Prove that $A$ is homeomorphic to $S^1times[1,2]$ where $A$ and $S^1$ are defined below:
let A = ${(x,y) in mathbb R^2 : 1 le sqrt{x^2+y^2} le 2 }$ . Prove that $A$ is homeomorphic to $S^1times[1,2]$ where $S^1 = {(x,y) in mathbb R^2 : x^2+y^2 =1 }$
general-topology
edited Nov 24 at 13:14
Tianlalu
3,13121038
asked Nov 24 at 13:03
Joy
12
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let A = ${(x,y) in mathbb R^2 : 1 le sqrt{x^2+y^2} le 2 }$ . Prove that $A$ is homeomorphic to $S^1times[1,2]$ where $S^1 = {(x,y) in mathbb R^2 : x^2+y^2 =1 }$
general-topology
edited Nov 24 at 13:14
Tianlalu
3,13121038
asked Nov 24 at 13:03
Joy
12
add a comment |
let A = ${(x,y) in mathbb R^2 : 1 le sqrt{x^2+y^2} le 2 }$ . Prove that $A$ is homeomorphic to $S^1times[1,2]$ where $S^1 = {(x,y) in mathbb R^2 : x^2+y^2 =1 }$
general-topology
edited Nov 24 at 13:14
Tianlalu
3,13121038
asked Nov 24 at 13:03
Joy
12
let A = ${(x,y) in mathbb R^2 : 1 le sqrt{x^2+y^2} le 2 }$ . Prove that $A$ is homeomorphic to $S^1times[1,2]$ where $S^1 = {(x,y) in mathbb R^2 : x^2+y^2 =1 }$
general-topology
general-topology
edited Nov 24 at 13:14
Tianlalu
3,13121038
asked Nov 24 at 13:03
Joy
12
edited Nov 24 at 13:14
Tianlalu
3,13121038
asked Nov 24 at 13:03
Joy
12
edited Nov 24 at 13:14
Tianlalu
3,13121038
edited Nov 24 at 13:14
Tianlalu
3,13121038
edited Nov 24 at 13:14
Tianlalu
3,13121038
3,13121038
asked Nov 24 at 13:03
Joy
12
asked Nov 24 at 13:03
Joy
12
asked Nov 24 at 13:03
Joy
12
12
add a comment |
add a comment |
1 Answer
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Notice that $A={(rcos theta,rsin theta): 1≤r≤2,0≤theta<2pi}$. Therefore required homeomorphism is $f:Arightarrow S^1×[1,2]$ is given by $f(rcos theta,rsin theta)=((costheta,sintheta),r)$.To show $f$ is homeomorphism first show it is bijective and continuous. Now $A$ is compact and $S^1×[1,2]$ is hausdorff i.e. $f^{-1}$ is also continuous.
answered Nov 24 at 13:12
UserS
1,5371112
I think you know that to show a function defined on a metric space is continuous it is enough to show that $x_nrightarrow ximplies f(x_n)rightarrow f(x)$.
– UserS
Nov 24 at 13:26
add a comment |
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1 Answer
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1 Answer
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Notice that $A={(rcos theta,rsin theta): 1≤r≤2,0≤theta<2pi}$. Therefore required homeomorphism is $f:Arightarrow S^1×[1,2]$ is given by $f(rcos theta,rsin theta)=((costheta,sintheta),r)$.To show $f$ is homeomorphism first show it is bijective and continuous. Now $A$ is compact and $S^1×[1,2]$ is hausdorff i.e. $f^{-1}$ is also continuous.
answered Nov 24 at 13:12
UserS
1,5371112
I think you know that to show a function defined on a metric space is continuous it is enough to show that $x_nrightarrow ximplies f(x_n)rightarrow f(x)$.
– UserS
Nov 24 at 13:26
add a comment |
Notice that $A={(rcos theta,rsin theta): 1≤r≤2,0≤theta<2pi}$. Therefore required homeomorphism is $f:Arightarrow S^1×[1,2]$ is given by $f(rcos theta,rsin theta)=((costheta,sintheta),r)$.To show $f$ is homeomorphism first show it is bijective and continuous. Now $A$ is compact and $S^1×[1,2]$ is hausdorff i.e. $f^{-1}$ is also continuous.
answered Nov 24 at 13:12
UserS
1,5371112
I think you know that to show a function defined on a metric space is continuous it is enough to show that $x_nrightarrow ximplies f(x_n)rightarrow f(x)$.
– UserS
Nov 24 at 13:26
add a comment |
Notice that $A={(rcos theta,rsin theta): 1≤r≤2,0≤theta<2pi}$. Therefore required homeomorphism is $f:Arightarrow S^1×[1,2]$ is given by $f(rcos theta,rsin theta)=((costheta,sintheta),r)$.To show $f$ is homeomorphism first show it is bijective and continuous. Now $A$ is compact and $S^1×[1,2]$ is hausdorff i.e. $f^{-1}$ is also continuous.
answered Nov 24 at 13:12
UserS
1,5371112
Notice that $A={(rcos theta,rsin theta): 1≤r≤2,0≤theta<2pi}$. Therefore required homeomorphism is $f:Arightarrow S^1×[1,2]$ is given by $f(rcos theta,rsin theta)=((costheta,sintheta),r)$.To show $f$ is homeomorphism first show it is bijective and continuous. Now $A$ is compact and $S^1×[1,2]$ is hausdorff i.e. $f^{-1}$ is also continuous.
answered Nov 24 at 13:12
UserS
1,5371112
edited Nov 24 at 13:18
answered Nov 24 at 13:12
UserS
1,5371112
answered Nov 24 at 13:12
UserS
1,5371112
answered Nov 24 at 13:12
UserS
1,5371112
1,5371112
I think you know that to show a function defined on a metric space is continuous it is enough to show that $x_nrightarrow ximplies f(x_n)rightarrow f(x)$.
– UserS
Nov 24 at 13:26
add a comment |
I think you know that to show a function defined on a metric space is continuous it is enough to show that $x_nrightarrow ximplies f(x_n)rightarrow f(x)$.
– UserS
Nov 24 at 13:26
I think you know that to show a function defined on a metric space is continuous it is enough to show that $x_nrightarrow ximplies f(x_n)rightarrow f(x)$.
– UserS
Nov 24 at 13:26
I think you know that to show a function defined on a metric space is continuous it is enough to show that $x_nrightarrow ximplies f(x_n)rightarrow f(x)$.
– UserS
Nov 24 at 13:26
add a comment |
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