Mixing
Sequential compactness and boundedness
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
The definition of sequential compactness states that set $K$ is sequentially compact if $forall$ sequences $a_n in K$, $exists a_n_i$, such that this subsequence converges to a point $p in K$.
I'm thinking of the corollary to the open cover definition of compactness, that set $K$ is closed and bounded, and trying to relate this to the definition of sequential compactness. I can see how sequential compactness would imply that $K$ is closed, since every sequence contains a limit point in $K$ can follow the definition of sequential compactness
My Question
How does the bounded property follow from the definition of sequential compactness? If we are given that set $K$ is sequentially compact then how could we show that $K$ is bounded? What restrictions or conditions apply to the definition of sequential compactness?
Thanks
real-analysis general-topology analysis proof-writing compactness
asked Jul 19 at 22:13
john fowles
1,093817
add a comment |Â
up vote
1
down vote
favorite
The definition of sequential compactness states that set $K$ is sequentially compact if $forall$ sequences $a_n in K$, $exists a_n_i$, such that this subsequence converges to a point $p in K$.
I'm thinking of the corollary to the open cover definition of compactness, that set $K$ is closed and bounded, and trying to relate this to the definition of sequential compactness. I can see how sequential compactness would imply that $K$ is closed, since every sequence contains a limit point in $K$ can follow the definition of sequential compactness
My Question
How does the bounded property follow from the definition of sequential compactness? If we are given that set $K$ is sequentially compact then how could we show that $K$ is bounded? What restrictions or conditions apply to the definition of sequential compactness?
Thanks
real-analysis general-topology analysis proof-writing compactness
asked Jul 19 at 22:13
john fowles
1,093817
Try an example: why is $mathbbR$ not sequentially compact?
– Eric Wofsey
Jul 19 at 22:18
well we could use the counterexample where we pick infinite sequence $a_i$, where $a_i=i$. so there do not exist any subsequences that converge since they all tend to $infty$
– john fowles
Jul 19 at 23:33
@EricWofsey and I read on wikipedia that sequential compactness implies compactness only for certain topological spaces. Could I treat the definitions are equivalent if I'm only dealing with sets in $mathbbR$, and what about $mathbbR^n$, and other common metric spaces? Could you tell me some spaces where the equivalence fails
– john fowles
Jul 19 at 23:40
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The definition of sequential compactness states that set $K$ is sequentially compact if $forall$ sequences $a_n in K$, $exists a_n_i$, such that this subsequence converges to a point $p in K$.
I'm thinking of the corollary to the open cover definition of compactness, that set $K$ is closed and bounded, and trying to relate this to the definition of sequential compactness. I can see how sequential compactness would imply that $K$ is closed, since every sequence contains a limit point in $K$ can follow the definition of sequential compactness
My Question
How does the bounded property follow from the definition of sequential compactness? If we are given that set $K$ is sequentially compact then how could we show that $K$ is bounded? What restrictions or conditions apply to the definition of sequential compactness?
Thanks
real-analysis general-topology analysis proof-writing compactness
asked Jul 19 at 22:13
john fowles
1,093817
The definition of sequential compactness states that set $K$ is sequentially compact if $forall$ sequences $a_n in K$, $exists a_n_i$, such that this subsequence converges to a point $p in K$.
I'm thinking of the corollary to the open cover definition of compactness, that set $K$ is closed and bounded, and trying to relate this to the definition of sequential compactness. I can see how sequential compactness would imply that $K$ is closed, since every sequence contains a limit point in $K$ can follow the definition of sequential compactness
My Question
How does the bounded property follow from the definition of sequential compactness? If we are given that set $K$ is sequentially compact then how could we show that $K$ is bounded? What restrictions or conditions apply to the definition of sequential compactness?
Thanks
real-analysis general-topology analysis proof-writing compactness
asked Jul 19 at 22:13
john fowles
1,093817
asked Jul 19 at 22:13
john fowles
1,093817
asked Jul 19 at 22:13
john fowles
1,093817
asked Jul 19 at 22:13
john fowles
1,093817
1,093817
Try an example: why is $mathbbR$ not sequentially compact?
– Eric Wofsey
Jul 19 at 22:18
well we could use the counterexample where we pick infinite sequence $a_i$, where $a_i=i$. so there do not exist any subsequences that converge since they all tend to $infty$
– john fowles
Jul 19 at 23:33
@EricWofsey and I read on wikipedia that sequential compactness implies compactness only for certain topological spaces. Could I treat the definitions are equivalent if I'm only dealing with sets in $mathbbR$, and what about $mathbbR^n$, and other common metric spaces? Could you tell me some spaces where the equivalence fails
– john fowles
Jul 19 at 23:40
add a comment |Â
Try an example: why is $mathbbR$ not sequentially compact?
