Prove a closed subset of a complete metric space is complete via contradiction
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I've seen some proofs by definition and just want to ask for proof verification on whether this is okay as well:
Given complete metric space $(X,d)$ and $A subset X$, $A$ closed. Prove $A$ is complete.
We know any cauchy sequence in $X$ converges, i.e. ${a_n} subset X$ and cauchy implies $a_n to a in X$.
Suppose for contradiction, $A$ is not complete, then $exists {a_n} subset A$ and cauchy such that $a_n$ does not converge in A. Since $A subset X$, this sequence converges to $a in X/A$.
Since $A$ closed, $X/A$ open, so $exists r > 0$ s.t. $B_r(a) subset X/A$. Then this contradicts that it is cauchy by picking any $epsilon < r$.
real-analysis proof-verification metric-spaces cauchy-sequences complete-spaces
asked Nov 28 '18 at 5:57
OneRaynyDayOneRaynyDay
306111
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add a comment |
$begingroup$
I've seen some proofs by definition and just want to ask for proof verification on whether this is okay as well:
Given complete metric space $(X,d)$ and $A subset X$, $A$ closed. Prove $A$ is complete.
We know any cauchy sequence in $X$ converges, i.e. ${a_n} subset X$ and cauchy implies $a_n to a in X$.
Suppose for contradiction, $A$ is not complete, then $exists {a_n} subset A$ and cauchy such that $a_n$ does not converge in A. Since $A subset X$, this sequence converges to $a in X/A$.
Since $A$ closed, $X/A$ open, so $exists r > 0$ s.t. $B_r(a) subset X/A$. Then this contradicts that it is cauchy by picking any $epsilon < r$.
real-analysis proof-verification metric-spaces cauchy-sequences complete-spaces
asked Nov 28 '18 at 5:57
OneRaynyDayOneRaynyDay
306111
$endgroup$
add a comment |
$begingroup$
I've seen some proofs by definition and just want to ask for proof verification on whether this is okay as well:
Given complete metric space $(X,d)$ and $A subset X$, $A$ closed. Prove $A$ is complete.
We know any cauchy sequence in $X$ converges, i.e. ${a_n} subset X$ and cauchy implies $a_n to a in X$.
Suppose for contradiction, $A$ is not complete, then $exists {a_n} subset A$ and cauchy such that $a_n$ does not converge in A. Since $A subset X$, this sequence converges to $a in X/A$.
Since $A$ closed, $X/A$ open, so $exists r > 0$ s.t. $B_r(a) subset X/A$. Then this contradicts that it is cauchy by picking any $epsilon < r$.
real-analysis proof-verification metric-spaces cauchy-sequences complete-spaces
asked Nov 28 '18 at 5:57
OneRaynyDayOneRaynyDay
306111
$endgroup$
I've seen some proofs by definition and just want to ask for proof verification on whether this is okay as well:
Given complete metric space $(X,d)$ and $A subset X$, $A$ closed. Prove $A$ is complete.
We know any cauchy sequence in $X$ converges, i.e. ${a_n} subset X$ and cauchy implies $a_n to a in X$.
Suppose for contradiction, $A$ is not complete, then $exists {a_n} subset A$ and cauchy such that $a_n$ does not converge in A. Since $A subset X$, this sequence converges to $a in X/A$.
Since $A$ closed, $X/A$ open, so $exists r > 0$ s.t. $B_r(a) subset X/A$. Then this contradicts that it is cauchy by picking any $epsilon < r$.
real-analysis proof-verification metric-spaces cauchy-sequences complete-spaces
real-analysis proof-verification metric-spaces cauchy-sequences complete-spaces
asked Nov 28 '18 at 5:57
OneRaynyDayOneRaynyDay
306111
asked Nov 28 '18 at 5:57
OneRaynyDayOneRaynyDay
306111
asked Nov 28 '18 at 5:57
OneRaynyDayOneRaynyDay
306111
asked Nov 28 '18 at 5:57
OneRaynyDayOneRaynyDay
306111
asked Nov 28 '18 at 5:57
OneRaynyDayOneRaynyDay
306111
306111
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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This is basically fine, but you don't really need to go by contradiction. $Asubset X$ is closed iff it contains all of its limit points. Just let an arbitrary Cauchy sequence, ${a_{n}}subset A$ be given, it must converge in $X$ by completeness, and then the closedness directly implies that this limit is in $A$, making $A$ a complete metric space in its own right.
answered Nov 28 '18 at 6:14
JWP_HTXJWP_HTX
410314
$endgroup$
$begingroup$
I see that in retrospect, but I always am pretty cautious when it comes to saying "directly implies" since I'm just starting to write longer proofs. I know that since the point $a$ is a limit point of ${a_n}$, it should be in $A$ since it's closed, but that is convergence in $X$. I was afraid it might not directly translate to convergence in $A$. I see that now, but for me it wasn't directly obvious so I tried to give another proof here :)
$endgroup$
– OneRaynyDay
Nov 28 '18 at 6:30
$begingroup$
No worries! Keep up the good work!
