Is there a continuous onto function $f:BbbD rightarrow [-1,1]$ ,
where $BbbD$ is a closed unit disk in $BbbR^2$ ?
I think the answer is no, otherwise $f(BbbD)=[-1,1]$, so removing one point in the domain is still connected but the image is not. Am I right? Any hint?
real-analysis continuity
asked Jul 27 at 5:36
Learning Mathematics
469213
7
Try $f(x,y) = x.$
– zhw.
Jul 27 at 5:40
1
Because you don't assume $f$ to be injective, "removing one point" in $[-1,1]$ may make you remove way more than just one point in $mathbbD$. It could for instance make you remove a diameter of the disk, which would then not be connected anymore, and your argument would fail.
– Suzet
Jul 27 at 5:42
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Is there a continuous onto function $f:BbbD rightarrow [-1,1]$ ,
where $BbbD$ is a closed unit disk in $BbbR^2$ ?
I think the answer is no, otherwise $f(BbbD)=[-1,1]$, so removing one point in the domain is still connected but the image is not. Am I right? Any hint?
real-analysis continuity
asked Jul 27 at 5:36
Learning Mathematics
469213
Is there a continuous onto function $f:BbbD rightarrow [-1,1]$ ,
where $BbbD$ is a closed unit disk in $BbbR^2$ ?
I think the answer is no, otherwise $f(BbbD)=[-1,1]$, so removing one point in the domain is still connected but the image is not. Am I right? Any hint?
real-analysis continuity
asked Jul 27 at 5:36
Learning Mathematics
469213
asked Jul 27 at 5:36
Learning Mathematics
469213
asked Jul 27 at 5:36
Learning Mathematics
469213
asked Jul 27 at 5:36
Learning Mathematics
469213
469213
7
Try $f(x,y) = x.$
– zhw.
Jul 27 at 5:40
1
Because you don't assume $f$ to be injective, "removing one point" in $[-1,1]$ may make you remove way more than just one point in $mathbbD$. It could for instance make you remove a diameter of the disk, which would then not be connected anymore, and your argument would fail.
– Suzet
Jul 27 at 5:42
add a comment |Â
7
Try $f(x,y) = x.$
– zhw.
Jul 27 at 5:40
1
Because you don't assume $f$ to be injective, "removing one point" in $[-1,1]$ may make you remove way more than just one point in $mathbbD$. It could for instance make you remove a diameter of the disk, which would then not be connected anymore, and your argument would fail.
– Suzet
Jul 27 at 5:42
7
7
Try $f(x,y) = x.$
– zhw.
Jul 27 at 5:40
Try $f(x,y) = x.$
– zhw.
Jul 27 at 5:40
1
1
Because you don't assume $f$ to be injective, "removing one point" in $[-1,1]$ may make you remove way more than just one point in $mathbbD$. It could for instance make you remove a diameter of the disk, which would then not be connected anymore, and your argument would fail.
– Suzet
Jul 27 at 5:42
Because you don't assume $f$ to be injective, "removing one point" in $[-1,1]$ may make you remove way more than just one point in $mathbbD$. It could for instance make you remove a diameter of the disk, which would then not be connected anymore, and your argument would fail.
– Suzet
Jul 27 at 5:42
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
Your argument fails.
The problem is, that because $f$ is not injective, we cannot assert that since $f(mathbb D) = [-1,1]$, we have for all $x in [-1,1]$ that $f(mathbb D backslash y) = [-1,1] backslash x$ for some $y$. Indeed, what may happen is that many points of $mathbb D$ might map to $x$, and when we remove those points, we may get exactly two connected components in whatever remains, thus preserving the number of connected components.
As you have seen, the first projection serves as an onto map.
If we assert that $f$ is injective, then your argument does work, because then we can say this : there is unique $x$ such that $f(x) = 0$, so we must have $f(D setminus x) = [-1,1] setminus 0$, and this is contradiction because $D setminus x$ is connected regardless of what $x$ is , but the image is disconnected.
In fact, if $f$ is injective, then since $mathbb D$ is compact and $[-1,1]$ is Hausdorff, then $f$ is actually a homeomorphism, from a common result.
Note that $mathbb D$ is a subset of $mathbb R^2$ and $[-1,1]$ is a subset of $mathbb R$. Because $2 neq 1$, we are inclined to believe that a homeomorphism is not possible.
This result, called the invariance of domain, holds in more generality : if a set in $mathbb R^m$ is homeomorphic to some other set in $mathbb R^n$, then $m=n$ must hold. We use this with $m=2,n=1$ to see your result, but the general result is more difficult to prove and requires better tools than connectedness.
answered Jul 27 at 5:51
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
31.7k22463
very clear explanation.Thanks!
– Learning Mathematics
Jul 27 at 5:57
You are welcome!
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jul 27 at 12:37
add a comment |Â
up vote
3
down vote
Sure.
