Why set of integer under indiscrete topology is compact?
Why set of integer under indiscrete topology is compact?
After looking this question I got surprised as In rudin I read in chapter 2 that compact space is independent of metric space chosen.
As metric space is special case of topology above should hold.
But where is I am making mistake .How to show that above is compact?
Any Help will be appreciated
real-analysis general-topology compactness
asked Jan 6 at 10:45
MathLoverMathLover
47910
add a comment |
Why set of integer under indiscrete topology is compact?
After looking this question I got surprised as In rudin I read in chapter 2 that compact space is independent of metric space chosen.
As metric space is special case of topology above should hold.
But where is I am making mistake .How to show that above is compact?
Any Help will be appreciated
real-analysis general-topology compactness
asked Jan 6 at 10:45
MathLoverMathLover
47910
In the indiscrete topology, think about what a cover of open sets must entail. There aren't many open sets, and the open covers are quite predictable. Can you show there's always a finite sub-cover?
– Theo Bendit
Jan 6 at 10:47
The indiscrete topology is not a metric topology.
– MJD
Jan 6 at 10:48
Where did you read that compactness is independent of the metric chosen? That's not true. There are many metrics where $mathbb{R}$ is compact.
– stressed out
Jan 6 at 11:10
He probably means that compactness does not depend on the embedding space, which is a very other thing.
– Math_QED
Jan 6 at 12:12
add a comment |
Why set of integer under indiscrete topology is compact?
After looking this question I got surprised as In rudin I read in chapter 2 that compact space is independent of metric space chosen.
As metric space is special case of topology above should hold.
But where is I am making mistake .How to show that above is compact?
Any Help will be appreciated
real-analysis general-topology compactness
asked Jan 6 at 10:45
MathLoverMathLover
47910
Why set of integer under indiscrete topology is compact?
After looking this question I got surprised as In rudin I read in chapter 2 that compact space is independent of metric space chosen.
As metric space is special case of topology above should hold.
But where is I am making mistake .How to show that above is compact?
Any Help will be appreciated
real-analysis general-topology compactness
real-analysis general-topology compactness
asked Jan 6 at 10:45
MathLoverMathLover
47910
asked Jan 6 at 10:45
MathLoverMathLover
47910
asked Jan 6 at 10:45
MathLoverMathLover
47910
asked Jan 6 at 10:45
MathLoverMathLover
47910
asked Jan 6 at 10:45
MathLoverMathLover
47910
47910
In the indiscrete topology, think about what a cover of open sets must entail. There aren't many open sets, and the open covers are quite predictable. Can you show there's always a finite sub-cover?
– Theo Bendit
Jan 6 at 10:47
The indiscrete topology is not a metric topology.
– MJD
Jan 6 at 10:48
Where did you read that compactness is independent of the metric chosen? That's not true. There are many metrics where $mathbb{R}$ is compact.
– stressed out
Jan 6 at 11:10
He probably means that compactness does not depend on the embedding space, which is a very other thing.
– Math_QED
Jan 6 at 12:12
add a comment |
In the indiscrete topology, think about what a cover of open sets must entail. There aren't many open sets, and the open covers are quite predictable. Can you show there's always a finite sub-cover?
– Theo Bendit
Jan 6 at 10:47
The indiscrete topology is not a metric topology.
– MJD
Jan 6 at 10:48
Where did you read that compactness is independent of the metric chosen? That's not true. There are many metrics where $mathbb{R}$ is compact.
– stressed out
Jan 6 at 11:10
He probably means that compactness does not depend on the embedding space, which is a very other thing.
– Math_QED
Jan 6 at 12:12
In the indiscrete topology, think about what a cover of open sets must entail. There aren't many open sets, and the open covers are quite predictable. Can you show there's always a finite sub-cover?
– Theo Bendit
Jan 6 at 10:47
In the indiscrete topology, think about what a cover of open sets must entail. There aren't many open sets, and the open covers are quite predictable. Can you show there's always a finite sub-cover?
– Theo Bendit
Jan 6 at 10:47
The indiscrete topology is not a metric topology.
