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Bolzano-Weistrass theorem true in any vector space? [duplicate]
This question already has an answer here:
In what spaces does the Bolzano-Weierstrass theorem hold?
3 answers
Let’s say I have a vector space $V$ endowed with an order relation : $leq$.
Then is Bolzano-Weirstrass theorem true in $V$ ?
In class we have seen Bolzano-Weirstrass theorem only in $mathbb{R}$, that’s why I am wondering if this theorem is true in arbitrary ordered structure like for example a vector space.
In the case where it’s not true for every vector space, is it possible to classify the vector space in which Bolzano-Weirstrass theorem holds ?
calculus real-analysis general-topology vector-spaces
asked Nov 21 '18 at 8:56
Interesting problems
13310
marked as duplicate by Community♦ Nov 21 '18 at 9:49
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
This question already has an answer here:
In what spaces does the Bolzano-Weierstrass theorem hold?
3 answers
Let’s say I have a vector space $V$ endowed with an order relation : $leq$.
Then is Bolzano-Weirstrass theorem true in $V$ ?
In class we have seen Bolzano-Weirstrass theorem only in $mathbb{R}$, that’s why I am wondering if this theorem is true in arbitrary ordered structure like for example a vector space.
In the case where it’s not true for every vector space, is it possible to classify the vector space in which Bolzano-Weirstrass theorem holds ?
calculus real-analysis general-topology vector-spaces
asked Nov 21 '18 at 8:56
Interesting problems
13310
marked as duplicate by Community♦ Nov 21 '18 at 9:49
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
This question already has an answer here:
In what spaces does the Bolzano-Weierstrass theorem hold?
3 answers
Let’s say I have a vector space $V$ endowed with an order relation : $leq$.
Then is Bolzano-Weirstrass theorem true in $V$ ?
In class we have seen Bolzano-Weirstrass theorem only in $mathbb{R}$, that’s why I am wondering if this theorem is true in arbitrary ordered structure like for example a vector space.
In the case where it’s not true for every vector space, is it possible to classify the vector space in which Bolzano-Weirstrass theorem holds ?
calculus real-analysis general-topology vector-spaces
asked Nov 21 '18 at 8:56
Interesting problems
13310
This question already has an answer here:
In what spaces does the Bolzano-Weierstrass theorem hold?
3 answers
Let’s say I have a vector space $V$ endowed with an order relation : $leq$.
Then is Bolzano-Weirstrass theorem true in $V$ ?
In class we have seen Bolzano-Weirstrass theorem only in $mathbb{R}$, that’s why I am wondering if this theorem is true in arbitrary ordered structure like for example a vector space.
In the case where it’s not true for every vector space, is it possible to classify the vector space in which Bolzano-Weirstrass theorem holds ?
This question already has an answer here:
In what spaces does the Bolzano-Weierstrass theorem hold?
3 answers
calculus real-analysis general-topology vector-spaces
calculus real-analysis general-topology vector-spaces
asked Nov 21 '18 at 8:56
Interesting problems
13310
asked Nov 21 '18 at 8:56
Interesting problems
13310
asked Nov 21 '18 at 8:56
Interesting problems
13310
asked Nov 21 '18 at 8:56
Interesting problems
13310
asked Nov 21 '18 at 8:56
Interesting problems
13310
13310
marked as duplicate by Community♦ Nov 21 '18 at 9:49
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Community♦ Nov 21 '18 at 9:49
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
add a comment |
1 Answer
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Bolzano-Weirstrass holds for a normed vector space $V$ if and only if $dim(V)<infty$.
Here's the proof of "$Leftarrow$":
Proof. If $V=mathbb{R}^n$ then you take a sequence $(a_m)subseteq V$. Write it down as $a_m=(a^1_m,ldots, a^n_m)$ where each coordinate is a sequence of reals. Now you use induction: pick a subsequence of first coordinate that is convergent (such exists by Bolzano-Weirstrass for reals). Then you take the subsequence of the whole sequence $a_m$ indexed by the same indexes as the first coordinate subsequence we've just picked. Then you take subsequence of second coordinate in the same manner, then third, fourth, and so on. Eventually, after finitely many steps you end up with a subsequence of $(a_m)$ such that each coordinate is convergent and thus so is $(a_m)$. $Box$
The "$Rightarrow$" is a bit more complicated. Here you can have a look at the sketch of the proof.
For a simple counterexample consider the direct product $V=mathbb{R}oplusmathbb{R}opluscdots$ with Euclidean norm. Then let $e_n=(0,0,ldots,0,1,0,ldots)$ where $1$ is on $n$-th coordinate, i.e. ${e_n}$ is the standard basis. Then the sequence $(e_n)$ is bounded but it does not have a convergent subsequence because $lVert e_n-e_mrVert=sqrt{2}$ for $nneq m$.
answered Nov 21 '18 at 9:21
freakish
11.5k1629
Thank you for your answer. It’s quite clear, yet I’ve asked an other question because it seems like I misunderstood something in the case where the vector space is infinite dimensional.
– Interesting problems
Nov 21 '18 at 10:14
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Bolzano-Weirstrass holds for a normed vector space $V$ if and only if $dim(V)<infty$.
Here's the proof of "$Leftarrow$":
Proof. If $V=mathbb{R}^n$ then you take a sequence $(a_m)subseteq V$. Write it down as $a_m=(a^1_m,ldots, a^n_m)$ where each coordinate is a sequence of reals. Now you use induction: pick a subsequence of first coordinate that is convergent (such exists by Bolzano-Weirstrass for reals). Then you take the subsequence of the whole sequence $a_m$ indexed by the same indexes as the first coordinate subsequence we've just picked. Then you take subsequence of second coordinate in the same manner, then third, fourth, and so on. Eventually, after finitely many steps you end up with a subsequence of $(a_m)$ such that each coordinate is convergent and thus so is $(a_m)$. $Box$
The "$Rightarrow$" is a bit more complicated. Here you can have a look at the sketch of the proof.
