Prove that the map $ zmapsto z^3 $ is a closed map.












1












$begingroup$



Prove that the map which winds the plane three times on itself given, in terms of complex numbers, by $ zmapsto z^3 $ is a closed map.




I was trying to solve problem 20 in chapter 2 from Basic Topology by Armstrong.



I have already proved that the map $ zmapsto z^3 $ is an open map, how am I supposed to show that it is also a closed map?



Can I directly show that $ mathbb{C}tomathbb{C}: zmapsto z^3 $ is a homeomorphism thus it is also a closed map?



Edit: A closed map takes closed sets to closed sets. Furthermore, a map is a continuous function.






general-topology continuity







share|cite|improve this question











$endgroup$












  • $begingroup$
    It can't be a homeomorphism as usually defined--- not one to one. BTW what definition of "closed map" do you use?
    $endgroup$
    – coffeemath
    Dec 24 '18 at 8:18












  • $begingroup$
    @coffeemath A closed map takes closed sets to closed sets. Furthermore, a map is a continuous function.
    $endgroup$
    – user549397
    Dec 24 '18 at 8:23
















1












$begingroup$



Prove that the map which winds the plane three times on itself given, in terms of complex numbers, by $ zmapsto z^3 $ is a closed map.




I was trying to solve problem 20 in chapter 2 from Basic Topology by Armstrong.



I have already proved that the map $ zmapsto z^3 $ is an open map, how am I supposed to show that it is also a closed map?



Can I directly show that $ mathbb{C}tomathbb{C}: zmapsto z^3 $ is a homeomorphism thus it is also a closed map?



Edit: A closed map takes closed sets to closed sets. Furthermore, a map is a continuous function.






general-topology continuity







share|cite|improve this question











$endgroup$












  • $begingroup$
    It can't be a homeomorphism as usually defined--- not one to one. BTW what definition of "closed map" do you use?
    $endgroup$
    – coffeemath
    Dec 24 '18 at 8:18












  • $begingroup$
    @coffeemath A closed map takes closed sets to closed sets. Furthermore, a map is a continuous function.
    $endgroup$
    – user549397
    Dec 24 '18 at 8:23














1












1








1


1



$begingroup$



Prove that the map which winds the plane three times on itself given, in terms of complex numbers, by $ zmapsto z^3 $ is a closed map.




I was trying to solve problem 20 in chapter 2 from Basic Topology by Armstrong.



I have already proved that the map $ zmapsto z^3 $ is an open map, how am I supposed to show that it is also a closed map?



Can I directly show that $ mathbb{C}tomathbb{C}: zmapsto z^3 $ is a homeomorphism thus it is also a closed map?



Edit: A closed map takes closed sets to closed sets. Furthermore, a map is a continuous function.






general-topology continuity







share|cite|improve this question











$endgroup$





Prove that the map which winds the plane three times on itself given, in terms of complex numbers, by $ zmapsto z^3 $ is a closed map.




I was trying to solve problem 20 in chapter 2 from Basic Topology by Armstrong.



I have already proved that the map $ zmapsto z^3 $ is an open map, how am I supposed to show that it is also a closed map?



Can I directly show that $ mathbb{C}tomathbb{C}: zmapsto z^3 $ is a homeomorphism thus it is also a closed map?



Edit: A closed map takes closed sets to closed sets. Furthermore, a map is a continuous function.






general-topology continuity




general-topology continuity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 24 '18 at 8:22







user549397

















asked Dec 24 '18 at 8:15









user549397user549397

1,6591418




1,6591418












  • $begingroup$
    It can't be a homeomorphism as usually defined--- not one to one. BTW what definition of "closed map" do you use?
    $endgroup$
    – coffeemath
    Dec 24 '18 at 8:18












  • $begingroup$
    @coffeemath A closed map takes closed sets to closed sets. Furthermore, a map is a continuous function.
    $endgroup$
    – user549397
    Dec 24 '18 at 8:23


















  • $begingroup$
    It can't be a homeomorphism as usually defined--- not one to one. BTW what definition of "closed map" do you use?
    $endgroup$
    – coffeemath
    Dec 24 '18 at 8:18












  • $begingroup$
    @coffeemath A closed map takes closed sets to closed sets. Furthermore, a map is a continuous function.
    $endgroup$
    – user549397
    Dec 24 '18 at 8:23
















$begingroup$
It can't be a homeomorphism as usually defined--- not one to one. BTW what definition of "closed map" do you use?
$endgroup$
– coffeemath
Dec 24 '18 at 8:18






$begingroup$
It can't be a homeomorphism as usually defined--- not one to one. BTW what definition of "closed map" do you use?
$endgroup$
– coffeemath
Dec 24 '18 at 8:18














$begingroup$
@coffeemath A closed map takes closed sets to closed sets. Furthermore, a map is a continuous function.
$endgroup$
– user549397
Dec 24 '18 at 8:23




$begingroup$
@coffeemath A closed map takes closed sets to closed sets. Furthermore, a map is a continuous function.
$endgroup$
– user549397
Dec 24 '18 at 8:23










1 Answer
1






active

oldest

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3












$begingroup$

$z to z^{3}$ is not one-to-one so it is not a homeomorphism. Let $C$ be a closed set and consider the image ${z^{3}:z in C}$. Suppose $z_n^{3} to zeta$. We have to show that $zeta$ can be written as $z^{3}$ for some $z$. Now ${z_n^{3}}$ is a bounded sequence which implies that ${z_n}$ is also bounded. There is a subsequence of this sequence converging to some $z$ and continuity of the map $z to z^{3}$ tells you that $zeta =z^{3}$.



