The radial plane is not locally compact
I'm trying to prove that the radial plane is not locally compact.
I assume on the contrary that is locally compact and take any arbitrary point in it (say 0). Now 0 has a compact neighborhood $K$. $Big[$Then $K$ contains a circle (or part of a circle) $C$ intersecting 0$Big]$. Since $C$ is closed and $K$ is compact, $K cap C=C$ is compact. But the subspace topology of $C$ is discrete in the radial plane. Thus ${{x}:xin C}$ is an open cover of $C$ that has not a finite subcover.
I could not prove nor disprove what in brackets, can anyone help?
If my claim is not right, then can we prove the theorem using a similar idea?
general-topology geometry differential-geometry differential-topology low-dimensional-topology
asked Nov 30 at 0:09
F.H.A
103
add a comment |
I'm trying to prove that the radial plane is not locally compact.
I assume on the contrary that is locally compact and take any arbitrary point in it (say 0). Now 0 has a compact neighborhood $K$. $Big[$Then $K$ contains a circle (or part of a circle) $C$ intersecting 0$Big]$. Since $C$ is closed and $K$ is compact, $K cap C=C$ is compact. But the subspace topology of $C$ is discrete in the radial plane. Thus ${{x}:xin C}$ is an open cover of $C$ that has not a finite subcover.
I could not prove nor disprove what in brackets, can anyone help?
If my claim is not right, then can we prove the theorem using a similar idea?
general-topology geometry differential-geometry differential-topology low-dimensional-topology
asked Nov 30 at 0:09
F.H.A
103
add a comment |
I'm trying to prove that the radial plane is not locally compact.
I assume on the contrary that is locally compact and take any arbitrary point in it (say 0). Now 0 has a compact neighborhood $K$. $Big[$Then $K$ contains a circle (or part of a circle) $C$ intersecting 0$Big]$. Since $C$ is closed and $K$ is compact, $K cap C=C$ is compact. But the subspace topology of $C$ is discrete in the radial plane. Thus ${{x}:xin C}$ is an open cover of $C$ that has not a finite subcover.
I could not prove nor disprove what in brackets, can anyone help?
If my claim is not right, then can we prove the theorem using a similar idea?
general-topology geometry differential-geometry differential-topology low-dimensional-topology
asked Nov 30 at 0:09
F.H.A
103
I'm trying to prove that the radial plane is not locally compact.
I assume on the contrary that is locally compact and take any arbitrary point in it (say 0). Now 0 has a compact neighborhood $K$. $Big[$Then $K$ contains a circle (or part of a circle) $C$ intersecting 0$Big]$. Since $C$ is closed and $K$ is compact, $K cap C=C$ is compact. But the subspace topology of $C$ is discrete in the radial plane. Thus ${{x}:xin C}$ is an open cover of $C$ that has not a finite subcover.
I could not prove nor disprove what in brackets, can anyone help?
If my claim is not right, then can we prove the theorem using a similar idea?
general-topology geometry differential-geometry differential-topology low-dimensional-topology
general-topology geometry differential-geometry differential-topology low-dimensional-topology
asked Nov 30 at 0:09
F.H.A
103
asked Nov 30 at 0:09
F.H.A
103
edited Nov 30 at 4:07
asked Nov 30 at 0:09
F.H.A
103
asked Nov 30 at 0:09
F.H.A
103
asked Nov 30 at 0:09
F.H.A
103
103
add a comment |
add a comment |
1 Answer
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Since for all $(x,y) in mathbb{R}^2$ and $r > 0$ the circle $$C_r (x,y) = { (u,v) in mathbb{R}^2 : (x-u)^2 + (y-v)^2 = r^2 }$$ of radius $r$ centered at $(x,y)$ is a closed discrete subset of the radial plane, it suffices to show that for each open set $U$ and each $(x,y) in U$ there is an $r > 0$ such that $C_r$ has infinite intersection with $U$. (Then if $K$ is a compact neighborhood of $(x,y)$ there is an $r>0$ such that $C_r(x,y) cap operatorname{Int} (K) subseteq C_r(x,y) cap K$ is infinite. This latter intersection cannot be compact, because it is infinite and discrete. But it must be compact, being a closed subset of the compact $K$. Because of this contradiction, $(x,y)$ cannot have a compact neighborhood.)
