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What are the compactifications of those spirals?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)How to describe the one point compactification of a spaceLooking for reference to a couple of proofs regarding the Stereographic Projection.Are a finite cylinder and the corresponding planes iso/homeomorphic?Show that the compactification is the Alexandroff-compactificationHow to show a function can or cannot be extended to a compactification?Show that a whole circle cannot be parametrized by a single mapCompactifications $Y_1geq Y_2$Question about the one point compactification $mathbbR^2 cup infty$ of $S^2$Homeomorphism from punctured sphere to horn torusQuestions about one-point compactification
$begingroup$
I want to find the explicit compactification (I don't mind if its a one-point compactification or not) of the spiral
$E_1=left left(e^tcost,e^tsintright):tinmathbbRright .$
and the spiral
$E_2=left left(fractt+1cost,fractt+1sintright):tinleft(0,+inftyright)right$
What I've managed to do is... for $E_1$:
First of all, I thought that I had to add $left(0,0right)$ because the limit when $trightarrow$-$infty$ doesn't reach it, so the $infty$ has to be linked to the origin, and because of the fact that $fleft(tright)=left(e^tcost,e^tsintright)$ is a parametrization, so it is an homeomorphism to $mathbbR^2$ so the compactification that comes to me is the one that comes with the map $fleft(tright)=begincases
left(e^tcost,e^tsintright) & tinmathbbR\
left(0,0right) & t=-infty
endcases$ but $partial E_1$ is what fails. So I don't think that I could the next one.
Thanks in advance!
general-topology
asked Apr 2 at 17:54
Jack TalionJack Talion
727
$endgroup$
add a comment |
$begingroup$
I want to find the explicit compactification (I don't mind if its a one-point compactification or not) of the spiral
$E_1=left left(e^tcost,e^tsintright):tinmathbbRright .$
and the spiral
$E_2=left left(fractt+1cost,fractt+1sintright):tinleft(0,+inftyright)right$
What I've managed to do is... for $E_1$:
First of all, I thought that I had to add $left(0,0right)$ because the limit when $trightarrow$-$infty$ doesn't reach it, so the $infty$ has to be linked to the origin, and because of the fact that $fleft(tright)=left(e^tcost,e^tsintright)$ is a parametrization, so it is an homeomorphism to $mathbbR^2$ so the compactification that comes to me is the one that comes with the map $fleft(tright)=begincases
left(e^tcost,e^tsintright) & tinmathbbR\
left(0,0right) & t=-infty
endcases$ but $partial E_1$ is what fails. So I don't think that I could the next one.
Thanks in advance!
general-topology
asked Apr 2 at 17:54
Jack TalionJack Talion
727
$endgroup$
1
$begingroup$
One of the possible approaches I see is following. Your curve $E_2$ is homeomorphic to half-line $R_+$ or equivalently $[0,1)$ see classification of 1-manifolds. Now the one point compactification of $[0,1)$ is $[0,1]$. Thus $E_2$ can be compactified intto $[0,1]$. Don't know if this helps. You only need to find the above homeomorphism.
$endgroup$
– ersh
Apr 2 at 18:16
$begingroup$
@ersh Thank you!! And for the other one?
$endgroup$
– Jack Talion
Apr 2 at 18:24
add a comment |
$begingroup$
I want to find the explicit compactification (I don't mind if its a one-point compactification or not) of the spiral
$E_1=left left(e^tcost,e^tsintright):tinmathbbRright .$
and the spiral
$E_2=left left(fractt+1cost,fractt+1sintright):tinleft(0,+inftyright)right$
What I've managed to do is... for $E_1$:
First of all, I thought that I had to add $left(0,0right)$ because the limit when $trightarrow$-$infty$ doesn't reach it, so the $infty$ has to be linked to the origin, and because of the fact that $fleft(tright)=left(e^tcost,e^tsintright)$ is a parametrization, so it is an homeomorphism to $mathbbR^2$ so the compactification that comes to me is the one that comes with the map $fleft(tright)=begincases
left(e^tcost,e^tsintright) & tinmathbbR\
left(0,0right) & t=-infty
endcases$ but $partial E_1$ is what fails. So I don't think that I could the next one.