– MJD
Jan 6 at 10:48
The indiscrete topology is not a metric topology.
– MJD
Jan 6 at 10:48
Where did you read that compactness is independent of the metric chosen? That's not true. There are many metrics where $mathbb{R}$ is compact.
– stressed out
Jan 6 at 11:10
Where did you read that compactness is independent of the metric chosen? That's not true. There are many metrics where $mathbb{R}$ is compact.
– stressed out
Jan 6 at 11:10
He probably means that compactness does not depend on the embedding space, which is a very other thing.
– Math_QED
Jan 6 at 12:12
He probably means that compactness does not depend on the embedding space, which is a very other thing.
– Math_QED
Jan 6 at 12:12
add a comment |
2 Answers
2
active
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Compactness is independent of the metric chosen among those that define the same topology (commonly termed equivalent metrics).
For instance, the closed interval $[0,1]$ is compact in the topology induced by the usual metric and all equivalent metrics. It is not compact under the discrete metric $delta(x,y)=1$ if $xne y$ and $delta(x,x)=0$: this metric induces a different topology, namely the discrete topology.
Any set endowed with the indiscrete topology is compact (for lack of infinite open covers), so this says nothing about compactness under different topologies.
answered Jan 6 at 11:21
egregegreg
179k1485202
add a comment |
As mentioned in the comments, if by indiscrete topology you mean the trivial topology ${0, X}$, then it's not metrizable. It's not Hausdorff, for example. This said, any subset of real numbers is compact in this topology because we have only two open sets in this topology!
Secondly, there are many metric topologies on $mathbb{R}$ that are compact, yet $mathbb{R}$ is not compact with its usual topology. I'm sure Rudin hasn't written anything like 'compactness is independent of the metric chosen'. You misinterpreted it.
On the other hand, compactness is an intrinsic property. Namely, it's not like openness or closedness that depends on the subspace topology. I think this is what Rudin has written.
answered Jan 6 at 11:13
stressed outstressed out
4,0811533
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Compactness is independent of the metric chosen among those that define the same topology (commonly termed equivalent metrics).
For instance, the closed interval $[0,1]$ is compact in the topology induced by the usual metric and all equivalent metrics. It is not compact under the discrete metric $delta(x,y)=1$ if $xne y$ and $delta(x,x)=0$: this metric induces a different topology, namely the discrete topology.
Any set endowed with the indiscrete topology is compact (for lack of infinite open covers), so this says nothing about compactness under different topologies.
answered Jan 6 at 11:21
egregegreg
179k1485202
add a comment |
Compactness is independent of the metric chosen among those that define the same topology (commonly termed equivalent metrics).
For instance, the closed interval $[0,1]$ is compact in the topology induced by the usual metric and all equivalent metrics. It is not compact under the discrete metric $delta(x,y)=1$ if $xne y$ and $delta(x,x)=0$: this metric induces a different topology, namely the discrete topology.
Any set endowed with the indiscrete topology is compact (for lack of infinite open covers), so this says nothing about compactness under different topologies.
answered Jan 6 at 11:21
egregegreg
179k1485202
add a comment |
Compactness is independent of the metric chosen among those that define the same topology (commonly termed equivalent metrics).
For instance, the closed interval $[0,1]$ is compact in the topology induced by the usual metric and all equivalent metrics. It is not compact under the discrete metric $delta(x,y)=1$ if $xne y$ and $delta(x,x)=0$: this metric induces a different topology, namely the discrete topology.
Any set endowed with the indiscrete topology is compact (for lack of infinite open covers), so this says nothing about compactness under different topologies.
answered Jan 6 at 11:21
egregegreg
179k1485202
Compactness is independent of the metric chosen among those that define the same topology (commonly termed equivalent metrics).
For instance, the closed interval $[0,1]$ is compact in the topology induced by the usual metric and all equivalent metrics. It is not compact under the discrete metric $delta(x,y)=1$ if $xne y$ and $delta(x,x)=0$: this metric induces a different topology, namely the discrete topology.