By definition of the topology, for each $theta in [0,2pi)$ there is an $r_theta > 0$ such that $(x+y) + t ( cos theta , sin theta ) in U$ for all $0 leq t < r_theta$. Then there must be an $n > 0$ such that ${ theta in [0,2pi) : frac{1}{n} < r_theta }$ is infinite (even uncountable). As $(x,y) + frac{1}{n} ( cos theta , sin theta ) in U$ for all of these $theta$ it is clear that the circle $C_{1/n} (x,y)$ has infinite (even uncountable) intersection with $U$.
answered Nov 30 at 7:54
stochastic randomness
40017
Thanks, that was helpful. But could you explain why ${theta:1/n < r_{theta} }$ is infinite?
– F.H.A
Nov 30 at 13:14
@F.H.A For $thetain[0,2pi)$ since $r_theta>0$ there must be an $n$ such that $1/n<r_theta$. This means that $bigcup_n { theta in [0,2pi) : frac{1}{n} < r_theta } = [0,2pi)$. Since $[0,2pi)$ is uncountable (and a countable union of countable sets is countable), for some $n$ the set ${ theta in [0,2pi) : frac{1}{n} < r_theta }$ must be uncountable.
– stochastic randomness
Nov 30 at 13:21
add a comment |
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Since for all $(x,y) in mathbb{R}^2$ and $r > 0$ the circle $$C_r (x,y) = { (u,v) in mathbb{R}^2 : (x-u)^2 + (y-v)^2 = r^2 }$$ of radius $r$ centered at $(x,y)$ is a closed discrete subset of the radial plane, it suffices to show that for each open set $U$ and each $(x,y) in U$ there is an $r > 0$ such that $C_r$ has infinite intersection with $U$. (Then if $K$ is a compact neighborhood of $(x,y)$ there is an $r>0$ such that $C_r(x,y) cap operatorname{Int} (K) subseteq C_r(x,y) cap K$ is infinite. This latter intersection cannot be compact, because it is infinite and discrete. But it must be compact, being a closed subset of the compact $K$. Because of this contradiction, $(x,y)$ cannot have a compact neighborhood.)
By definition of the topology, for each $theta in [0,2pi)$ there is an $r_theta > 0$ such that $(x+y) + t ( cos theta , sin theta ) in U$ for all $0 leq t < r_theta$. Then there must be an $n > 0$ such that ${ theta in [0,2pi) : frac{1}{n} < r_theta }$ is infinite (even uncountable). As $(x,y) + frac{1}{n} ( cos theta , sin theta ) in U$ for all of these $theta$ it is clear that the circle $C_{1/n} (x,y)$ has infinite (even uncountable) intersection with $U$.
answered Nov 30 at 7:54
stochastic randomness
40017
Thanks, that was helpful. But could you explain why ${theta:1/n < r_{theta} }$ is infinite?
– F.H.A
Nov 30 at 13:14
@F.H.A For $thetain[0,2pi)$ since $r_theta>0$ there must be an $n$ such that $1/n<r_theta$. This means that $bigcup_n { theta in [0,2pi) : frac{1}{n} < r_theta } = [0,2pi)$. Since $[0,2pi)$ is uncountable (and a countable union of countable sets is countable), for some $n$ the set ${ theta in [0,2pi) : frac{1}{n} < r_theta }$ must be uncountable.