As requested by the OP I am providing a second proof that does not use compactness. There an open sector containing $zeta$ in which $z to z^{3}$ is a homeomorphism. [In any sector of the type $|arg(theta) - arg(theta_0)| <pi /3$ the given map is a homeomorphism]. We can apply the inverse map $ z to z^{1/3} $ to conclude that $z_n to zeta ^{1/3}$ so $zeta ^{1/3} in C$.






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    Note that $z_n^{3} to zeta$ implies $z_{n_k}^{3} to zeta$. So if $z_{n_k} to z$ then $z_{n_k}^{3} to z^{3}$ and $z_{n_k}^{3} to zeta$ which implies $zeta=z^{3}$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 24 '18 at 8:33












  • $begingroup$
    Nice! But at this point in the book, we haven't learned compactness. So I am thinking that there should be another way without using the compactness.
    $endgroup$
    – user549397
    Dec 24 '18 at 8:38










  • $begingroup$
    @Philip I have added another proof to my answer.
    $endgroup$
    – Kavi Rama Murthy
    Dec 24 '18 at 8:48












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1 Answer
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1 Answer
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active

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3












$begingroup$

$z to z^{3}$ is not one-to-one so it is not a homeomorphism. Let $C$ be a closed set and consider the image ${z^{3}:z in C}$. Suppose $z_n^{3} to zeta$. We have to show that $zeta$ can be written as $z^{3}$ for some $z$. Now ${z_n^{3}}$ is a bounded sequence which implies that ${z_n}$ is also bounded. There is a subsequence of this sequence converging to some $z$ and continuity of the map $z to z^{3}$ tells you that $zeta =z^{3}$.



As requested by the OP I am providing a second proof that does not use compactness. There an open sector containing $zeta$ in which $z to z^{3}$ is a homeomorphism. [In any sector of the type $|arg(theta) - arg(theta_0)| <pi /3$ the given map is a homeomorphism]. We can apply the inverse map $ z to z^{1/3} $ to conclude that $z_n to zeta ^{1/3}$ so $zeta ^{1/3} in C$.






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    Note that $z_n^{3} to zeta$ implies $z_{n_k}^{3} to zeta$. So if $z_{n_k} to z$ then $z_{n_k}^{3} to z^{3}$ and $z_{n_k}^{3} to zeta$ which implies $zeta=z^{3}$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 24 '18 at 8:33












  • $begingroup$
    Nice! But at this point in the book, we haven't learned compactness. So I am thinking that there should be another way without using the compactness.
    $endgroup$
    – user549397
    Dec 24 '18 at 8:38










  • $begingroup$
    @Philip I have added another proof to my answer.
    $endgroup$
    – Kavi Rama Murthy
    Dec 24 '18 at 8:48
















3












$begingroup$

$z to z^{3}$ is not one-to-one so it is not a homeomorphism. Let $C$ be a closed set and consider the image ${z^{3}:z in C}$. Suppose $z_n^{3} to zeta$. We have to show that $zeta$ can be written as $z^{3}$ for some $z$. Now ${z_n^{3}}$ is a bounded sequence which implies that ${z_n}$ is also bounded. There is a subsequence of this sequence converging to some $z$ and continuity of the map $z to z^{3}$ tells you that $zeta =z^{3}$.



As requested by the OP I am providing a second proof that does not use compactness. There an open sector containing $zeta$ in which $z to z^{3}$ is a homeomorphism. [In any sector of the type $|arg(theta) - arg(theta_0)| <pi /3$ the given map is a homeomorphism]. We can apply the inverse map $ z to z^{1/3} $ to conclude that $z_n to zeta ^{1/3}$ so $zeta ^{1/3} in C$.






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    Note that $z_n^{3} to zeta$ implies $z_{n_k}^{3} to zeta$. So if $z_{n_k} to z$ then $z_{n_k}^{3} to z^{3}$ and $z_{n_k}^{3} to zeta$ which implies $zeta=z^{3}$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 24 '18 at 8:33












  • $begingroup$
    Nice! But at this point in the book, we haven't learned compactness. So I am thinking that there should be another way without using the compactness.
    $endgroup$
    – user549397
    Dec 24 '18 at 8:38










  • $begingroup$
    @Philip I have added another proof to my answer.
    $endgroup$
    – Kavi Rama Murthy
    Dec 24 '18 at 8:48














3












3








3





$begingroup$

$z to z^{3}$ is not one-to-one so it is not a homeomorphism. Let $C$ be a closed set and consider the image ${z^{3}:z in C}$. Suppose $z_n^{3} to zeta$. We have to show that $zeta$ can be written as $z^{3}$ for some $z$. Now ${z_n^{3}}$ is a bounded sequence which implies that ${z_n}$ is also bounded. There is a subsequence of this sequence converging to some $z$ and continuity of the map $z to z^{3}$ tells you that $zeta =z^{3}$.



