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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 2 days ago
M. A. SARKAR
1,9401618
add a comment |
up vote
0
down vote
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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 2 days ago
M. A. SARKAR
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
add a comment |
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
real-analysis general-topology
asked 2 days ago
M. A. SARKAR
1,9401618
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 2 days ago
M. A. SARKAR
1,9401618
asked 2 days ago
M. A. SARKAR
1,9401618
asked 2 days ago
M. A. SARKAR
1,9401618
asked 2 days ago
M. A. SARKAR
1,9401618
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
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.
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
add a comment |
1 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.
answered 2 days ago
Kavi Rama Murthy
40.8k31751
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 2 days ago
Kavi Rama Murthy
40.8k31751
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 2 days ago
Kavi Rama Murthy
40.8k31751
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 2 days ago
Kavi Rama Murthy
40.8k31751
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 2 days ago
Kavi Rama Murthy
40.8k31751
answered 2 days ago
Kavi Rama Murthy
40.8k31751
answered 2 days ago
Kavi Rama Murthy
40.8k31751
answered 2 days ago
Kavi Rama Murthy
40.8k31751
40.8k31751
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.
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