– stochastic randomness
Nov 30 at 13:21
add a comment |
Since for all $(x,y) in mathbb{R}^2$ and $r > 0$ the circle $$C_r (x,y) = { (u,v) in mathbb{R}^2 : (x-u)^2 + (y-v)^2 = r^2 }$$ of radius $r$ centered at $(x,y)$ is a closed discrete subset of the radial plane, it suffices to show that for each open set $U$ and each $(x,y) in U$ there is an $r > 0$ such that $C_r$ has infinite intersection with $U$. (Then if $K$ is a compact neighborhood of $(x,y)$ there is an $r>0$ such that $C_r(x,y) cap operatorname{Int} (K) subseteq C_r(x,y) cap K$ is infinite. This latter intersection cannot be compact, because it is infinite and discrete. But it must be compact, being a closed subset of the compact $K$. Because of this contradiction, $(x,y)$ cannot have a compact neighborhood.)
By definition of the topology, for each $theta in [0,2pi)$ there is an $r_theta > 0$ such that $(x+y) + t ( cos theta , sin theta ) in U$ for all $0 leq t < r_theta$. Then there must be an $n > 0$ such that ${ theta in [0,2pi) : frac{1}{n} < r_theta }$ is infinite (even uncountable). As $(x,y) + frac{1}{n} ( cos theta , sin theta ) in U$ for all of these $theta$ it is clear that the circle $C_{1/n} (x,y)$ has infinite (even uncountable) intersection with $U$.
answered Nov 30 at 7:54
stochastic randomness
40017
Thanks, that was helpful. But could you explain why ${theta:1/n < r_{theta} }$ is infinite?
– F.H.A
Nov 30 at 13:14
@F.H.A For $thetain[0,2pi)$ since $r_theta>0$ there must be an $n$ such that $1/n<r_theta$. This means that $bigcup_n { theta in [0,2pi) : frac{1}{n} < r_theta } = [0,2pi)$. Since $[0,2pi)$ is uncountable (and a countable union of countable sets is countable), for some $n$ the set ${ theta in [0,2pi) : frac{1}{n} < r_theta }$ must be uncountable.
– stochastic randomness
Nov 30 at 13:21
add a comment |
Since for all $(x,y) in mathbb{R}^2$ and $r > 0$ the circle $$C_r (x,y) = { (u,v) in mathbb{R}^2 : (x-u)^2 + (y-v)^2 = r^2 }$$ of radius $r$ centered at $(x,y)$ is a closed discrete subset of the radial plane, it suffices to show that for each open set $U$ and each $(x,y) in U$ there is an $r > 0$ such that $C_r$ has infinite intersection with $U$. (Then if $K$ is a compact neighborhood of $(x,y)$ there is an $r>0$ such that $C_r(x,y) cap operatorname{Int} (K) subseteq C_r(x,y) cap K$ is infinite. This latter intersection cannot be compact, because it is infinite and discrete. But it must be compact, being a closed subset of the compact $K$. Because of this contradiction, $(x,y)$ cannot have a compact neighborhood.)
By definition of the topology, for each $theta in [0,2pi)$ there is an $r_theta > 0$ such that $(x+y) + t ( cos theta , sin theta ) in U$ for all $0 leq t < r_theta$. Then there must be an $n > 0$ such that ${ theta in [0,2pi) : frac{1}{n} < r_theta }$ is infinite (even uncountable). As $(x,y) + frac{1}{n} ( cos theta , sin theta ) in U$ for all of these $theta$ it is clear that the circle $C_{1/n} (x,y)$ has infinite (even uncountable) intersection with $U$.
answered Nov 30 at 7:54
stochastic randomness
40017
Since for all $(x,y) in mathbb{R}^2$ and $r > 0$ the circle $$C_r (x,y) = { (u,v) in mathbb{R}^2 : (x-u)^2 + (y-v)^2 = r^2 }$$ of radius $r$ centered at $(x,y)$ is a closed discrete subset of the radial plane, it suffices to show that for each open set $U$ and each $(x,y) in U$ there is an $r > 0$ such that $C_r$ has infinite intersection with $U$. (Then if $K$ is a compact neighborhood of $(x,y)$ there is an $r>0$ such that $C_r(x,y) cap operatorname{Int} (K) subseteq C_r(x,y) cap K$ is infinite. This latter intersection cannot be compact, because it is infinite and discrete. But it must be compact, being a closed subset of the compact $K$. Because of this contradiction, $(x,y)$ cannot have a compact neighborhood.)
