Mixing
Is finding the closure of a space a method of the completion of a metric space?
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Suppose we have a metric space $(X,d)$. I wonder whether finding the closure of it is the completion of the metric space?
My thinking
I do know the defination of the completion of a metric space. It seems it's more general and complex than just find the closure. But why we need such a complex defination if clusure can make the space complete?
functional-analysis metric-spaces
asked yesterday
maple
8112922
add a comment |Â
up vote
1
down vote
favorite
Suppose we have a metric space $(X,d)$. I wonder whether finding the closure of it is the completion of the metric space?
My thinking
I do know the defination of the completion of a metric space. It seems it's more general and complex than just find the closure. But why we need such a complex defination if clusure can make the space complete?
functional-analysis metric-spaces
asked yesterday
maple
8112922
1
By axiom, a topological space is itself always closed (and also open). So the closure of $X$ is $X$, unless I am misunderstanding something...
– Suzet
yesterday
For example, a open ball is not closed. Is there anything misleading in my stating of the problem?
– maple
yesterday
The open ball is not closed when you look at it inside an ambient space, say $mathbbR^n$ for instance. This is what is usually meant. However, the open ball surely is closed for the induced topology on it. In your statement, I think you need to mention that $(X,d)$ embeds inside an ambient (metric) space. Then however, I am no expert to answer your question.
– Suzet
yesterday
Just a little bit confuse. For example, what if we don't mention the ambient space of $mathbbQ$? Some sequence still converge out of the set, how to describe this situation?
– maple
yesterday
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Suppose we have a metric space $(X,d)$. I wonder whether finding the closure of it is the completion of the metric space?
My thinking
I do know the defination of the completion of a metric space. It seems it's more general and complex than just find the closure. But why we need such a complex defination if clusure can make the space complete?
functional-analysis metric-spaces
asked yesterday
maple
8112922
Suppose we have a metric space $(X,d)$. I wonder whether finding the closure of it is the completion of the metric space?
My thinking
I do know the defination of the completion of a metric space. It seems it's more general and complex than just find the closure. But why we need such a complex defination if clusure can make the space complete?
functional-analysis metric-spaces
asked yesterday
maple
8112922
asked yesterday
maple
8112922
asked yesterday
maple
8112922
asked yesterday
maple
8112922
8112922
1
By axiom, a topological space is itself always closed (and also open). So the closure of $X$ is $X$, unless I am misunderstanding something...
– Suzet
yesterday
For example, a open ball is not closed. Is there anything misleading in my stating of the problem?
– maple
yesterday
The open ball is not closed when you look at it inside an ambient space, say $mathbbR^n$ for instance. This is what is usually meant. However, the open ball surely is closed for the induced topology on it. In your statement, I think you need to mention that $(X,d)$ embeds inside an ambient (metric) space. Then however, I am no expert to answer your question.
– Suzet
yesterday
Just a little bit confuse. For example, what if we don't mention the ambient space of $mathbbQ$? Some sequence still converge out of the set, how to describe this situation?
– maple
yesterday
add a comment |Â
1
By axiom, a topological space is itself always closed (and also open). So the closure of $X$ is $X$, unless I am misunderstanding something...
– Suzet
yesterday
For example, a open ball is not closed. Is there anything misleading in my stating of the problem?
– maple
yesterday
The open ball is not closed when you look at it inside an ambient space, say $mathbbR^n$ for instance. This is what is usually meant. However, the open ball surely is closed for the induced topology on it. In your statement, I think you need to mention that $(X,d)$ embeds inside an ambient (metric) space. Then however, I am no expert to answer your question.
– Suzet
yesterday
Just a little bit confuse. For example, what if we don't mention the ambient space of $mathbbQ$? Some sequence still converge out of the set, how to describe this situation?
– maple
yesterday
1
1
By axiom, a topological space is itself always closed (and also open). So the closure of $X$ is $X$, unless I am misunderstanding something...
– Suzet
yesterday
By axiom, a topological space is itself always closed (and also open). So the closure of $X$ is $X$, unless I am misunderstanding something...
– Suzet
yesterday
For example, a open ball is not closed. Is there anything misleading in my stating of the problem?
– maple
yesterday
For example, a open ball is not closed. Is there anything misleading in my stating of the problem?
– maple
yesterday
The open ball is not closed when you look at it inside an ambient space, say $mathbbR^n$ for instance. This is what is usually meant. However, the open ball surely is closed for the induced topology on it. In your statement, I think you need to mention that $(X,d)$ embeds inside an ambient (metric) space. Then however, I am no expert to answer your question.