Any set endowed with the indiscrete topology is compact (for lack of infinite open covers), so this says nothing about compactness under different topologies.
answered Jan 6 at 11:21
egregegreg
179k1485202
answered Jan 6 at 11:21
egregegreg
179k1485202
answered Jan 6 at 11:21
egregegreg
179k1485202
answered Jan 6 at 11:21
egregegreg
179k1485202
179k1485202
add a comment |
add a comment |
As mentioned in the comments, if by indiscrete topology you mean the trivial topology ${0, X}$, then it's not metrizable. It's not Hausdorff, for example. This said, any subset of real numbers is compact in this topology because we have only two open sets in this topology!
Secondly, there are many metric topologies on $mathbb{R}$ that are compact, yet $mathbb{R}$ is not compact with its usual topology. I'm sure Rudin hasn't written anything like 'compactness is independent of the metric chosen'. You misinterpreted it.
On the other hand, compactness is an intrinsic property. Namely, it's not like openness or closedness that depends on the subspace topology. I think this is what Rudin has written.
answered Jan 6 at 11:13
stressed outstressed out
4,0811533
add a comment |
As mentioned in the comments, if by indiscrete topology you mean the trivial topology ${0, X}$, then it's not metrizable. It's not Hausdorff, for example. This said, any subset of real numbers is compact in this topology because we have only two open sets in this topology!
Secondly, there are many metric topologies on $mathbb{R}$ that are compact, yet $mathbb{R}$ is not compact with its usual topology. I'm sure Rudin hasn't written anything like 'compactness is independent of the metric chosen'. You misinterpreted it.
On the other hand, compactness is an intrinsic property. Namely, it's not like openness or closedness that depends on the subspace topology. I think this is what Rudin has written.
answered Jan 6 at 11:13
stressed outstressed out
4,0811533
add a comment |
As mentioned in the comments, if by indiscrete topology you mean the trivial topology ${0, X}$, then it's not metrizable. It's not Hausdorff, for example. This said, any subset of real numbers is compact in this topology because we have only two open sets in this topology!
Secondly, there are many metric topologies on $mathbb{R}$ that are compact, yet $mathbb{R}$ is not compact with its usual topology. I'm sure Rudin hasn't written anything like 'compactness is independent of the metric chosen'. You misinterpreted it.
On the other hand, compactness is an intrinsic property. Namely, it's not like openness or closedness that depends on the subspace topology. I think this is what Rudin has written.
answered Jan 6 at 11:13
stressed outstressed out
4,0811533
As mentioned in the comments, if by indiscrete topology you mean the trivial topology ${0, X}$, then it's not metrizable. It's not Hausdorff, for example. This said, any subset of real numbers is compact in this topology because we have only two open sets in this topology!
Secondly, there are many metric topologies on $mathbb{R}$ that are compact, yet $mathbb{R}$ is not compact with its usual topology. I'm sure Rudin hasn't written anything like 'compactness is independent of the metric chosen'. You misinterpreted it.
On the other hand, compactness is an intrinsic property. Namely, it's not like openness or closedness that depends on the subspace topology. I think this is what Rudin has written.
answered Jan 6 at 11:13
stressed outstressed out
4,0811533
answered Jan 6 at 11:13
stressed outstressed out
4,0811533
answered Jan 6 at 11:13
stressed outstressed out
4,0811533
answered Jan 6 at 11:13
stressed outstressed out
4,0811533
4,0811533
add a comment |
add a comment |
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In the indiscrete topology, think about what a cover of open sets must entail. There aren't many open sets, and the open covers are quite predictable. Can you show there's always a finite sub-cover?
– Theo Bendit
Jan 6 at 10:47
The indiscrete topology is not a metric topology.
– MJD
Jan 6 at 10:48
Where did you read that compactness is independent of the metric chosen? That's not true. There are many metrics where $mathbb{R}$ is compact.
– stressed out
Jan 6 at 11:10
He probably means that compactness does not depend on the embedding space, which is a very other thing.
– Math_QED
Jan 6 at 12:12