Mixing
Let I denote the unit interval $[0, 1].$ Which of the following statements are true?
$begingroup$
Let $B := {(x, y) in mathbb{R}^2
: x^2 + y^2 le 1}$be the closed ball in $mathbb{R^2}$ with center at the origin.
Let I denote the unit interval $[0, 1].$ Which of the following statements are true?
Which of the following statements are true?
$(a)$ There exists a continuous function $f : B rightarrow mathbb{R}$ which is one-one
$(b)$ There exists a continuous function $f : B rightarrow mathbb{R}$ which is onto.
$(c)$ There exists a continuous function $f : B rightarrow I × I$ which is one-one.
$(d)$ There exists a continuous function $f : B rightarrow I × I$ which is onto.
I thinks none of option will be correct
option $a)$ and option $b)$ is false Just using the logics of compactness, that is $mathbb{R}$ is not compacts
option c) and option d) is false just using the logic of connectedness that is $B-{0}$ is not connected but $I × I-{0}$ is connectedness
Is my logics is correct or not ?
Any hints/solution will be appreciated
thanks u
general-topology
asked Jan 11 at 17:40
jasminejasmine
1,706417
$endgroup$
|
show 3 more comments
$begingroup$
Let $B := {(x, y) in mathbb{R}^2
: x^2 + y^2 le 1}$be the closed ball in $mathbb{R^2}$ with center at the origin.
Let I denote the unit interval $[0, 1].$ Which of the following statements are true?
Which of the following statements are true?
$(a)$ There exists a continuous function $f : B rightarrow mathbb{R}$ which is one-one
$(b)$ There exists a continuous function $f : B rightarrow mathbb{R}$ which is onto.
$(c)$ There exists a continuous function $f : B rightarrow I × I$ which is one-one.
$(d)$ There exists a continuous function $f : B rightarrow I × I$ which is onto.
I thinks none of option will be correct
option $a)$ and option $b)$ is false Just using the logics of compactness, that is $mathbb{R}$ is not compacts
option c) and option d) is false just using the logic of connectedness that is $B-{0}$ is not connected but $I × I-{0}$ is connectedness
Is my logics is correct or not ?
Any hints/solution will be appreciated
thanks u
general-topology
asked Jan 11 at 17:40
jasminejasmine
1,706417
$endgroup$
1
$begingroup$
why does compactness of $B$ help in (a)? $f$ is not onto there. Also, $B-{0}$ is certainly connected.
$endgroup$
– Randall
Jan 11 at 17:43
$begingroup$
@Randall B is a circle , cut the circle it will disconnect
$endgroup$
– jasmine
Jan 11 at 17:45
2
$begingroup$
I have no idea what you're saying. $B$ is a solid disk. If you poke a hole in a disk it is still connected.
$endgroup$
– Randall
Jan 11 at 17:46
1
$begingroup$
I don't know....
$endgroup$
– Randall
Jan 11 at 17:52
1
$begingroup$
@Randall compactness does help if you know dimension theory too: if $f: B to mathbb{R}$ were continuous and 1-1, $f[B]$ would be homeomorphic to $B$ by compactness. But $dim f[B] le 1$ while $dim B=2$. Bit overkill though.
$endgroup$
– Henno Brandsma
Jan 11 at 18:41
|
show 3 more comments
$begingroup$
Let $B := {(x, y) in mathbb{R}^2
: x^2 + y^2 le 1}$be the closed ball in $mathbb{R^2}$ with center at the origin.
Let I denote the unit interval $[0, 1].$ Which of the following statements are true?
Which of the following statements are true?
$(a)$ There exists a continuous function $f : B rightarrow mathbb{R}$ which is one-one
$(b)$ There exists a continuous function $f : B rightarrow mathbb{R}$ which is onto.
$(c)$ There exists a continuous function $f : B rightarrow I × I$ which is one-one.
$(d)$ There exists a continuous function $f : B rightarrow I × I$ which is onto.
I thinks none of option will be correct
option $a)$ and option $b)$ is false Just using the logics of compactness, that is $mathbb{R}$ is not compacts
option c) and option d) is false just using the logic of connectedness that is $B-{0}$ is not connected but $I × I-{0}$ is connectedness
Is my logics is correct or not ?
Any hints/solution will be appreciated
thanks u
general-topology
asked Jan 11 at 17:40
jasminejasmine
1,706417
$endgroup$
Let $B := {(x, y) in mathbb{R}^2
: x^2 + y^2 le 1}$be the closed ball in $mathbb{R^2}$ with center at the origin.