– Eric Wofsey
Jul 19 at 22:18
well we could use the counterexample where we pick infinite sequence $a_i$, where $a_i=i$. so there do not exist any subsequences that converge since they all tend to $infty$
– john fowles
Jul 19 at 23:33
@EricWofsey and I read on wikipedia that sequential compactness implies compactness only for certain topological spaces. Could I treat the definitions are equivalent if I'm only dealing with sets in $mathbbR$, and what about $mathbbR^n$, and other common metric spaces? Could you tell me some spaces where the equivalence fails
– john fowles
Jul 19 at 23:40
Try an example: why is $mathbbR$ not sequentially compact?
– Eric Wofsey
Jul 19 at 22:18
Try an example: why is $mathbbR$ not sequentially compact?
– Eric Wofsey
Jul 19 at 22:18
well we could use the counterexample where we pick infinite sequence $a_i$, where $a_i=i$. so there do not exist any subsequences that converge since they all tend to $infty$
– john fowles
Jul 19 at 23:33
well we could use the counterexample where we pick infinite sequence $a_i$, where $a_i=i$. so there do not exist any subsequences that converge since they all tend to $infty$
– john fowles
Jul 19 at 23:33
@EricWofsey and I read on wikipedia that sequential compactness implies compactness only for certain topological spaces. Could I treat the definitions are equivalent if I'm only dealing with sets in $mathbbR$, and what about $mathbbR^n$, and other common metric spaces? Could you tell me some spaces where the equivalence fails
– john fowles
Jul 19 at 23:40
@EricWofsey and I read on wikipedia that sequential compactness implies compactness only for certain topological spaces. Could I treat the definitions are equivalent if I'm only dealing with sets in $mathbbR$, and what about $mathbbR^n$, and other common metric spaces? Could you tell me some spaces where the equivalence fails
– john fowles
Jul 19 at 23:40
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
Suppose it were not bounded. Then we see $X$ is not contained in any balls of finite radius. Pick one point in $X$ (this is possible as the empty set is bounded), say $x_0$ then choose a $x_1 notin B_1(x_0)$. This is possible as otherwise we would get a bounded $X$. Now choose a $x_2 notin B_2(x_1) cup B_1(x_1)$ as otherwise we could bound $X$. Keep repeating this and you'll get a sequence that cannot obtain a convergent subsequence, which is your desired contradiction.
answered Jul 19 at 22:58
Raymond Chu
1,03719
1
Did you mean $B_1(x_1) cup B_2(x_1)$, instead of $B_2(x_1) cup B_2(x_1)$?
– john fowles
Jul 19 at 23:16
Yes, my bad. I fixed the typo
– Raymond Chu
Jul 19 at 23:24
add a comment |Â
up vote
1
down vote
Hint: suppose $K$ is not bounded. This means that there is a point $a_0in K$, and for each $i$, a point $a_iin K$ such that the distance between $a_0$ and $a_i$ is larger than $i$. Can $K$ be sequentially compact?
answered Jul 19 at 22:21
Anonymous
4,8033940
What definition of bounded are you using?
– john fowles
Jul 19 at 22:39
1
@johnfowles there exists a (finite) "diameter" D such that the distance between any two elements of the set is less than D.
– Anonymous
Jul 19 at 22:48
So to negate sequential compactness of $K$ I use the sequence $a_i$ that's defined as $|a_0-a_i|>i$, where $i=1,2,...,n$, and need to show that $forall$ subsequences of $a_i$ they converge to a point that's outside of $K$. I'm not sure how to show that last part
– john fowles
Jul 19 at 23:10
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Suppose it were not bounded. Then we see $X$ is not contained in any balls of finite radius. Pick one point in $X$ (this is possible as the empty set is bounded), say $x_0$ then choose a $x_1 notin B_1(x_0)$. This is possible as otherwise we would get a bounded $X$. Now choose a $x_2 notin B_2(x_1) cup B_1(x_1)$ as otherwise we could bound $X$. Keep repeating this and you'll get a sequence that cannot obtain a convergent subsequence, which is your desired contradiction.
answered Jul 19 at 22:58
Raymond Chu
1,03719
1
Did you mean $B_1(x_1) cup B_2(x_1)$, instead of $B_2(x_1) cup B_2(x_1)$?
