A set F is closed if and only if every convergent sequence in F converges to a limit in F
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$F subseteq X$, where $X$ is a metric space.
Then $F$ is closed if and only if every convergent sequence in F converges to a limit in $F$.
Attempt
$ implies$ If $F$ is closed, $F= overline F$, but $overline F=F cup F'$ where $F'$ is the set of all accumulation points of $F$.
Thus, $F' subseteq F$ which means that $F$ contains all of its limit points which means that every sequence in F converges to a limit that is in F.
Is this a valid proof for this direction?
Edit
After reading the comments, I should have written every convergent sequence in F converges to a limit in F.
real-analysis proof-verification metric-spaces
asked Nov 29 at 4:16
Snop D.
285
add a comment |
up vote
0
down vote
favorite
$F subseteq X$, where $X$ is a metric space.
Then $F$ is closed if and only if every convergent sequence in F converges to a limit in $F$.
Attempt
$ implies$ If $F$ is closed, $F= overline F$, but $overline F=F cup F'$ where $F'$ is the set of all accumulation points of $F$.
Thus, $F' subseteq F$ which means that $F$ contains all of its limit points which means that every sequence in F converges to a limit that is in F.
Is this a valid proof for this direction?
Edit
After reading the comments, I should have written every convergent sequence in F converges to a limit in F.
real-analysis proof-verification metric-spaces
asked Nov 29 at 4:16
Snop D.
285
You may find your answer here: math.stackexchange.com/questions/882876/…
– Aniruddha Deshmukh
Nov 29 at 4:20
1
If F is the reals, your conclusion that every sequence within F converges to a limit of F, is false.
– William Elliot
Nov 29 at 4:26
So if F contains all its limit points, it does not imply that every sequence in F converges to a limit in F?
– Snop D.
Nov 29 at 4:31
1
@SnopD.: You can say for every point in $F'$ there exist a sequence in $F$ such that it converges to that point.
– Yadati Kiran
Nov 29 at 4:36
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$F subseteq X$, where $X$ is a metric space.
Then $F$ is closed if and only if every convergent sequence in F converges to a limit in $F$.
Attempt
$ implies$ If $F$ is closed, $F= overline F$, but $overline F=F cup F'$ where $F'$ is the set of all accumulation points of $F$.
Thus, $F' subseteq F$ which means that $F$ contains all of its limit points which means that every sequence in F converges to a limit that is in F.
Is this a valid proof for this direction?
Edit
After reading the comments, I should have written every convergent sequence in F converges to a limit in F.
real-analysis proof-verification metric-spaces
asked Nov 29 at 4:16
Snop D.
285
$F subseteq X$, where $X$ is a metric space.
Then $F$ is closed if and only if every convergent sequence in F converges to a limit in $F$.
Attempt
$ implies$ If $F$ is closed, $F= overline F$, but $overline F=F cup F'$ where $F'$ is the set of all accumulation points of $F$.
Thus, $F' subseteq F$ which means that $F$ contains all of its limit points which means that every sequence in F converges to a limit that is in F.
Is this a valid proof for this direction?
Edit
After reading the comments, I should have written every convergent sequence in F converges to a limit in F.
real-analysis proof-verification metric-spaces
real-analysis proof-verification metric-spaces
asked Nov 29 at 4:16
Snop D.
285
asked Nov 29 at 4:16
Snop D.
285
edited Nov 29 at 5:01
asked Nov 29 at 4:16
Snop D.
285
asked Nov 29 at 4:16
Snop D.
285
asked Nov 29 at 4:16
Snop D.
285
285
You may find your answer here: math.stackexchange.com/questions/882876/…
– Aniruddha Deshmukh
Nov 29 at 4:20
1
If F is the reals, your conclusion that every sequence within F converges to a limit of F, is false.
– William Elliot
Nov 29 at 4:26
So if F contains all its limit points, it does not imply that every sequence in F converges to a limit in F?
– Snop D.
Nov 29 at 4:31
1
@SnopD.: You can say for every point in $F'$ there exist a sequence in $F$ such that it converges to that point.