By definition of the topology, for each $theta in [0,2pi)$ there is an $r_theta > 0$ such that $(x+y) + t ( cos theta , sin theta ) in U$ for all $0 leq t < r_theta$. Then there must be an $n > 0$ such that ${ theta in [0,2pi) : frac{1}{n} < r_theta }$ is infinite (even uncountable). As $(x,y) + frac{1}{n} ( cos theta , sin theta ) in U$ for all of these $theta$ it is clear that the circle $C_{1/n} (x,y)$ has infinite (even uncountable) intersection with $U$.
answered Nov 30 at 7:54
stochastic randomness
40017
edited Nov 30 at 12:06
answered Nov 30 at 7:54
stochastic randomness
40017
answered Nov 30 at 7:54
stochastic randomness
40017
answered Nov 30 at 7:54
stochastic randomness
40017
40017
Thanks, that was helpful. But could you explain why ${theta:1/n < r_{theta} }$ is infinite?
– F.H.A
Nov 30 at 13:14
@F.H.A For $thetain[0,2pi)$ since $r_theta>0$ there must be an $n$ such that $1/n<r_theta$. This means that $bigcup_n { theta in [0,2pi) : frac{1}{n} < r_theta } = [0,2pi)$. Since $[0,2pi)$ is uncountable (and a countable union of countable sets is countable), for some $n$ the set ${ theta in [0,2pi) : frac{1}{n} < r_theta }$ must be uncountable.
– stochastic randomness
Nov 30 at 13:21
add a comment |
Thanks, that was helpful. But could you explain why ${theta:1/n < r_{theta} }$ is infinite?
– F.H.A
Nov 30 at 13:14
@F.H.A For $thetain[0,2pi)$ since $r_theta>0$ there must be an $n$ such that $1/n<r_theta$. This means that $bigcup_n { theta in [0,2pi) : frac{1}{n} < r_theta } = [0,2pi)$. Since $[0,2pi)$ is uncountable (and a countable union of countable sets is countable), for some $n$ the set ${ theta in [0,2pi) : frac{1}{n} < r_theta }$ must be uncountable.
– stochastic randomness
Nov 30 at 13:21
Thanks, that was helpful. But could you explain why ${theta:1/n < r_{theta} }$ is infinite?
– F.H.A
Nov 30 at 13:14
Thanks, that was helpful. But could you explain why ${theta:1/n < r_{theta} }$ is infinite?
– F.H.A
Nov 30 at 13:14
@F.H.A For $thetain[0,2pi)$ since $r_theta>0$ there must be an $n$ such that $1/n<r_theta$. This means that $bigcup_n { theta in [0,2pi) : frac{1}{n} < r_theta } = [0,2pi)$. Since $[0,2pi)$ is uncountable (and a countable union of countable sets is countable), for some $n$ the set ${ theta in [0,2pi) : frac{1}{n} < r_theta }$ must be uncountable.
– stochastic randomness
Nov 30 at 13:21
@F.H.A For $thetain[0,2pi)$ since $r_theta>0$ there must be an $n$ such that $1/n<r_theta$. This means that $bigcup_n { theta in [0,2pi) : frac{1}{n} < r_theta } = [0,2pi)$. Since $[0,2pi)$ is uncountable (and a countable union of countable sets is countable), for some $n$ the set ${ theta in [0,2pi) : frac{1}{n} < r_theta }$ must be uncountable.
– stochastic randomness
Nov 30 at 13:21
add a comment |
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