Thanks in advance!
general-topology
asked Apr 2 at 17:54
Jack TalionJack Talion
727
$endgroup$
I want to find the explicit compactification (I don't mind if its a one-point compactification or not) of the spiral
$E_1=left left(e^tcost,e^tsintright):tinmathbbRright .$
and the spiral
$E_2=left left(fractt+1cost,fractt+1sintright):tinleft(0,+inftyright)right$
What I've managed to do is... for $E_1$:
First of all, I thought that I had to add $left(0,0right)$ because the limit when $trightarrow$-$infty$ doesn't reach it, so the $infty$ has to be linked to the origin, and because of the fact that $fleft(tright)=left(e^tcost,e^tsintright)$ is a parametrization, so it is an homeomorphism to $mathbbR^2$ so the compactification that comes to me is the one that comes with the map $fleft(tright)=begincases
left(e^tcost,e^tsintright) & tinmathbbR\
left(0,0right) & t=-infty
endcases$ but $partial E_1$ is what fails. So I don't think that I could the next one.
Thanks in advance!
general-topology
general-topology
asked Apr 2 at 17:54
Jack TalionJack Talion
727
asked Apr 2 at 17:54
Jack TalionJack Talion
727
edited Apr 2 at 18:23
Jack Talion
asked Apr 2 at 17:54
Jack TalionJack Talion
727
asked Apr 2 at 17:54
Jack TalionJack Talion
727
asked Apr 2 at 17:54
Jack TalionJack Talion
727
727
1
$begingroup$
One of the possible approaches I see is following. Your curve $E_2$ is homeomorphic to half-line $R_+$ or equivalently $[0,1)$ see classification of 1-manifolds. Now the one point compactification of $[0,1)$ is $[0,1]$. Thus $E_2$ can be compactified intto $[0,1]$. Don't know if this helps. You only need to find the above homeomorphism.
$endgroup$
– ersh
Apr 2 at 18:16
$begingroup$
@ersh Thank you!! And for the other one?
$endgroup$
– Jack Talion
Apr 2 at 18:24
add a comment |
1
$begingroup$
One of the possible approaches I see is following. Your curve $E_2$ is homeomorphic to half-line $R_+$ or equivalently $[0,1)$ see classification of 1-manifolds. Now the one point compactification of $[0,1)$ is $[0,1]$. Thus $E_2$ can be compactified intto $[0,1]$. Don't know if this helps. You only need to find the above homeomorphism.
$endgroup$
– ersh
Apr 2 at 18:16
$begingroup$
@ersh Thank you!! And for the other one?
$endgroup$
– Jack Talion
Apr 2 at 18:24
1
1
$begingroup$
One of the possible approaches I see is following. Your curve $E_2$ is homeomorphic to half-line $R_+$ or equivalently $[0,1)$ see classification of 1-manifolds. Now the one point compactification of $[0,1)$ is $[0,1]$. Thus $E_2$ can be compactified intto $[0,1]$. Don't know if this helps. You only need to find the above homeomorphism.
$endgroup$
– ersh
Apr 2 at 18:16
$begingroup$
One of the possible approaches I see is following. Your curve $E_2$ is homeomorphic to half-line $R_+$ or equivalently $[0,1)$ see classification of 1-manifolds. Now the one point compactification of $[0,1)$ is $[0,1]$. Thus $E_2$ can be compactified intto $[0,1]$. Don't know if this helps. You only need to find the above homeomorphism.
$endgroup$
– ersh
Apr 2 at 18:16
$begingroup$
@ersh Thank you!! And for the other one?
$endgroup$
– Jack Talion
Apr 2 at 18:24
$begingroup$
@ersh Thank you!! And for the other one?
$endgroup$
– Jack Talion
Apr 2 at 18:24
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A compactification of $E_2$ is $E_2 cup S^1.$
answered Apr 2 at 21:47
William ElliotWilliam Elliot
9,2562820
$endgroup$
$begingroup$
+1... To the proposer: Closed bounded subsets of $Bbb R^2$ are compact so if $E$ is any bounded subset of $Bbb R^2$ then $id_E:Eto overline E$ is a compactification of $E.$ Also, BTW,$; E_2$ is a locally compact Tychonoff space so it has a $1$-point compactification.