Let I denote the unit interval $[0, 1].$ Which of the following statements are true?
Which of the following statements are true?
$(a)$ There exists a continuous function $f : B rightarrow mathbb{R}$ which is one-one
$(b)$ There exists a continuous function $f : B rightarrow mathbb{R}$ which is onto.
$(c)$ There exists a continuous function $f : B rightarrow I × I$ which is one-one.
$(d)$ There exists a continuous function $f : B rightarrow I × I$ which is onto.
I thinks none of option will be correct
option $a)$ and option $b)$ is false Just using the logics of compactness, that is $mathbb{R}$ is not compacts
option c) and option d) is false just using the logic of connectedness that is $B-{0}$ is not connected but $I × I-{0}$ is connectedness
Is my logics is correct or not ?
Any hints/solution will be appreciated
thanks u
general-topology
general-topology
asked Jan 11 at 17:40
jasminejasmine
1,706417
asked Jan 11 at 17:40
jasminejasmine
1,706417
asked Jan 11 at 17:40
jasminejasmine
1,706417
asked Jan 11 at 17:40
jasminejasmine
1,706417
asked Jan 11 at 17:40
jasminejasmine
1,706417
1,706417
1
$begingroup$
why does compactness of $B$ help in (a)? $f$ is not onto there. Also, $B-{0}$ is certainly connected.
$endgroup$
– Randall
Jan 11 at 17:43
$begingroup$
@Randall B is a circle , cut the circle it will disconnect
$endgroup$
– jasmine
Jan 11 at 17:45
2
$begingroup$
I have no idea what you're saying. $B$ is a solid disk. If you poke a hole in a disk it is still connected.
$endgroup$
– Randall
Jan 11 at 17:46
1
$begingroup$
I don't know....
$endgroup$
– Randall
Jan 11 at 17:52
1
$begingroup$
@Randall compactness does help if you know dimension theory too: if $f: B to mathbb{R}$ were continuous and 1-1, $f[B]$ would be homeomorphic to $B$ by compactness. But $dim f[B] le 1$ while $dim B=2$. Bit overkill though.
$endgroup$
– Henno Brandsma
Jan 11 at 18:41
|
show 3 more comments
1
$begingroup$
why does compactness of $B$ help in (a)? $f$ is not onto there. Also, $B-{0}$ is certainly connected.
$endgroup$
– Randall
Jan 11 at 17:43
$begingroup$
@Randall B is a circle , cut the circle it will disconnect
$endgroup$
– jasmine
Jan 11 at 17:45
2
$begingroup$
I have no idea what you're saying. $B$ is a solid disk. If you poke a hole in a disk it is still connected.
$endgroup$
– Randall
Jan 11 at 17:46
1
$begingroup$
I don't know....
$endgroup$
– Randall
Jan 11 at 17:52
1
$begingroup$
@Randall compactness does help if you know dimension theory too: if $f: B to mathbb{R}$ were continuous and 1-1, $f[B]$ would be homeomorphic to $B$ by compactness. But $dim f[B] le 1$ while $dim B=2$. Bit overkill though.
$endgroup$
– Henno Brandsma
Jan 11 at 18:41
1
1
$begingroup$
why does compactness of $B$ help in (a)? $f$ is not onto there. Also, $B-{0}$ is certainly connected.
$endgroup$
– Randall
Jan 11 at 17:43
$begingroup$
why does compactness of $B$ help in (a)? $f$ is not onto there. Also, $B-{0}$ is certainly connected.
$endgroup$
– Randall
Jan 11 at 17:43
$begingroup$
@Randall B is a circle , cut the circle it will disconnect
$endgroup$
– jasmine
Jan 11 at 17:45
$begingroup$
@Randall B is a circle , cut the circle it will disconnect
$endgroup$
– jasmine
Jan 11 at 17:45
2
2
$begingroup$
I have no idea what you're saying. $B$ is a solid disk. If you poke a hole in a disk it is still connected.
$endgroup$
– Randall
Jan 11 at 17:46
$begingroup$
I have no idea what you're saying. $B$ is a solid disk. If you poke a hole in a disk it is still connected.
$endgroup$
– Randall
Jan 11 at 17:46
1
1
$begingroup$
I don't know....
$endgroup$
– Randall
Jan 11 at 17:52
$begingroup$
I don't know....