– Suzet
yesterday
The open ball is not closed when you look at it inside an ambient space, say $mathbbR^n$ for instance. This is what is usually meant. However, the open ball surely is closed for the induced topology on it. In your statement, I think you need to mention that $(X,d)$ embeds inside an ambient (metric) space. Then however, I am no expert to answer your question.
– Suzet
yesterday
Just a little bit confuse. For example, what if we don't mention the ambient space of $mathbbQ$? Some sequence still converge out of the set, how to describe this situation?
– maple
yesterday
Just a little bit confuse. For example, what if we don't mention the ambient space of $mathbbQ$? Some sequence still converge out of the set, how to describe this situation?
– maple
yesterday
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
Consider $(0,1) cap mathbbQ$. Its closure in $mathbbQ$ is $[0,1] cap mathbbQ$, but this space is not complete. Its completion would be $[0,1]$, as a subset of real numbers.
Similarly, $mathbbQ$, $(0,1)$, an open ball in any metric space, etc. are closed in themselves, but are not equal to their completion. Recall that a space $Y subset X$ is closed in $X$ if the complement $X backslash Y$ is open. If $Y=X$, then this complement is the empty set, and the empty set is certainly open.
answered yesterday
Sambo
1,2501327
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Consider $(0,1) cap mathbbQ$. Its closure in $mathbbQ$ is $[0,1] cap mathbbQ$, but this space is not complete. Its completion would be $[0,1]$, as a subset of real numbers.
Similarly, $mathbbQ$, $(0,1)$, an open ball in any metric space, etc. are closed in themselves, but are not equal to their completion. Recall that a space $Y subset X$ is closed in $X$ if the complement $X backslash Y$ is open. If $Y=X$, then this complement is the empty set, and the empty set is certainly open.
answered yesterday
Sambo
1,2501327
add a comment |Â
up vote
0
down vote
Consider $(0,1) cap mathbbQ$. Its closure in $mathbbQ$ is $[0,1] cap mathbbQ$, but this space is not complete. Its completion would be $[0,1]$, as a subset of real numbers.
Similarly, $mathbbQ$, $(0,1)$, an open ball in any metric space, etc. are closed in themselves, but are not equal to their completion. Recall that a space $Y subset X$ is closed in $X$ if the complement $X backslash Y$ is open. If $Y=X$, then this complement is the empty set, and the empty set is certainly open.
answered yesterday
Sambo
1,2501327
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Consider $(0,1) cap mathbbQ$. Its closure in $mathbbQ$ is $[0,1] cap mathbbQ$, but this space is not complete. Its completion would be $[0,1]$, as a subset of real numbers.
Similarly, $mathbbQ$, $(0,1)$, an open ball in any metric space, etc. are closed in themselves, but are not equal to their completion. Recall that a space $Y subset X$ is closed in $X$ if the complement $X backslash Y$ is open. If $Y=X$, then this complement is the empty set, and the empty set is certainly open.
answered yesterday
Sambo
1,2501327
Consider $(0,1) cap mathbbQ$. Its closure in $mathbbQ$ is $[0,1] cap mathbbQ$, but this space is not complete. Its completion would be $[0,1]$, as a subset of real numbers.
Similarly, $mathbbQ$, $(0,1)$, an open ball in any metric space, etc. are closed in themselves, but are not equal to their completion. Recall that a space $Y subset X$ is closed in $X$ if the complement $X backslash Y$ is open. If $Y=X$, then this complement is the empty set, and the empty set is certainly open.
answered yesterday
Sambo
1,2501327
answered yesterday
Sambo
1,2501327
answered yesterday
Sambo
1,2501327
answered yesterday
Sambo
1,2501327
1,2501327
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2872593%2fis-finding-the-closure-of-a-space-a-method-of-the-completion-of-a-metric-space%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
By axiom, a topological space is itself always closed (and also open). So the closure of $X$ is $X$, unless I am misunderstanding something...
– Suzet
yesterday
For example, a open ball is not closed. Is there anything misleading in my stating of the problem?
– maple
yesterday
The open ball is not closed when you look at it inside an ambient space, say $mathbbR^n$ for instance. This is what is usually meant. However, the open ball surely is closed for the induced topology on it. In your statement, I think you need to mention that $(X,d)$ embeds inside an ambient (metric) space. Then however, I am no expert to answer your question.
– Suzet
yesterday
Just a little bit confuse. For example, what if we don't mention the ambient space of $mathbbQ$? Some sequence still converge out of the set, how to describe this situation?
– maple
yesterday