Mixing
Continuous function from a closed disk to real line
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Is there a continuous function $f:B rightarrow BbbR$ which is one-one ?, where $B=(x,y) in BbbR^2: x^2+y^2 leq 1$
Suppose it were, then $f(B)=[a,b] subset BbbR$. So consider $z in B$ with $f(z) neq a,b$. Then $f(B setminus z)=[a,b] setminus f(z)$, we have a contradiction. since one is connected the other is not! Is this correct?
Any help?
real-analysis
asked Jul 20 at 15:17
Learning Mathematics
522313
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up vote
1
down vote
favorite
Is there a continuous function $f:B rightarrow BbbR$ which is one-one ?, where $B=(x,y) in BbbR^2: x^2+y^2 leq 1$
Suppose it were, then $f(B)=[a,b] subset BbbR$. So consider $z in B$ with $f(z) neq a,b$. Then $f(B setminus z)=[a,b] setminus f(z)$, we have a contradiction. since one is connected the other is not! Is this correct?
Any help?
real-analysis
asked Jul 20 at 15:17
Learning Mathematics
522313
1
Yes, this is correct
– User
Jul 20 at 15:21
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Is there a continuous function $f:B rightarrow BbbR$ which is one-one ?, where $B=(x,y) in BbbR^2: x^2+y^2 leq 1$
Suppose it were, then $f(B)=[a,b] subset BbbR$. So consider $z in B$ with $f(z) neq a,b$. Then $f(B setminus z)=[a,b] setminus f(z)$, we have a contradiction. since one is connected the other is not! Is this correct?
Any help?
real-analysis
asked Jul 20 at 15:17
Learning Mathematics
522313
Is there a continuous function $f:B rightarrow BbbR$ which is one-one ?, where $B=(x,y) in BbbR^2: x^2+y^2 leq 1$
Suppose it were, then $f(B)=[a,b] subset BbbR$. So consider $z in B$ with $f(z) neq a,b$. Then $f(B setminus z)=[a,b] setminus f(z)$, we have a contradiction. since one is connected the other is not! Is this correct?
Any help?
real-analysis
asked Jul 20 at 15:17
Learning Mathematics
522313
asked Jul 20 at 15:17
Learning Mathematics
522313
asked Jul 20 at 15:17
Learning Mathematics
522313
asked Jul 20 at 15:17
Learning Mathematics
522313
522313
1
Yes, this is correct
– User
Jul 20 at 15:21
add a comment |Â
1
Yes, this is correct
– User
Jul 20 at 15:21
1
1
Yes, this is correct
– User
Jul 20 at 15:21
Yes, this is correct
– User
Jul 20 at 15:21
add a comment |Â
1 Answer
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It is correct, assuming you know that $f(B)$ is of the form $[a,b]$ with $a ne b$.
To show it, note that $B$ is compact and connected so $f(B)$ is also compact and connected in $mathbbR$. Hence $f(B)$ is a segment $[a,b]$. Since $f$ is injective and $B$ is uncountable, $f(B)$ cannot be a singleton so $a ne b$.
answered Jul 20 at 15:22
mechanodroid
22.2k52041
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
It is correct, assuming you know that $f(B)$ is of the form $[a,b]$ with $a ne b$.
To show it, note that $B$ is compact and connected so $f(B)$ is also compact and connected in $mathbbR$. Hence $f(B)$ is a segment $[a,b]$. Since $f$ is injective and $B$ is uncountable, $f(B)$ cannot be a singleton so $a ne b$.
answered Jul 20 at 15:22
mechanodroid
22.2k52041
add a comment |Â
up vote
3
down vote
accepted
It is correct, assuming you know that $f(B)$ is of the form $[a,b]$ with $a ne b$.
To show it, note that $B$ is compact and connected so $f(B)$ is also compact and connected in $mathbbR$. Hence $f(B)$ is a segment $[a,b]$. Since $f$ is injective and $B$ is uncountable, $f(B)$ cannot be a singleton so $a ne b$.
answered Jul 20 at 15:22
mechanodroid
22.2k52041
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
It is correct, assuming you know that $f(B)$ is of the form $[a,b]$ with $a ne b$.
To show it, note that $B$ is compact and connected so $f(B)$ is also compact and connected in $mathbbR$. Hence $f(B)$ is a segment $[a,b]$. Since $f$ is injective and $B$ is uncountable, $f(B)$ cannot be a singleton so $a ne b$.
answered Jul 20 at 15:22
mechanodroid
22.2k52041
It is correct, assuming you know that $f(B)$ is of the form $[a,b]$ with $a ne b$.
To show it, note that $B$ is compact and connected so $f(B)$ is also compact and connected in $mathbbR$. Hence $f(B)$ is a segment $[a,b]$. Since $f$ is injective and $B$ is uncountable, $f(B)$ cannot be a singleton so $a ne b$.
answered Jul 20 at 15:22
mechanodroid
22.2k52041
answered Jul 20 at 15:22
mechanodroid
22.2k52041
answered Jul 20 at 15:22
mechanodroid
22.2k52041
answered Jul 20 at 15:22
mechanodroid
22.2k52041
22.2k52041
add a comment |Â
add a comment |Â
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1
Yes, this is correct
– User
Jul 20 at 15:21