$endgroup$
– Randall
Jan 11 at 17:52
1
1
$begingroup$
@Randall compactness does help if you know dimension theory too: if $f: B to mathbb{R}$ were continuous and 1-1, $f[B]$ would be homeomorphic to $B$ by compactness. But $dim f[B] le 1$ while $dim B=2$. Bit overkill though.
$endgroup$
– Henno Brandsma
Jan 11 at 18:41
$begingroup$
@Randall compactness does help if you know dimension theory too: if $f: B to mathbb{R}$ were continuous and 1-1, $f[B]$ would be homeomorphic to $B$ by compactness. But $dim f[B] le 1$ while $dim B=2$. Bit overkill though.
$endgroup$
– Henno Brandsma
Jan 11 at 18:41
|
show 3 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Option a) is false because we cannot even find such a map from $S^1$, the unit circle by the simplest version of Borsuk-Ulam: there are already points $x$ and $-x$ on the boundary of $B$ that have the same value.
Option b) is indeed most easily disproved by noting that $f[B]$ is compact and the reals are not.
Options c) and d) are true: $B$ is homeomorphic to $ I times I$, as is well-known. A homeomorphism will fulfill both. Note that $Bsetminus{0}$ is actually connected so your proposed argument doesn’t work.
answered Jan 11 at 18:34
Henno BrandsmaHenno Brandsma
108k347114
$endgroup$
add a comment |
$begingroup$
(a) is false
As $B$ is a compact and $f$ is continuous, $f$ is an homeomorphism from the compact $B$ to $f[B]$. As $B$ is connected, $f[B]$ is also connected and is therefore an interval. However $Bsetminus {0}$ is connected and $f[Bsetminus {0}]$ cannot be connected. That can’t be as $f$ is an homeomorphism.
(b) is false
The image of the compact $B$ must be compact and $mathbb R$ isn’t.
(c) and (d) are true
Consider the application that transform a ray of the unit ball into the line segment joining the origin of $B$ to the point of square aligned with the original ray.
answered Jan 11 at 18:37
mathcounterexamples.netmathcounterexamples.net
26.7k22157
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Option a) is false because we cannot even find such a map from $S^1$, the unit circle by the simplest version of Borsuk-Ulam: there are already points $x$ and $-x$ on the boundary of $B$ that have the same value.
Option b) is indeed most easily disproved by noting that $f[B]$ is compact and the reals are not.
Options c) and d) are true: $B$ is homeomorphic to $ I times I$, as is well-known. A homeomorphism will fulfill both. Note that $Bsetminus{0}$ is actually connected so your proposed argument doesn’t work.
answered Jan 11 at 18:34
Henno BrandsmaHenno Brandsma
108k347114
$endgroup$
add a comment |
$begingroup$
Option a) is false because we cannot even find such a map from $S^1$, the unit circle by the simplest version of Borsuk-Ulam: there are already points $x$ and $-x$ on the boundary of $B$ that have the same value.
Option b) is indeed most easily disproved by noting that $f[B]$ is compact and the reals are not.
Options c) and d) are true: $B$ is homeomorphic to $ I times I$, as is well-known. A homeomorphism will fulfill both. Note that $Bsetminus{0}$ is actually connected so your proposed argument doesn’t work.
answered Jan 11 at 18:34
Henno BrandsmaHenno Brandsma
108k347114
$endgroup$
add a comment |
$begingroup$
Option a) is false because we cannot even find such a map from $S^1$, the unit circle by the simplest version of Borsuk-Ulam: there are already points $x$ and $-x$ on the boundary of $B$ that have the same value.
Option b) is indeed most easily disproved by noting that $f[B]$ is compact and the reals are not.
Options c) and d) are true: $B$ is homeomorphic to $ I times I$, as is well-known. A homeomorphism will fulfill both. Note that $Bsetminus{0}$ is actually connected so your proposed argument doesn’t work.
answered Jan 11 at 18:34
Henno BrandsmaHenno Brandsma
108k347114
$endgroup$
Option a) is false because we cannot even find such a map from $S^1$, the unit circle by the simplest version of Borsuk-Ulam: there are already points $x$ and $-x$ on the boundary of $B$ that have the same value.
Option b) is indeed most easily disproved by noting that $f[B]$ is compact and the reals are not.
