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A metric for which set of rational numbers complete [closed]
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Does there exist a metric on $mathbbQ$ which is equivalent
to the standard metric but $(mathbbQ, d)$ is complete.??
My attempt :
We know that with respect to standard metric, each singleton is closed subsets.
Since we know that a countable union of nowhere dense sets in a metric space need not be nowhere dense.
As for example, consider the set of rational numbers as a subset of $mathbbR$, we can write $mathbbQ$ as the union of singleton set, which of course nowhere dense sets in $mathbbR$.
But, of closure ($mathbbQ$) is $mathbbR$ and $mathbbQ$ is everywhere dense in $mathbbR$ and hence, $mathbbQ$ is not nowhere dense set in $mathbbR$.
Therefore, there exist a metric s. t $(mathbbQ, d)$ is complete.
Is my arguments true.??
Please help me. Thanks
metric-spaces
edited Jul 31 at 14:49
Cornman
2,29021027
asked Jul 31 at 14:42
Golam Biswas
63
closed as off-topic by Andrés E. Caicedo, user21820, Shailesh, José Carlos Santos, John Ma Jul 31 at 18:15
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РAndr̩s E. Caicedo, Shailesh, Jos̩ Carlos Santos, John Ma
add a comment |Â
up vote
0
down vote
favorite
Does there exist a metric on $mathbbQ$ which is equivalent
to the standard metric but $(mathbbQ, d)$ is complete.??
My attempt :
We know that with respect to standard metric, each singleton is closed subsets.
Since we know that a countable union of nowhere dense sets in a metric space need not be nowhere dense.
As for example, consider the set of rational numbers as a subset of $mathbbR$, we can write $mathbbQ$ as the union of singleton set, which of course nowhere dense sets in $mathbbR$.
But, of closure ($mathbbQ$) is $mathbbR$ and $mathbbQ$ is everywhere dense in $mathbbR$ and hence, $mathbbQ$ is not nowhere dense set in $mathbbR$.
Therefore, there exist a metric s. t $(mathbbQ, d)$ is complete.
Is my arguments true.??
Please help me. Thanks
metric-spaces
edited Jul 31 at 14:49
Cornman
2,29021027
asked Jul 31 at 14:42
Golam Biswas
63
closed as off-topic by Andrés E. Caicedo, user21820, Shailesh, José Carlos Santos, John Ma Jul 31 at 18:15
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РAndr̩s E. Caicedo, Shailesh, Jos̩ Carlos Santos, John Ma
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Does there exist a metric on $mathbbQ$ which is equivalent
to the standard metric but $(mathbbQ, d)$ is complete.??
My attempt :
We know that with respect to standard metric, each singleton is closed subsets.
Since we know that a countable union of nowhere dense sets in a metric space need not be nowhere dense.
As for example, consider the set of rational numbers as a subset of $mathbbR$, we can write $mathbbQ$ as the union of singleton set, which of course nowhere dense sets in $mathbbR$.
But, of closure ($mathbbQ$) is $mathbbR$ and $mathbbQ$ is everywhere dense in $mathbbR$ and hence, $mathbbQ$ is not nowhere dense set in $mathbbR$.
Therefore, there exist a metric s. t $(mathbbQ, d)$ is complete.
Is my arguments true.??
Please help me. Thanks
metric-spaces
edited Jul 31 at 14:49
Cornman
2,29021027
asked Jul 31 at 14:42
Golam Biswas
63
Does there exist a metric on $mathbbQ$ which is equivalent
to the standard metric but $(mathbbQ, d)$ is complete.??
My attempt :
We know that with respect to standard metric, each singleton is closed subsets.
Since we know that a countable union of nowhere dense sets in a metric space need not be nowhere dense.
As for example, consider the set of rational numbers as a subset of $mathbbR$, we can write $mathbbQ$ as the union of singleton set, which of course nowhere dense sets in $mathbbR$.
But, of closure ($mathbbQ$) is $mathbbR$ and $mathbbQ$ is everywhere dense in $mathbbR$ and hence, $mathbbQ$ is not nowhere dense set in $mathbbR$.
Therefore, there exist a metric s. t $(mathbbQ, d)$ is complete.
Is my arguments true.??
Please help me. Thanks
metric-spaces
edited Jul 31 at 14:49
Cornman
2,29021027
asked Jul 31 at 14:42
Golam Biswas
63
edited Jul 31 at 14:49
Cornman
2,29021027
edited Jul 31 at 14:49
Cornman
2,29021027
edited Jul 31 at 14:49
Cornman
2,29021027
2,29021027
asked Jul 31 at 14:42
Golam Biswas
63
asked Jul 31 at 14:42
Golam Biswas
63
asked Jul 31 at 14:42
Golam Biswas
63
63
closed as off-topic by Andrés E. Caicedo, user21820, Shailesh, José Carlos Santos, John Ma Jul 31 at 18:15
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РAndr̩s E. Caicedo, Shailesh, Jos̩ Carlos Santos, John Ma
closed as off-topic by Andrés E. Caicedo, user21820, Shailesh, José Carlos Santos, John Ma Jul 31 at 18:15
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РAndr̩s E. Caicedo, Shailesh, Jos̩ Carlos Santos, John Ma
add a comment |Â
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1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
By BCT, a countable complete metric space is not a union of nowhere dense sets; every singleton is nowhere dense; $Bbb Q$ is countable.
Also your argument does not actually establish a metric on $Bbb Q$ that makes $Bbb Q$ complete. You talked about nowhere dense sets and I don't know what you want.
answered Jul 31 at 14:44
Kenny Lau
17.7k2156
Then what can you say about your explanation?