$endgroup$
– DanielWainfleet
Apr 3 at 5:36
add a comment |
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$begingroup$
A compactification of $E_2$ is $E_2 cup S^1.$
answered Apr 2 at 21:47
William ElliotWilliam Elliot
9,2562820
$endgroup$
$begingroup$
+1... To the proposer: Closed bounded subsets of $Bbb R^2$ are compact so if $E$ is any bounded subset of $Bbb R^2$ then $id_E:Eto overline E$ is a compactification of $E.$ Also, BTW,$; E_2$ is a locally compact Tychonoff space so it has a $1$-point compactification.
$endgroup$
– DanielWainfleet
Apr 3 at 5:36
add a comment |
$begingroup$
A compactification of $E_2$ is $E_2 cup S^1.$
answered Apr 2 at 21:47
William ElliotWilliam Elliot
9,2562820
$endgroup$
$begingroup$
+1... To the proposer: Closed bounded subsets of $Bbb R^2$ are compact so if $E$ is any bounded subset of $Bbb R^2$ then $id_E:Eto overline E$ is a compactification of $E.$ Also, BTW,$; E_2$ is a locally compact Tychonoff space so it has a $1$-point compactification.
$endgroup$
– DanielWainfleet
Apr 3 at 5:36
add a comment |
$begingroup$
A compactification of $E_2$ is $E_2 cup S^1.$
answered Apr 2 at 21:47
William ElliotWilliam Elliot
9,2562820
$endgroup$
A compactification of $E_2$ is $E_2 cup S^1.$
answered Apr 2 at 21:47
William ElliotWilliam Elliot
9,2562820
answered Apr 2 at 21:47
William ElliotWilliam Elliot
9,2562820
answered Apr 2 at 21:47
William ElliotWilliam Elliot
9,2562820
answered Apr 2 at 21:47
William ElliotWilliam Elliot
9,2562820
9,2562820
$begingroup$
+1... To the proposer: Closed bounded subsets of $Bbb R^2$ are compact so if $E$ is any bounded subset of $Bbb R^2$ then $id_E:Eto overline E$ is a compactification of $E.$ Also, BTW,$; E_2$ is a locally compact Tychonoff space so it has a $1$-point compactification.
$endgroup$
– DanielWainfleet
Apr 3 at 5:36
add a comment |
$begingroup$
+1... To the proposer: Closed bounded subsets of $Bbb R^2$ are compact so if $E$ is any bounded subset of $Bbb R^2$ then $id_E:Eto overline E$ is a compactification of $E.$ Also, BTW,$; E_2$ is a locally compact Tychonoff space so it has a $1$-point compactification.
$endgroup$
– DanielWainfleet
Apr 3 at 5:36
$begingroup$
+1... To the proposer: Closed bounded subsets of $Bbb R^2$ are compact so if $E$ is any bounded subset of $Bbb R^2$ then $id_E:Eto overline E$ is a compactification of $E.$ Also, BTW,$; E_2$ is a locally compact Tychonoff space so it has a $1$-point compactification.
$endgroup$
– DanielWainfleet
Apr 3 at 5:36
$begingroup$
+1... To the proposer: Closed bounded subsets of $Bbb R^2$ are compact so if $E$ is any bounded subset of $Bbb R^2$ then $id_E:Eto overline E$ is a compactification of $E.$ Also, BTW,$; E_2$ is a locally compact Tychonoff space so it has a $1$-point compactification.
$endgroup$
– DanielWainfleet
Apr 3 at 5:36
add a comment |
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$begingroup$
One of the possible approaches I see is following. Your curve $E_2$ is homeomorphic to half-line $R_+$ or equivalently $[0,1)$ see classification of 1-manifolds. Now the one point compactification of $[0,1)$ is $[0,1]$. Thus $E_2$ can be compactified intto $[0,1]$. Don't know if this helps. You only need to find the above homeomorphism.
$endgroup$
– ersh
Apr 2 at 18:16
$begingroup$
@ersh Thank you!! And for the other one?
$endgroup$
– Jack Talion
Apr 2 at 18:24