As requested by the OP I am providing a second proof that does not use compactness. There an open sector containing $zeta$ in which $z to z^{3}$ is a homeomorphism. [In any sector of the type $|arg(theta) - arg(theta_0)| <pi /3$ the given map is a homeomorphism]. We can apply the inverse map $ z to z^{1/3} $ to conclude that $z_n to zeta ^{1/3}$ so $zeta ^{1/3} in C$.






share|cite|improve this answer











$endgroup$



$z to z^{3}$ is not one-to-one so it is not a homeomorphism. Let $C$ be a closed set and consider the image ${z^{3}:z in C}$. Suppose $z_n^{3} to zeta$. We have to show that $zeta$ can be written as $z^{3}$ for some $z$. Now ${z_n^{3}}$ is a bounded sequence which implies that ${z_n}$ is also bounded. There is a subsequence of this sequence converging to some $z$ and continuity of the map $z to z^{3}$ tells you that $zeta =z^{3}$.



As requested by the OP I am providing a second proof that does not use compactness. There an open sector containing $zeta$ in which $z to z^{3}$ is a homeomorphism. [In any sector of the type $|arg(theta) - arg(theta_0)| <pi /3$ the given map is a homeomorphism]. We can apply the inverse map $ z to z^{1/3} $ to conclude that $z_n to zeta ^{1/3}$ so $zeta ^{1/3} in C$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 24 '18 at 8:58









user549397

1,6591418




1,6591418










answered Dec 24 '18 at 8:21









Kavi Rama MurthyKavi Rama Murthy

72.2k53170




72.2k53170








  • 2




    $begingroup$
    Note that $z_n^{3} to zeta$ implies $z_{n_k}^{3} to zeta$. So if $z_{n_k} to z$ then $z_{n_k}^{3} to z^{3}$ and $z_{n_k}^{3} to zeta$ which implies $zeta=z^{3}$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 24 '18 at 8:33












  • $begingroup$
    Nice! But at this point in the book, we haven't learned compactness. So I am thinking that there should be another way without using the compactness.
    $endgroup$
    – user549397
    Dec 24 '18 at 8:38










  • $begingroup$
    @Philip I have added another proof to my answer.
    $endgroup$
    – Kavi Rama Murthy
    Dec 24 '18 at 8:48














  • 2




    $begingroup$
    Note that $z_n^{3} to zeta$ implies $z_{n_k}^{3} to zeta$. So if $z_{n_k} to z$ then $z_{n_k}^{3} to z^{3}$ and $z_{n_k}^{3} to zeta$ which implies $zeta=z^{3}$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 24 '18 at 8:33












  • $begingroup$
    Nice! But at this point in the book, we haven't learned compactness. So I am thinking that there should be another way without using the compactness.
    $endgroup$
    – user549397
    Dec 24 '18 at 8:38










  • $begingroup$
    @Philip I have added another proof to my answer.
    $endgroup$
    – Kavi Rama Murthy
    Dec 24 '18 at 8:48








2




2




$begingroup$
Note that $z_n^{3} to zeta$ implies $z_{n_k}^{3} to zeta$. So if $z_{n_k} to z$ then $z_{n_k}^{3} to z^{3}$ and $z_{n_k}^{3} to zeta$ which implies $zeta=z^{3}$.
$endgroup$
– Kavi Rama Murthy
Dec 24 '18 at 8:33






$begingroup$
Note that $z_n^{3} to zeta$ implies $z_{n_k}^{3} to zeta$. So if $z_{n_k} to z$ then $z_{n_k}^{3} to z^{3}$ and $z_{n_k}^{3} to zeta$ which implies $zeta=z^{3}$.
$endgroup$
– Kavi Rama Murthy
Dec 24 '18 at 8:33














$begingroup$
Nice! But at this point in the book, we haven't learned compactness. So I am thinking that there should be another way without using the compactness.
$endgroup$
– user549397
Dec 24 '18 at 8:38




$begingroup$
Nice! But at this point in the book, we haven't learned compactness. So I am thinking that there should be another way without using the compactness.
$endgroup$
– user549397
Dec 24 '18 at 8:38












$begingroup$
@Philip I have added another proof to my answer.
$endgroup$
– Kavi Rama Murthy
Dec 24 '18 at 8:48




$begingroup$
@Philip I have added another proof to my answer.
$endgroup$
– Kavi Rama Murthy
Dec 24 '18 at 8:48


















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