If you map each point to its radius $(x,y)mapstosqrtx^2+y^2$, you have a continuous function $mathbbDto[0,1]$. A slight modification to $(x,y)mapsto2sqrtx^2+y^2-1$ should be what you want.
answered Jul 27 at 5:46
Nick Peterson
25.5k23758
Thank you sir! this one helps too!
– Learning Mathematics
Jul 27 at 5:56
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Your argument fails.
The problem is, that because $f$ is not injective, we cannot assert that since $f(mathbb D) = [-1,1]$, we have for all $x in [-1,1]$ that $f(mathbb D backslash y) = [-1,1] backslash x$ for some $y$. Indeed, what may happen is that many points of $mathbb D$ might map to $x$, and when we remove those points, we may get exactly two connected components in whatever remains, thus preserving the number of connected components.
As you have seen, the first projection serves as an onto map.
If we assert that $f$ is injective, then your argument does work, because then we can say this : there is unique $x$ such that $f(x) = 0$, so we must have $f(D setminus x) = [-1,1] setminus 0$, and this is contradiction because $D setminus x$ is connected regardless of what $x$ is , but the image is disconnected.
In fact, if $f$ is injective, then since $mathbb D$ is compact and $[-1,1]$ is Hausdorff, then $f$ is actually a homeomorphism, from a common result.
Note that $mathbb D$ is a subset of $mathbb R^2$ and $[-1,1]$ is a subset of $mathbb R$. Because $2 neq 1$, we are inclined to believe that a homeomorphism is not possible.
This result, called the invariance of domain, holds in more generality : if a set in $mathbb R^m$ is homeomorphic to some other set in $mathbb R^n$, then $m=n$ must hold. We use this with $m=2,n=1$ to see your result, but the general result is more difficult to prove and requires better tools than connectedness.
answered Jul 27 at 5:51
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
31.7k22463
very clear explanation.Thanks!
– Learning Mathematics
Jul 27 at 5:57
You are welcome!
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jul 27 at 12:37
add a comment |Â
up vote
3
down vote
accepted
Your argument fails.
The problem is, that because $f$ is not injective, we cannot assert that since $f(mathbb D) = [-1,1]$, we have for all $x in [-1,1]$ that $f(mathbb D backslash y) = [-1,1] backslash x$ for some $y$. Indeed, what may happen is that many points of $mathbb D$ might map to $x$, and when we remove those points, we may get exactly two connected components in whatever remains, thus preserving the number of connected components.
As you have seen, the first projection serves as an onto map.
If we assert that $f$ is injective, then your argument does work, because then we can say this : there is unique $x$ such that $f(x) = 0$, so we must have $f(D setminus x) = [-1,1] setminus 0$, and this is contradiction because $D setminus x$ is connected regardless of what $x$ is , but the image is disconnected.
In fact, if $f$ is injective, then since $mathbb D$ is compact and $[-1,1]$ is Hausdorff, then $f$ is actually a homeomorphism, from a common result.
Note that $mathbb D$ is a subset of $mathbb R^2$ and $[-1,1]$ is a subset of $mathbb R$. Because $2 neq 1$, we are inclined to believe that a homeomorphism is not possible.
This result, called the invariance of domain, holds in more generality : if a set in $mathbb R^m$ is homeomorphic to some other set in $mathbb R^n$, then $m=n$ must hold. We use this with $m=2,n=1$ to see your result, but the general result is more difficult to prove and requires better tools than connectedness.
answered Jul 27 at 5:51
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
31.7k22463
very clear explanation.Thanks!
– Learning Mathematics
Jul 27 at 5:57
You are welcome!
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jul 27 at 12:37
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Your argument fails.
The problem is, that because $f$ is not injective, we cannot assert that since $f(mathbb D) = [-1,1]$, we have for all $x in [-1,1]$ that $f(mathbb D backslash y) = [-1,1] backslash x$ for some $y$. Indeed, what may happen is that many points of $mathbb D$ might map to $x$, and when we remove those points, we may get exactly two connected components in whatever remains, thus preserving the number of connected components.
As you have seen, the first projection serves as an onto map.
If we assert that $f$ is injective, then your argument does work, because then we can say this : there is unique $x$ such that $f(x) = 0$, so we must have $f(D setminus x) = [-1,1] setminus 0$, and this is contradiction because $D setminus x$ is connected regardless of what $x$ is , but the image is disconnected.
In fact, if $f$ is injective, then since $mathbb D$ is compact and $[-1,1]$ is Hausdorff, then $f$ is actually a homeomorphism, from a common result.
Note that $mathbb D$ is a subset of $mathbb R^2$ and $[-1,1]$ is a subset of $mathbb R$. Because $2 neq 1$, we are inclined to believe that a homeomorphism is not possible.