– john fowles
Jul 19 at 23:16
Yes, my bad. I fixed the typo
– Raymond Chu
Jul 19 at 23:24
add a comment |Â
up vote
4
down vote
accepted
Suppose it were not bounded. Then we see $X$ is not contained in any balls of finite radius. Pick one point in $X$ (this is possible as the empty set is bounded), say $x_0$ then choose a $x_1 notin B_1(x_0)$. This is possible as otherwise we would get a bounded $X$. Now choose a $x_2 notin B_2(x_1) cup B_1(x_1)$ as otherwise we could bound $X$. Keep repeating this and you'll get a sequence that cannot obtain a convergent subsequence, which is your desired contradiction.
answered Jul 19 at 22:58
Raymond Chu
1,03719
1
Did you mean $B_1(x_1) cup B_2(x_1)$, instead of $B_2(x_1) cup B_2(x_1)$?
– john fowles
Jul 19 at 23:16
Yes, my bad. I fixed the typo
– Raymond Chu
Jul 19 at 23:24
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Suppose it were not bounded. Then we see $X$ is not contained in any balls of finite radius. Pick one point in $X$ (this is possible as the empty set is bounded), say $x_0$ then choose a $x_1 notin B_1(x_0)$. This is possible as otherwise we would get a bounded $X$. Now choose a $x_2 notin B_2(x_1) cup B_1(x_1)$ as otherwise we could bound $X$. Keep repeating this and you'll get a sequence that cannot obtain a convergent subsequence, which is your desired contradiction.
answered Jul 19 at 22:58
Raymond Chu
1,03719
Suppose it were not bounded. Then we see $X$ is not contained in any balls of finite radius. Pick one point in $X$ (this is possible as the empty set is bounded), say $x_0$ then choose a $x_1 notin B_1(x_0)$. This is possible as otherwise we would get a bounded $X$. Now choose a $x_2 notin B_2(x_1) cup B_1(x_1)$ as otherwise we could bound $X$. Keep repeating this and you'll get a sequence that cannot obtain a convergent subsequence, which is your desired contradiction.
answered Jul 19 at 22:58
Raymond Chu
1,03719
edited Jul 19 at 23:24
answered Jul 19 at 22:58
Raymond Chu
1,03719
answered Jul 19 at 22:58
Raymond Chu
1,03719
answered Jul 19 at 22:58
Raymond Chu
1,03719
1,03719
1
Did you mean $B_1(x_1) cup B_2(x_1)$, instead of $B_2(x_1) cup B_2(x_1)$?
– john fowles
Jul 19 at 23:16
Yes, my bad. I fixed the typo
– Raymond Chu
Jul 19 at 23:24
add a comment |Â
1
Did you mean $B_1(x_1) cup B_2(x_1)$, instead of $B_2(x_1) cup B_2(x_1)$?
– john fowles
Jul 19 at 23:16
Yes, my bad. I fixed the typo
– Raymond Chu
Jul 19 at 23:24
1
1
Did you mean $B_1(x_1) cup B_2(x_1)$, instead of $B_2(x_1) cup B_2(x_1)$?
– john fowles
Jul 19 at 23:16
Did you mean $B_1(x_1) cup B_2(x_1)$, instead of $B_2(x_1) cup B_2(x_1)$?
– john fowles
Jul 19 at 23:16
Yes, my bad. I fixed the typo
– Raymond Chu
Jul 19 at 23:24
Yes, my bad. I fixed the typo
– Raymond Chu
Jul 19 at 23:24
add a comment |Â
up vote
1
down vote
Hint: suppose $K$ is not bounded. This means that there is a point $a_0in K$, and for each $i$, a point $a_iin K$ such that the distance between $a_0$ and $a_i$ is larger than $i$. Can $K$ be sequentially compact?
answered Jul 19 at 22:21
Anonymous
4,8033940
What definition of bounded are you using?
– john fowles
Jul 19 at 22:39
1
@johnfowles there exists a (finite) "diameter" D such that the distance between any two elements of the set is less than D.
– Anonymous
Jul 19 at 22:48
So to negate sequential compactness of $K$ I use the sequence $a_i$ that's defined as $|a_0-a_i|>i$, where $i=1,2,...,n$, and need to show that $forall$ subsequences of $a_i$ they converge to a point that's outside of $K$. I'm not sure how to show that last part
– john fowles
Jul 19 at 23:10
add a comment |Â
up vote
1
down vote
Hint: suppose $K$ is not bounded. This means that there is a point $a_0in K$, and for each $i$, a point $a_iin K$ such that the distance between $a_0$ and $a_i$ is larger than $i$. Can $K$ be sequentially compact?
answered Jul 19 at 22:21
Anonymous
4,8033940
What definition of bounded are you using?