$endgroup$
– JWP_HTX
Nov 28 '18 at 6:48
add a comment |
$begingroup$
Sure, this proof works!
Just a few comments, though: firstly, take the sentence
We know any cauchy sequence in $X$ converges, . . .
In my opinion, it would be preferable to say that
We know any Cauchy sequence in $X$ converges in $X$, . . .
just for the sake of more clarity, even though you have explained what this means by adding an "i.e.".
Secondly,
Then this contradicts that it is cauchy by picking any $epsilon < r$.
It will be helpful to explain how the contradiction arises. At first glance, I can see that there will be a contradiction to the fact that ${ a_n }$ converges to $a$, not that ${ a_n }$ is Cauchy. But, the two ideas are closely related enough that I am still confident that you're not necessarily wrong. So, adding more details here would be better, in my opinion.
answered Nov 28 '18 at 6:06
BrahadeeshBrahadeesh
6,22742361
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
This is basically fine, but you don't really need to go by contradiction. $Asubset X$ is closed iff it contains all of its limit points. Just let an arbitrary Cauchy sequence, ${a_{n}}subset A$ be given, it must converge in $X$ by completeness, and then the closedness directly implies that this limit is in $A$, making $A$ a complete metric space in its own right.
answered Nov 28 '18 at 6:14
JWP_HTXJWP_HTX
410314
$endgroup$
$begingroup$
I see that in retrospect, but I always am pretty cautious when it comes to saying "directly implies" since I'm just starting to write longer proofs. I know that since the point $a$ is a limit point of ${a_n}$, it should be in $A$ since it's closed, but that is convergence in $X$. I was afraid it might not directly translate to convergence in $A$. I see that now, but for me it wasn't directly obvious so I tried to give another proof here :)
$endgroup$
– OneRaynyDay
Nov 28 '18 at 6:30
$begingroup$
No worries! Keep up the good work!
$endgroup$
– JWP_HTX
Nov 28 '18 at 6:48
add a comment |
$begingroup$
This is basically fine, but you don't really need to go by contradiction. $Asubset X$ is closed iff it contains all of its limit points. Just let an arbitrary Cauchy sequence, ${a_{n}}subset A$ be given, it must converge in $X$ by completeness, and then the closedness directly implies that this limit is in $A$, making $A$ a complete metric space in its own right.
answered Nov 28 '18 at 6:14
JWP_HTXJWP_HTX
410314
$endgroup$
$begingroup$
I see that in retrospect, but I always am pretty cautious when it comes to saying "directly implies" since I'm just starting to write longer proofs. I know that since the point $a$ is a limit point of ${a_n}$, it should be in $A$ since it's closed, but that is convergence in $X$. I was afraid it might not directly translate to convergence in $A$. I see that now, but for me it wasn't directly obvious so I tried to give another proof here :)
$endgroup$
– OneRaynyDay
Nov 28 '18 at 6:30
$begingroup$
No worries! Keep up the good work!
$endgroup$
– JWP_HTX
Nov 28 '18 at 6:48
add a comment |
$begingroup$
This is basically fine, but you don't really need to go by contradiction. $Asubset X$ is closed iff it contains all of its limit points. Just let an arbitrary Cauchy sequence, ${a_{n}}subset A$ be given, it must converge in $X$ by completeness, and then the closedness directly implies that this limit is in $A$, making $A$ a complete metric space in its own right.
answered Nov 28 '18 at 6:14
JWP_HTXJWP_HTX
410314
$endgroup$
This is basically fine, but you don't really need to go by contradiction. $Asubset X$ is closed iff it contains all of its limit points. Just let an arbitrary Cauchy sequence, ${a_{n}}subset A$ be given, it must converge in $X$ by completeness, and then the closedness directly implies that this limit is in $A$, making $A$ a complete metric space in its own right.
answered Nov 28 '18 at 6:14
JWP_HTXJWP_HTX
410314
answered Nov 28 '18 at 6:14
JWP_HTXJWP_HTX
410314
answered Nov 28 '18 at 6:14
JWP_HTXJWP_HTX
410314
answered Nov 28 '18 at 6:14
JWP_HTXJWP_HTX
410314
410314
$begingroup$
I see that in retrospect, but I always am pretty cautious when it comes to saying "directly implies" since I'm just starting to write longer proofs. I know that since the point $a$ is a limit point of ${a_n}$, it should be in $A$ since it's closed, but that is convergence in $X$. I was afraid it might not directly translate to convergence in $A$. I see that now, but for me it wasn't directly obvious so I tried to give another proof here :)
$endgroup$
– OneRaynyDay
Nov 28 '18 at 6:30
$begingroup$
No worries! Keep up the good work!
$endgroup$
– JWP_HTX
Nov 28 '18 at 6:48
add a comment |
$begingroup$
I see that in retrospect, but I always am pretty cautious when it comes to saying "directly implies" since I'm just starting to write longer proofs. I know that since the point $a$ is a limit point of ${a_n}$, it should be in $A$ since it's closed, but that is convergence in $X$. I was afraid it might not directly translate to convergence in $A$. I see that now, but for me it wasn't directly obvious so I tried to give another proof here :)
$endgroup$
– OneRaynyDay
Nov 28 '18 at 6:30
$begingroup$
No worries! Keep up the good work!
