Show that there is a $ rho>0$ such that $ mathcal{F}$ covers the set
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Suppose that the family $ mathcal{F}$ is a family of open sets in $ mathbb{R}^2$ which covers the circle $C={(x,y): x^2+y^2 = 1 }$. Show that there is a $ rho>0$ such that $ mathcal{F}$ covers the set
$A={(x,y): (1-rho)^2 leq x^2+y^2 leq (1+rho)^2 }$.
Answer:
If we show that the family $ mathcal{F}$ covers the set $ {(x,y): 0 leq x^2+y^2 leq (1+rho)^2 }$, then it will be sufficient.
Since the family $ mathcal{F}$ covers the unit circle, the covers contains sets of radius $ 1+ rho $ for some $rho>0$ (without loss of generality).
Thus $ mathcal{F}$ covers the set $A$.
But I think, it is not appropriate .
Help me
real-analysis general-topology
asked Nov 19 at 22:44
M. A. SARKAR
2,1251619
add a comment |
up vote
0
down vote
favorite
Suppose that the family $ mathcal{F}$ is a family of open sets in $ mathbb{R}^2$ which covers the circle $C={(x,y): x^2+y^2 = 1 }$. Show that there is a $ rho>0$ such that $ mathcal{F}$ covers the set
$A={(x,y): (1-rho)^2 leq x^2+y^2 leq (1+rho)^2 }$.
Answer:
If we show that the family $ mathcal{F}$ covers the set $ {(x,y): 0 leq x^2+y^2 leq (1+rho)^2 }$, then it will be sufficient.
Since the family $ mathcal{F}$ covers the unit circle, the covers contains sets of radius $ 1+ rho $ for some $rho>0$ (without loss of generality).
Thus $ mathcal{F}$ covers the set $A$.
But I think, it is not appropriate .
Help me
real-analysis general-topology
asked Nov 19 at 22:44
M. A. SARKAR
2,1251619
What do you mean by the cover containing sets of radius $1+ rho$? We could very well cover the circle by balls of radius $0 < varepsilon << 1$ centered at each point of the circle.
– Guido A.
Nov 19 at 23:05
You can assume without loss of generality that $mathcal{F}$ is a family of open balls (since open balls form a basis of the topology on $mathbb{R}^2$). Since $C$ is compact, $mathcal{F}$ will have a finite subcover. You can probably find $rho$ by looking at this finite cover of open balls.
– suchan
Nov 19 at 23:09
add a comment |
up vote
0
down vote
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up vote
0
down vote
favorite
Suppose that the family $ mathcal{F}$ is a family of open sets in $ mathbb{R}^2$ which covers the circle $C={(x,y): x^2+y^2 = 1 }$. Show that there is a $ rho>0$ such that $ mathcal{F}$ covers the set
$A={(x,y): (1-rho)^2 leq x^2+y^2 leq (1+rho)^2 }$.
Answer:
If we show that the family $ mathcal{F}$ covers the set $ {(x,y): 0 leq x^2+y^2 leq (1+rho)^2 }$, then it will be sufficient.
Since the family $ mathcal{F}$ covers the unit circle, the covers contains sets of radius $ 1+ rho $ for some $rho>0$ (without loss of generality).
Thus $ mathcal{F}$ covers the set $A$.
But I think, it is not appropriate .
Help me
real-analysis general-topology
asked Nov 19 at 22:44
M. A. SARKAR
2,1251619
Suppose that the family $ mathcal{F}$ is a family of open sets in $ mathbb{R}^2$ which covers the circle $C={(x,y): x^2+y^2 = 1 }$. Show that there is a $ rho>0$ such that $ mathcal{F}$ covers the set
$A={(x,y): (1-rho)^2 leq x^2+y^2 leq (1+rho)^2 }$.
Answer:
If we show that the family $ mathcal{F}$ covers the set $ {(x,y): 0 leq x^2+y^2 leq (1+rho)^2 }$, then it will be sufficient.
Since the family $ mathcal{F}$ covers the unit circle, the covers contains sets of radius $ 1+ rho $ for some $rho>0$ (without loss of generality).
Thus $ mathcal{F}$ covers the set $A$.
But I think, it is not appropriate .
Help me
real-analysis general-topology
real-analysis general-topology
asked Nov 19 at 22:44
M. A. SARKAR
2,1251619
asked Nov 19 at 22:44
M. A. SARKAR
2,1251619
asked Nov 19 at 22:44
M. A. SARKAR
2,1251619
asked Nov 19 at 22:44
M. A. SARKAR
2,1251619
asked Nov 19 at 22:44
M. A. SARKAR
2,1251619
2,1251619
What do you mean by the cover containing sets of radius $1+ rho$? We could very well cover the circle by balls of radius $0 < varepsilon << 1$ centered at each point of the circle.
– Guido A.
Nov 19 at 23:05
You can assume without loss of generality that $mathcal{F}$ is a family of open balls (since open balls form a basis of the topology on $mathbb{R}^2$). Since $C$ is compact, $mathcal{F}$ will have a finite subcover. You can probably find $rho$ by looking at this finite cover of open balls.
– suchan
Nov 19 at 23:09
add a comment |
What do you mean by the cover containing sets of radius $1+ rho$? We could very well cover the circle by balls of radius $0 < varepsilon << 1$ centered at each point of the circle.
– Guido A.
Nov 19 at 23:05
You can assume without loss of generality that $mathcal{F}$ is a family of open balls (since open balls form a basis of the topology on $mathbb{R}^2$). Since $C$ is compact, $mathcal{F}$ will have a finite subcover. You can probably find $rho$ by looking at this finite cover of open balls.