Options c) and d) are true: $B$ is homeomorphic to $ I times I$, as is well-known. A homeomorphism will fulfill both. Note that $Bsetminus{0}$ is actually connected so your proposed argument doesn’t work.
answered Jan 11 at 18:34
Henno BrandsmaHenno Brandsma
108k347114
answered Jan 11 at 18:34
Henno BrandsmaHenno Brandsma
108k347114
answered Jan 11 at 18:34
Henno BrandsmaHenno Brandsma
108k347114
answered Jan 11 at 18:34
Henno BrandsmaHenno Brandsma
108k347114
108k347114
add a comment |
add a comment |
$begingroup$
(a) is false
As $B$ is a compact and $f$ is continuous, $f$ is an homeomorphism from the compact $B$ to $f[B]$. As $B$ is connected, $f[B]$ is also connected and is therefore an interval. However $Bsetminus {0}$ is connected and $f[Bsetminus {0}]$ cannot be connected. That can’t be as $f$ is an homeomorphism.
(b) is false
The image of the compact $B$ must be compact and $mathbb R$ isn’t.
(c) and (d) are true
Consider the application that transform a ray of the unit ball into the line segment joining the origin of $B$ to the point of square aligned with the original ray.
answered Jan 11 at 18:37
mathcounterexamples.netmathcounterexamples.net
26.7k22157
$endgroup$
add a comment |
$begingroup$
(a) is false
As $B$ is a compact and $f$ is continuous, $f$ is an homeomorphism from the compact $B$ to $f[B]$. As $B$ is connected, $f[B]$ is also connected and is therefore an interval. However $Bsetminus {0}$ is connected and $f[Bsetminus {0}]$ cannot be connected. That can’t be as $f$ is an homeomorphism.
(b) is false
The image of the compact $B$ must be compact and $mathbb R$ isn’t.
(c) and (d) are true
Consider the application that transform a ray of the unit ball into the line segment joining the origin of $B$ to the point of square aligned with the original ray.
answered Jan 11 at 18:37
mathcounterexamples.netmathcounterexamples.net
26.7k22157
$endgroup$
add a comment |
$begingroup$
(a) is false
As $B$ is a compact and $f$ is continuous, $f$ is an homeomorphism from the compact $B$ to $f[B]$. As $B$ is connected, $f[B]$ is also connected and is therefore an interval. However $Bsetminus {0}$ is connected and $f[Bsetminus {0}]$ cannot be connected. That can’t be as $f$ is an homeomorphism.
(b) is false
The image of the compact $B$ must be compact and $mathbb R$ isn’t.
(c) and (d) are true
Consider the application that transform a ray of the unit ball into the line segment joining the origin of $B$ to the point of square aligned with the original ray.
answered Jan 11 at 18:37
mathcounterexamples.netmathcounterexamples.net
26.7k22157
$endgroup$
(a) is false
As $B$ is a compact and $f$ is continuous, $f$ is an homeomorphism from the compact $B$ to $f[B]$. As $B$ is connected, $f[B]$ is also connected and is therefore an interval. However $Bsetminus {0}$ is connected and $f[Bsetminus {0}]$ cannot be connected. That can’t be as $f$ is an homeomorphism.
(b) is false
The image of the compact $B$ must be compact and $mathbb R$ isn’t.
(c) and (d) are true
Consider the application that transform a ray of the unit ball into the line segment joining the origin of $B$ to the point of square aligned with the original ray.
answered Jan 11 at 18:37
mathcounterexamples.netmathcounterexamples.net
26.7k22157
answered Jan 11 at 18:37
mathcounterexamples.netmathcounterexamples.net
26.7k22157
answered Jan 11 at 18:37
mathcounterexamples.netmathcounterexamples.net
26.7k22157
answered Jan 11 at 18:37
mathcounterexamples.netmathcounterexamples.net
26.7k22157
26.7k22157
add a comment |
add a comment |
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1
$begingroup$
why does compactness of $B$ help in (a)? $f$ is not onto there. Also, $B-{0}$ is certainly connected.
$endgroup$
– Randall
Jan 11 at 17:43
$begingroup$
@Randall B is a circle , cut the circle it will disconnect
$endgroup$
– jasmine
Jan 11 at 17:45
2
$begingroup$
I have no idea what you're saying. $B$ is a solid disk. If you poke a hole in a disk it is still connected.
$endgroup$
– Randall
Jan 11 at 17:46
1
$begingroup$
I don't know....
$endgroup$
– Randall
Jan 11 at 17:52
1
$begingroup$
@Randall compactness does help if you know dimension theory too: if $f: B to mathbb{R}$ were continuous and 1-1, $f[B]$ would be homeomorphic to $B$ by compactness. But $dim f[B] le 1$ while $dim B=2$. Bit overkill though.
$endgroup$
– Henno Brandsma
Jan 11 at 18:41