– john fowles
Jul 19 at 22:39
1
@johnfowles there exists a (finite) "diameter" D such that the distance between any two elements of the set is less than D.
– Anonymous
Jul 19 at 22:48
So to negate sequential compactness of $K$ I use the sequence $a_i$ that's defined as $|a_0-a_i|>i$, where $i=1,2,...,n$, and need to show that $forall$ subsequences of $a_i$ they converge to a point that's outside of $K$. I'm not sure how to show that last part
– john fowles
Jul 19 at 23:10
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Hint: suppose $K$ is not bounded. This means that there is a point $a_0in K$, and for each $i$, a point $a_iin K$ such that the distance between $a_0$ and $a_i$ is larger than $i$. Can $K$ be sequentially compact?
answered Jul 19 at 22:21
Anonymous
4,8033940
Hint: suppose $K$ is not bounded. This means that there is a point $a_0in K$, and for each $i$, a point $a_iin K$ such that the distance between $a_0$ and $a_i$ is larger than $i$. Can $K$ be sequentially compact?
answered Jul 19 at 22:21
Anonymous
4,8033940
answered Jul 19 at 22:21
Anonymous
4,8033940
answered Jul 19 at 22:21
Anonymous
4,8033940
answered Jul 19 at 22:21
Anonymous
4,8033940
4,8033940
What definition of bounded are you using?
– john fowles
Jul 19 at 22:39
1
@johnfowles there exists a (finite) "diameter" D such that the distance between any two elements of the set is less than D.
– Anonymous
Jul 19 at 22:48
So to negate sequential compactness of $K$ I use the sequence $a_i$ that's defined as $|a_0-a_i|>i$, where $i=1,2,...,n$, and need to show that $forall$ subsequences of $a_i$ they converge to a point that's outside of $K$. I'm not sure how to show that last part
– john fowles
Jul 19 at 23:10
add a comment |Â
What definition of bounded are you using?
– john fowles
Jul 19 at 22:39
1
@johnfowles there exists a (finite) "diameter" D such that the distance between any two elements of the set is less than D.
– Anonymous
Jul 19 at 22:48
So to negate sequential compactness of $K$ I use the sequence $a_i$ that's defined as $|a_0-a_i|>i$, where $i=1,2,...,n$, and need to show that $forall$ subsequences of $a_i$ they converge to a point that's outside of $K$. I'm not sure how to show that last part
– john fowles
Jul 19 at 23:10
What definition of bounded are you using?
– john fowles
Jul 19 at 22:39
What definition of bounded are you using?
– john fowles
Jul 19 at 22:39
1
1
@johnfowles there exists a (finite) "diameter" D such that the distance between any two elements of the set is less than D.
– Anonymous
Jul 19 at 22:48
@johnfowles there exists a (finite) "diameter" D such that the distance between any two elements of the set is less than D.
– Anonymous
Jul 19 at 22:48
So to negate sequential compactness of $K$ I use the sequence $a_i$ that's defined as $|a_0-a_i|>i$, where $i=1,2,...,n$, and need to show that $forall$ subsequences of $a_i$ they converge to a point that's outside of $K$. I'm not sure how to show that last part
– john fowles
Jul 19 at 23:10
So to negate sequential compactness of $K$ I use the sequence $a_i$ that's defined as $|a_0-a_i|>i$, where $i=1,2,...,n$, and need to show that $forall$ subsequences of $a_i$ they converge to a point that's outside of $K$. I'm not sure how to show that last part
– john fowles
Jul 19 at 23:10
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%2f2857086%2fsequential-compactness-and-boundedness%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
Try an example: why is $mathbbR$ not sequentially compact?
– Eric Wofsey
Jul 19 at 22:18
well we could use the counterexample where we pick infinite sequence $a_i$, where $a_i=i$. so there do not exist any subsequences that converge since they all tend to $infty$
– john fowles
Jul 19 at 23:33
@EricWofsey and I read on wikipedia that sequential compactness implies compactness only for certain topological spaces. Could I treat the definitions are equivalent if I'm only dealing with sets in $mathbbR$, and what about $mathbbR^n$, and other common metric spaces? Could you tell me some spaces where the equivalence fails
– john fowles
Jul 19 at 23:40