– suchan
Nov 19 at 23:09
What do you mean by the cover containing sets of radius $1+ rho$? We could very well cover the circle by balls of radius $0 < varepsilon << 1$ centered at each point of the circle.
– Guido A.
Nov 19 at 23:05
What do you mean by the cover containing sets of radius $1+ rho$? We could very well cover the circle by balls of radius $0 < varepsilon << 1$ centered at each point of the circle.
– Guido A.
Nov 19 at 23:05
You can assume without loss of generality that $mathcal{F}$ is a family of open balls (since open balls form a basis of the topology on $mathbb{R}^2$). Since $C$ is compact, $mathcal{F}$ will have a finite subcover. You can probably find $rho$ by looking at this finite cover of open balls.
– suchan
Nov 19 at 23:09
You can assume without loss of generality that $mathcal{F}$ is a family of open balls (since open balls form a basis of the topology on $mathbb{R}^2$). Since $C$ is compact, $mathcal{F}$ will have a finite subcover. You can probably find $rho$ by looking at this finite cover of open balls.
– suchan
Nov 19 at 23:09
add a comment |
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There is no reason why points close to the origin are covered, so your argument is not correct. Let us prove this by contradiction. Suppose for any $rho$ there exists poits in the annulus $A$ which are not covered. Then there is a sequence $(x_n,y_n)$ with $x_n^{2}+y_n^{2} to 1$ such that no $(x_n,y_n)$ belongs to any of the open sets in our open cover. By Bolzano - Weirstrass Theorem there is a subsequence which converges to some point $(x,y)$. Since $x^{2}+y^{2}=1$ the point $(x,y)$ belongs to some open set $U$ in the open cover. But then $(x_n,y_n)$ is also in $U$ for $n$ sufficiently large. This is a contradiction.
answered Nov 19 at 23:24
Kavi Rama Murthy
46.9k31854
add a comment |
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There is no reason why points close to the origin are covered, so your argument is not correct. Let us prove this by contradiction. Suppose for any $rho$ there exists poits in the annulus $A$ which are not covered. Then there is a sequence $(x_n,y_n)$ with $x_n^{2}+y_n^{2} to 1$ such that no $(x_n,y_n)$ belongs to any of the open sets in our open cover. By Bolzano - Weirstrass Theorem there is a subsequence which converges to some point $(x,y)$. Since $x^{2}+y^{2}=1$ the point $(x,y)$ belongs to some open set $U$ in the open cover. But then $(x_n,y_n)$ is also in $U$ for $n$ sufficiently large. This is a contradiction.
answered Nov 19 at 23:24
Kavi Rama Murthy
46.9k31854
add a comment |
up vote
2
down vote
There is no reason why points close to the origin are covered, so your argument is not correct. Let us prove this by contradiction. Suppose for any $rho$ there exists poits in the annulus $A$ which are not covered. Then there is a sequence $(x_n,y_n)$ with $x_n^{2}+y_n^{2} to 1$ such that no $(x_n,y_n)$ belongs to any of the open sets in our open cover. By Bolzano - Weirstrass Theorem there is a subsequence which converges to some point $(x,y)$. Since $x^{2}+y^{2}=1$ the point $(x,y)$ belongs to some open set $U$ in the open cover. But then $(x_n,y_n)$ is also in $U$ for $n$ sufficiently large. This is a contradiction.
answered Nov 19 at 23:24
Kavi Rama Murthy
46.9k31854
add a comment |
up vote
2
down vote
up vote
2
down vote
There is no reason why points close to the origin are covered, so your argument is not correct. Let us prove this by contradiction. Suppose for any $rho$ there exists poits in the annulus $A$ which are not covered. Then there is a sequence $(x_n,y_n)$ with $x_n^{2}+y_n^{2} to 1$ such that no $(x_n,y_n)$ belongs to any of the open sets in our open cover. By Bolzano - Weirstrass Theorem there is a subsequence which converges to some point $(x,y)$. Since $x^{2}+y^{2}=1$ the point $(x,y)$ belongs to some open set $U$ in the open cover. But then $(x_n,y_n)$ is also in $U$ for $n$ sufficiently large. This is a contradiction.
answered Nov 19 at 23:24
Kavi Rama Murthy
46.9k31854
There is no reason why points close to the origin are covered, so your argument is not correct. Let us prove this by contradiction. Suppose for any $rho$ there exists poits in the annulus $A$ which are not covered. Then there is a sequence $(x_n,y_n)$ with $x_n^{2}+y_n^{2} to 1$ such that no $(x_n,y_n)$ belongs to any of the open sets in our open cover. By Bolzano - Weirstrass Theorem there is a subsequence which converges to some point $(x,y)$. Since $x^{2}+y^{2}=1$ the point $(x,y)$ belongs to some open set $U$ in the open cover. But then $(x_n,y_n)$ is also in $U$ for $n$ sufficiently large. This is a contradiction.
answered Nov 19 at 23:24
Kavi Rama Murthy
46.9k31854
answered Nov 19 at 23:24
Kavi Rama Murthy
46.9k31854
answered Nov 19 at 23:24
Kavi Rama Murthy
46.9k31854
answered Nov 19 at 23:24
Kavi Rama Murthy
46.9k31854
46.9k31854
add a comment |
add a comment |
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What do you mean by the cover containing sets of radius $1+ rho$? We could very well cover the circle by balls of radius $0 < varepsilon << 1$ centered at each point of the circle.
– Guido A.
Nov 19 at 23:05
You can assume without loss of generality that $mathcal{F}$ is a family of open balls (since open balls form a basis of the topology on $mathbb{R}^2$). Since $C$ is compact, $mathcal{F}$ will have a finite subcover. You can probably find $rho$ by looking at this finite cover of open balls.
– suchan
Nov 19 at 23:09