– Yadati Kiran
Nov 29 at 4:36
add a comment |
You may find your answer here: math.stackexchange.com/questions/882876/…
– Aniruddha Deshmukh
Nov 29 at 4:20
1
If F is the reals, your conclusion that every sequence within F converges to a limit of F, is false.
– William Elliot
Nov 29 at 4:26
So if F contains all its limit points, it does not imply that every sequence in F converges to a limit in F?
– Snop D.
Nov 29 at 4:31
1
@SnopD.: You can say for every point in $F'$ there exist a sequence in $F$ such that it converges to that point.
– Yadati Kiran
Nov 29 at 4:36
You may find your answer here: math.stackexchange.com/questions/882876/…
– Aniruddha Deshmukh
Nov 29 at 4:20
You may find your answer here: math.stackexchange.com/questions/882876/…
– Aniruddha Deshmukh
Nov 29 at 4:20
1
1
If F is the reals, your conclusion that every sequence within F converges to a limit of F, is false.
– William Elliot
Nov 29 at 4:26
If F is the reals, your conclusion that every sequence within F converges to a limit of F, is false.
– William Elliot
Nov 29 at 4:26
So if F contains all its limit points, it does not imply that every sequence in F converges to a limit in F?
– Snop D.
Nov 29 at 4:31
So if F contains all its limit points, it does not imply that every sequence in F converges to a limit in F?
– Snop D.
Nov 29 at 4:31
1
1
@SnopD.: You can say for every point in $F'$ there exist a sequence in $F$ such that it converges to that point.
– Yadati Kiran
Nov 29 at 4:36
@SnopD.: You can say for every point in $F'$ there exist a sequence in $F$ such that it converges to that point.
– Yadati Kiran
Nov 29 at 4:36
add a comment |
1 Answer
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up vote
2
down vote
accepted
That's a valid proof (but you wrote every sequence where you should have wrote every convergent sequence).
For the converse:
If every convergent sequence in $A$ converges to a point in $A$, then $A'subset A$. Thus A is closed.
answered Nov 29 at 4:38
Chris Custer
10.4k3724
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
That's a valid proof (but you wrote every sequence where you should have wrote every convergent sequence).
For the converse:
If every convergent sequence in $A$ converges to a point in $A$, then $A'subset A$. Thus A is closed.
answered Nov 29 at 4:38
Chris Custer
10.4k3724
add a comment |
up vote
2
down vote
accepted
That's a valid proof (but you wrote every sequence where you should have wrote every convergent sequence).
For the converse:
If every convergent sequence in $A$ converges to a point in $A$, then $A'subset A$. Thus A is closed.
answered Nov 29 at 4:38
Chris Custer
10.4k3724
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
That's a valid proof (but you wrote every sequence where you should have wrote every convergent sequence).
For the converse:
If every convergent sequence in $A$ converges to a point in $A$, then $A'subset A$. Thus A is closed.
answered Nov 29 at 4:38
Chris Custer
10.4k3724
That's a valid proof (but you wrote every sequence where you should have wrote every convergent sequence).
For the converse:
If every convergent sequence in $A$ converges to a point in $A$, then $A'subset A$. Thus A is closed.
answered Nov 29 at 4:38
Chris Custer
10.4k3724
answered Nov 29 at 4:38
Chris Custer
10.4k3724
answered Nov 29 at 4:38
Chris Custer
10.4k3724
answered Nov 29 at 4:38
Chris Custer
10.4k3724
10.4k3724
add a comment |
add a comment |
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You may find your answer here: math.stackexchange.com/questions/882876/…
– Aniruddha Deshmukh
Nov 29 at 4:20
1
If F is the reals, your conclusion that every sequence within F converges to a limit of F, is false.
– William Elliot
Nov 29 at 4:26
So if F contains all its limit points, it does not imply that every sequence in F converges to a limit in F?
– Snop D.
Nov 29 at 4:31
1
@SnopD.: You can say for every point in $F'$ there exist a sequence in $F$ such that it converges to that point.
– Yadati Kiran
Nov 29 at 4:36