– Golam Biswas
Jul 31 at 14:46
No. Finite union of nowhere dense set is nowhere dense but not countable number.
– Golam Biswas
Jul 31 at 14:48
That still has nothing to do with a metric on $Bbb Q$ making it complete.
– Kenny Lau
Jul 31 at 14:48
# Kenny, I know that Q has been expressed as a countable union of singleton sets.
– Golam Biswas
Jul 31 at 14:55
You can check your result.
– Golam Biswas
Jul 31 at 14:55
 |Â
show 6 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
By BCT, a countable complete metric space is not a union of nowhere dense sets; every singleton is nowhere dense; $Bbb Q$ is countable.
Also your argument does not actually establish a metric on $Bbb Q$ that makes $Bbb Q$ complete. You talked about nowhere dense sets and I don't know what you want.
answered Jul 31 at 14:44
Kenny Lau
17.7k2156
Then what can you say about your explanation?
– Golam Biswas
Jul 31 at 14:46
No. Finite union of nowhere dense set is nowhere dense but not countable number.
– Golam Biswas
Jul 31 at 14:48
That still has nothing to do with a metric on $Bbb Q$ making it complete.
– Kenny Lau
Jul 31 at 14:48
# Kenny, I know that Q has been expressed as a countable union of singleton sets.
– Golam Biswas
Jul 31 at 14:55
You can check your result.
– Golam Biswas
Jul 31 at 14:55
 |Â
show 6 more comments
up vote
3
down vote
accepted
By BCT, a countable complete metric space is not a union of nowhere dense sets; every singleton is nowhere dense; $Bbb Q$ is countable.
Also your argument does not actually establish a metric on $Bbb Q$ that makes $Bbb Q$ complete. You talked about nowhere dense sets and I don't know what you want.
answered Jul 31 at 14:44
Kenny Lau
17.7k2156
Then what can you say about your explanation?
– Golam Biswas
Jul 31 at 14:46
No. Finite union of nowhere dense set is nowhere dense but not countable number.
– Golam Biswas
Jul 31 at 14:48
That still has nothing to do with a metric on $Bbb Q$ making it complete.
– Kenny Lau
Jul 31 at 14:48
# Kenny, I know that Q has been expressed as a countable union of singleton sets.
– Golam Biswas
Jul 31 at 14:55
You can check your result.
– Golam Biswas
Jul 31 at 14:55
 |Â
show 6 more comments
up vote
3
down vote
accepted
up vote
3
down vote
accepted
By BCT, a countable complete metric space is not a union of nowhere dense sets; every singleton is nowhere dense; $Bbb Q$ is countable.
Also your argument does not actually establish a metric on $Bbb Q$ that makes $Bbb Q$ complete. You talked about nowhere dense sets and I don't know what you want.
answered Jul 31 at 14:44
Kenny Lau
17.7k2156
By BCT, a countable complete metric space is not a union of nowhere dense sets; every singleton is nowhere dense; $Bbb Q$ is countable.
Also your argument does not actually establish a metric on $Bbb Q$ that makes $Bbb Q$ complete. You talked about nowhere dense sets and I don't know what you want.
answered Jul 31 at 14:44
Kenny Lau
17.7k2156
edited Jul 31 at 14:55
answered Jul 31 at 14:44
Kenny Lau
17.7k2156
answered Jul 31 at 14:44
Kenny Lau
17.7k2156
answered Jul 31 at 14:44
Kenny Lau
17.7k2156
17.7k2156
Then what can you say about your explanation?
– Golam Biswas
Jul 31 at 14:46
No. Finite union of nowhere dense set is nowhere dense but not countable number.
– Golam Biswas
Jul 31 at 14:48
That still has nothing to do with a metric on $Bbb Q$ making it complete.
– Kenny Lau
Jul 31 at 14:48
# Kenny, I know that Q has been expressed as a countable union of singleton sets.
– Golam Biswas
Jul 31 at 14:55
You can check your result.
– Golam Biswas
Jul 31 at 14:55
 |Â
show 6 more comments
Then what can you say about your explanation?
– Golam Biswas
Jul 31 at 14:46
No. Finite union of nowhere dense set is nowhere dense but not countable number.
– Golam Biswas
Jul 31 at 14:48
That still has nothing to do with a metric on $Bbb Q$ making it complete.
– Kenny Lau
Jul 31 at 14:48
# Kenny, I know that Q has been expressed as a countable union of singleton sets.
– Golam Biswas
Jul 31 at 14:55
You can check your result.
– Golam Biswas
Jul 31 at 14:55
Then what can you say about your explanation?
– Golam Biswas
Jul 31 at 14:46
Then what can you say about your explanation?
– Golam Biswas
Jul 31 at 14:46
No. Finite union of nowhere dense set is nowhere dense but not countable number.
– Golam Biswas
Jul 31 at 14:48
No. Finite union of nowhere dense set is nowhere dense but not countable number.
– Golam Biswas
Jul 31 at 14:48
That still has nothing to do with a metric on $Bbb Q$ making it complete.
– Kenny Lau
Jul 31 at 14:48
That still has nothing to do with a metric on $Bbb Q$ making it complete.
– Kenny Lau
Jul 31 at 14:48
# Kenny, I know that Q has been expressed as a countable union of singleton sets.
– Golam Biswas
Jul 31 at 14:55
# Kenny, I know that Q has been expressed as a countable union of singleton sets.
– Golam Biswas
Jul 31 at 14:55
You can check your result.
– Golam Biswas
Jul 31 at 14:55
You can check your result.
– Golam Biswas
Jul 31 at 14:55
 |Â
show 6 more comments