Inner of metric completion
$begingroup$
I am reading a proof where we have the following:
$M$ a metric space
$N$ its metric completion
It is then stated: for $yin N$ let ($y_s$) be a Cauchy sequence converging to y with $y_sin M$.
I do not understand why can we choose $y_sin M$.
Do we have $yin Nsetminus M Rightarrow yinpartial M$ ?
Edit
It is my understanding a property of the completion is being the smallest complete space $N$ such that $Msubset N$, and so the intuition tells me the above should hold. But I did not manage proving it yet.
analysis
asked Jan 25 at 10:00
natenate
1186
$endgroup$
add a comment |
$begingroup$
I am reading a proof where we have the following:
$M$ a metric space
$N$ its metric completion
It is then stated: for $yin N$ let ($y_s$) be a Cauchy sequence converging to y with $y_sin M$.
I do not understand why can we choose $y_sin M$.
Do we have $yin Nsetminus M Rightarrow yinpartial M$ ?
Edit
It is my understanding a property of the completion is being the smallest complete space $N$ such that $Msubset N$, and so the intuition tells me the above should hold. But I did not manage proving it yet.
analysis
asked Jan 25 at 10:00
natenate
1186
$endgroup$
add a comment |
$begingroup$
I am reading a proof where we have the following:
$M$ a metric space
$N$ its metric completion
It is then stated: for $yin N$ let ($y_s$) be a Cauchy sequence converging to y with $y_sin M$.
I do not understand why can we choose $y_sin M$.
Do we have $yin Nsetminus M Rightarrow yinpartial M$ ?
Edit
It is my understanding a property of the completion is being the smallest complete space $N$ such that $Msubset N$, and so the intuition tells me the above should hold. But I did not manage proving it yet.
analysis
asked Jan 25 at 10:00
natenate
1186
$endgroup$
I am reading a proof where we have the following:
$M$ a metric space
$N$ its metric completion
It is then stated: for $yin N$ let ($y_s$) be a Cauchy sequence converging to y with $y_sin M$.
I do not understand why can we choose $y_sin M$.
Do we have $yin Nsetminus M Rightarrow yinpartial M$ ?
Edit
It is my understanding a property of the completion is being the smallest complete space $N$ such that $Msubset N$, and so the intuition tells me the above should hold. But I did not manage proving it yet.
analysis
analysis
asked Jan 25 at 10:00
natenate
1186
asked Jan 25 at 10:00
natenate
1186
edited Jan 25 at 10:04
nate
asked Jan 25 at 10:00
natenate
1186
asked Jan 25 at 10:00
natenate
1186
asked Jan 25 at 10:00
natenate
1186
1186
add a comment |
add a comment |
1 Answer
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$begingroup$
Yes, $M$ is dense in $N$ by the very definition of completion. Ref. https://en.wikipedia.org/wiki/Complete_metric_space
answered Jan 25 at 10:01
Kavi Rama MurthyKavi Rama Murthy
66.3k42867
$endgroup$
add a comment |
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1 Answer
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active
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$begingroup$
Yes, $M$ is dense in $N$ by the very definition of completion. Ref. https://en.wikipedia.org/wiki/Complete_metric_space
answered Jan 25 at 10:01
Kavi Rama MurthyKavi Rama Murthy
66.3k42867
$endgroup$
add a comment |
$begingroup$
Yes, $M$ is dense in $N$ by the very definition of completion. Ref. https://en.wikipedia.org/wiki/Complete_metric_space
answered Jan 25 at 10:01
Kavi Rama MurthyKavi Rama Murthy
66.3k42867
$endgroup$
add a comment |
$begingroup$
Yes, $M$ is dense in $N$ by the very definition of completion. Ref. https://en.wikipedia.org/wiki/Complete_metric_space
answered Jan 25 at 10:01
Kavi Rama MurthyKavi Rama Murthy
66.3k42867
$endgroup$
Yes, $M$ is dense in $N$ by the very definition of completion. Ref. https://en.wikipedia.org/wiki/Complete_metric_space
answered Jan 25 at 10:01
Kavi Rama MurthyKavi Rama Murthy
66.3k42867
answered Jan 25 at 10:01
Kavi Rama MurthyKavi Rama Murthy
66.3k42867
answered Jan 25 at 10:01
Kavi Rama MurthyKavi Rama Murthy
66.3k42867
answered Jan 25 at 10:01
Kavi Rama MurthyKavi Rama Murthy
66.3k42867
66.3k42867
add a comment |
add a comment |
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