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










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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.
    2 days ago










  • 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
    2 days ago

















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










share|cite|improve this question






















  • 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.
    2 days ago










  • 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
    2 days ago















up vote
0
down vote

favorite









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










share|cite|improve this question













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






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asked 2 days ago









M. A. SARKAR

1,9401618




1,9401618












  • 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.
    2 days ago










  • 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
    2 days ago




















  • 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.
    2 days ago










  • 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
    2 days ago


















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.
2 days ago




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.
2 days ago












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
2 days ago






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
2 days ago












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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.






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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.






    share|cite|improve this answer

























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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.






      share|cite|improve this answer























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        up vote
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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.






        share|cite|improve this answer












        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.







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        answered 2 days ago









        Kavi Rama Murthy

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