$endgroup$
– JWP_HTX
Nov 28 '18 at 6:48
$begingroup$
I see that in retrospect, but I always am pretty cautious when it comes to saying "directly implies" since I'm just starting to write longer proofs. I know that since the point $a$ is a limit point of ${a_n}$, it should be in $A$ since it's closed, but that is convergence in $X$. I was afraid it might not directly translate to convergence in $A$. I see that now, but for me it wasn't directly obvious so I tried to give another proof here :)
$endgroup$
– OneRaynyDay
Nov 28 '18 at 6:30
$begingroup$
I see that in retrospect, but I always am pretty cautious when it comes to saying "directly implies" since I'm just starting to write longer proofs. I know that since the point $a$ is a limit point of ${a_n}$, it should be in $A$ since it's closed, but that is convergence in $X$. I was afraid it might not directly translate to convergence in $A$. I see that now, but for me it wasn't directly obvious so I tried to give another proof here :)
$endgroup$
– OneRaynyDay
Nov 28 '18 at 6:30
$begingroup$
No worries! Keep up the good work!
$endgroup$
– JWP_HTX
Nov 28 '18 at 6:48
$begingroup$
No worries! Keep up the good work!
$endgroup$
– JWP_HTX
Nov 28 '18 at 6:48
add a comment |
$begingroup$
Sure, this proof works!
Just a few comments, though: firstly, take the sentence
We know any cauchy sequence in $X$ converges, . . .
In my opinion, it would be preferable to say that
We know any Cauchy sequence in $X$ converges in $X$, . . .
just for the sake of more clarity, even though you have explained what this means by adding an "i.e.".
Secondly,
Then this contradicts that it is cauchy by picking any $epsilon < r$.
It will be helpful to explain how the contradiction arises. At first glance, I can see that there will be a contradiction to the fact that ${ a_n }$ converges to $a$, not that ${ a_n }$ is Cauchy. But, the two ideas are closely related enough that I am still confident that you're not necessarily wrong. So, adding more details here would be better, in my opinion.
answered Nov 28 '18 at 6:06
BrahadeeshBrahadeesh
6,22742361
$endgroup$
add a comment |
$begingroup$
Sure, this proof works!
Just a few comments, though: firstly, take the sentence
We know any cauchy sequence in $X$ converges, . . .
In my opinion, it would be preferable to say that
We know any Cauchy sequence in $X$ converges in $X$, . . .
just for the sake of more clarity, even though you have explained what this means by adding an "i.e.".
Secondly,
Then this contradicts that it is cauchy by picking any $epsilon < r$.
It will be helpful to explain how the contradiction arises. At first glance, I can see that there will be a contradiction to the fact that ${ a_n }$ converges to $a$, not that ${ a_n }$ is Cauchy. But, the two ideas are closely related enough that I am still confident that you're not necessarily wrong. So, adding more details here would be better, in my opinion.
answered Nov 28 '18 at 6:06
BrahadeeshBrahadeesh
6,22742361
$endgroup$
add a comment |
$begingroup$
Sure, this proof works!
Just a few comments, though: firstly, take the sentence
We know any cauchy sequence in $X$ converges, . . .
In my opinion, it would be preferable to say that
We know any Cauchy sequence in $X$ converges in $X$, . . .
just for the sake of more clarity, even though you have explained what this means by adding an "i.e.".
Secondly,
Then this contradicts that it is cauchy by picking any $epsilon < r$.
It will be helpful to explain how the contradiction arises. At first glance, I can see that there will be a contradiction to the fact that ${ a_n }$ converges to $a$, not that ${ a_n }$ is Cauchy. But, the two ideas are closely related enough that I am still confident that you're not necessarily wrong. So, adding more details here would be better, in my opinion.
answered Nov 28 '18 at 6:06
BrahadeeshBrahadeesh
6,22742361
$endgroup$
Sure, this proof works!
Just a few comments, though: firstly, take the sentence
We know any cauchy sequence in $X$ converges, . . .
In my opinion, it would be preferable to say that
We know any Cauchy sequence in $X$ converges in $X$, . . .
just for the sake of more clarity, even though you have explained what this means by adding an "i.e.".
Secondly,
Then this contradicts that it is cauchy by picking any $epsilon < r$.
It will be helpful to explain how the contradiction arises. At first glance, I can see that there will be a contradiction to the fact that ${ a_n }$ converges to $a$, not that ${ a_n }$ is Cauchy. But, the two ideas are closely related enough that I am still confident that you're not necessarily wrong. So, adding more details here would be better, in my opinion.
answered Nov 28 '18 at 6:06
BrahadeeshBrahadeesh
6,22742361
answered Nov 28 '18 at 6:06
BrahadeeshBrahadeesh
6,22742361
answered Nov 28 '18 at 6:06
BrahadeeshBrahadeesh
6,22742361
answered Nov 28 '18 at 6:06
BrahadeeshBrahadeesh
6,22742361
6,22742361
add a comment |
add a comment |
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