This result, called the invariance of domain, holds in more generality : if a set in $mathbb R^m$ is homeomorphic to some other set in $mathbb R^n$, then $m=n$ must hold. We use this with $m=2,n=1$ to see your result, but the general result is more difficult to prove and requires better tools than connectedness.
answered Jul 27 at 5:51
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
31.7k22463
Your argument fails.
The problem is, that because $f$ is not injective, we cannot assert that since $f(mathbb D) = [-1,1]$, we have for all $x in [-1,1]$ that $f(mathbb D backslash y) = [-1,1] backslash x$ for some $y$. Indeed, what may happen is that many points of $mathbb D$ might map to $x$, and when we remove those points, we may get exactly two connected components in whatever remains, thus preserving the number of connected components.
As you have seen, the first projection serves as an onto map.
If we assert that $f$ is injective, then your argument does work, because then we can say this : there is unique $x$ such that $f(x) = 0$, so we must have $f(D setminus x) = [-1,1] setminus 0$, and this is contradiction because $D setminus x$ is connected regardless of what $x$ is , but the image is disconnected.
In fact, if $f$ is injective, then since $mathbb D$ is compact and $[-1,1]$ is Hausdorff, then $f$ is actually a homeomorphism, from a common result.
Note that $mathbb D$ is a subset of $mathbb R^2$ and $[-1,1]$ is a subset of $mathbb R$. Because $2 neq 1$, we are inclined to believe that a homeomorphism is not possible.
This result, called the invariance of domain, holds in more generality : if a set in $mathbb R^m$ is homeomorphic to some other set in $mathbb R^n$, then $m=n$ must hold. We use this with $m=2,n=1$ to see your result, but the general result is more difficult to prove and requires better tools than connectedness.
answered Jul 27 at 5:51
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
31.7k22463
answered Jul 27 at 5:51
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
31.7k22463
answered Jul 27 at 5:51
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
31.7k22463
answered Jul 27 at 5:51
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
31.7k22463
31.7k22463
very clear explanation.Thanks!
– Learning Mathematics
Jul 27 at 5:57
You are welcome!
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jul 27 at 12:37
add a comment |Â
very clear explanation.Thanks!
– Learning Mathematics
Jul 27 at 5:57
You are welcome!
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jul 27 at 12:37
very clear explanation.Thanks!
– Learning Mathematics
Jul 27 at 5:57
very clear explanation.Thanks!
– Learning Mathematics
Jul 27 at 5:57
You are welcome!
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jul 27 at 12:37
You are welcome!
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jul 27 at 12:37
add a comment |Â
up vote
3
down vote
Sure.
If you map each point to its radius $(x,y)mapstosqrtx^2+y^2$, you have a continuous function $mathbbDto[0,1]$. A slight modification to $(x,y)mapsto2sqrtx^2+y^2-1$ should be what you want.
answered Jul 27 at 5:46
Nick Peterson
25.5k23758
Thank you sir! this one helps too!
– Learning Mathematics
Jul 27 at 5:56
add a comment |Â
up vote
3
down vote
Sure.
If you map each point to its radius $(x,y)mapstosqrtx^2+y^2$, you have a continuous function $mathbbDto[0,1]$. A slight modification to $(x,y)mapsto2sqrtx^2+y^2-1$ should be what you want.
answered Jul 27 at 5:46
Nick Peterson
25.5k23758
Thank you sir! this one helps too!
– Learning Mathematics
Jul 27 at 5:56
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Sure.
If you map each point to its radius $(x,y)mapstosqrtx^2+y^2$, you have a continuous function $mathbbDto[0,1]$. A slight modification to $(x,y)mapsto2sqrtx^2+y^2-1$ should be what you want.
answered Jul 27 at 5:46
Nick Peterson
25.5k23758
Sure.
If you map each point to its radius $(x,y)mapstosqrtx^2+y^2$, you have a continuous function $mathbbDto[0,1]$. A slight modification to $(x,y)mapsto2sqrtx^2+y^2-1$ should be what you want.
answered Jul 27 at 5:46
Nick Peterson
25.5k23758
answered Jul 27 at 5:46
Nick Peterson
25.5k23758
answered Jul 27 at 5:46
Nick Peterson
25.5k23758
answered Jul 27 at 5:46
Nick Peterson
25.5k23758
25.5k23758
Thank you sir! this one helps too!
– Learning Mathematics
Jul 27 at 5:56
add a comment |Â
Thank you sir! this one helps too!
– Learning Mathematics
Jul 27 at 5:56
Thank you sir! this one helps too!
– Learning Mathematics
Jul 27 at 5:56
Thank you sir! this one helps too!
– Learning Mathematics
Jul 27 at 5:56
add a comment |Â
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7
Try $f(x,y) = x.$
– zhw.
Jul 27 at 5:40
1
Because you don't assume $f$ to be injective, "removing one point" in $[-1,1]$ may make you remove way more than just one point in $mathbbD$. It could for instance make you remove a diameter of the disk, which would then not be connected anymore, and your argument would fail.
– Suzet
Jul 27 at 5:42