For a simple counterexample consider the direct product $V=mathbb{R}oplusmathbb{R}opluscdots$ with Euclidean norm. Then let $e_n=(0,0,ldots,0,1,0,ldots)$ where $1$ is on $n$-th coordinate, i.e. ${e_n}$ is the standard basis. Then the sequence $(e_n)$ is bounded but it does not have a convergent subsequence because $lVert e_n-e_mrVert=sqrt{2}$ for $nneq m$.
answered Nov 21 '18 at 9:21
freakish
11.5k1629
Thank you for your answer. It’s quite clear, yet I’ve asked an other question because it seems like I misunderstood something in the case where the vector space is infinite dimensional.
– Interesting problems
Nov 21 '18 at 10:14
add a comment |
Bolzano-Weirstrass holds for a normed vector space $V$ if and only if $dim(V)<infty$.
Here's the proof of "$Leftarrow$":
Proof. If $V=mathbb{R}^n$ then you take a sequence $(a_m)subseteq V$. Write it down as $a_m=(a^1_m,ldots, a^n_m)$ where each coordinate is a sequence of reals. Now you use induction: pick a subsequence of first coordinate that is convergent (such exists by Bolzano-Weirstrass for reals). Then you take the subsequence of the whole sequence $a_m$ indexed by the same indexes as the first coordinate subsequence we've just picked. Then you take subsequence of second coordinate in the same manner, then third, fourth, and so on. Eventually, after finitely many steps you end up with a subsequence of $(a_m)$ such that each coordinate is convergent and thus so is $(a_m)$. $Box$
The "$Rightarrow$" is a bit more complicated. Here you can have a look at the sketch of the proof.
For a simple counterexample consider the direct product $V=mathbb{R}oplusmathbb{R}opluscdots$ with Euclidean norm. Then let $e_n=(0,0,ldots,0,1,0,ldots)$ where $1$ is on $n$-th coordinate, i.e. ${e_n}$ is the standard basis. Then the sequence $(e_n)$ is bounded but it does not have a convergent subsequence because $lVert e_n-e_mrVert=sqrt{2}$ for $nneq m$.
answered Nov 21 '18 at 9:21
freakish
11.5k1629
Thank you for your answer. It’s quite clear, yet I’ve asked an other question because it seems like I misunderstood something in the case where the vector space is infinite dimensional.
– Interesting problems
Nov 21 '18 at 10:14
add a comment |
Bolzano-Weirstrass holds for a normed vector space $V$ if and only if $dim(V)<infty$.
Here's the proof of "$Leftarrow$":
Proof. If $V=mathbb{R}^n$ then you take a sequence $(a_m)subseteq V$. Write it down as $a_m=(a^1_m,ldots, a^n_m)$ where each coordinate is a sequence of reals. Now you use induction: pick a subsequence of first coordinate that is convergent (such exists by Bolzano-Weirstrass for reals). Then you take the subsequence of the whole sequence $a_m$ indexed by the same indexes as the first coordinate subsequence we've just picked. Then you take subsequence of second coordinate in the same manner, then third, fourth, and so on. Eventually, after finitely many steps you end up with a subsequence of $(a_m)$ such that each coordinate is convergent and thus so is $(a_m)$. $Box$
The "$Rightarrow$" is a bit more complicated. Here you can have a look at the sketch of the proof.
For a simple counterexample consider the direct product $V=mathbb{R}oplusmathbb{R}opluscdots$ with Euclidean norm. Then let $e_n=(0,0,ldots,0,1,0,ldots)$ where $1$ is on $n$-th coordinate, i.e. ${e_n}$ is the standard basis. Then the sequence $(e_n)$ is bounded but it does not have a convergent subsequence because $lVert e_n-e_mrVert=sqrt{2}$ for $nneq m$.
answered Nov 21 '18 at 9:21
freakish
11.5k1629
Bolzano-Weirstrass holds for a normed vector space $V$ if and only if $dim(V)<infty$.
Here's the proof of "$Leftarrow$":
Proof. If $V=mathbb{R}^n$ then you take a sequence $(a_m)subseteq V$. Write it down as $a_m=(a^1_m,ldots, a^n_m)$ where each coordinate is a sequence of reals. Now you use induction: pick a subsequence of first coordinate that is convergent (such exists by Bolzano-Weirstrass for reals). Then you take the subsequence of the whole sequence $a_m$ indexed by the same indexes as the first coordinate subsequence we've just picked. Then you take subsequence of second coordinate in the same manner, then third, fourth, and so on. Eventually, after finitely many steps you end up with a subsequence of $(a_m)$ such that each coordinate is convergent and thus so is $(a_m)$. $Box$
The "$Rightarrow$" is a bit more complicated. Here you can have a look at the sketch of the proof.
For a simple counterexample consider the direct product $V=mathbb{R}oplusmathbb{R}opluscdots$ with Euclidean norm. Then let $e_n=(0,0,ldots,0,1,0,ldots)$ where $1$ is on $n$-th coordinate, i.e. ${e_n}$ is the standard basis. Then the sequence $(e_n)$ is bounded but it does not have a convergent subsequence because $lVert e_n-e_mrVert=sqrt{2}$ for $nneq m$.
answered Nov 21 '18 at 9:21
freakish
11.5k1629
edited Nov 21 '18 at 9:32
answered Nov 21 '18 at 9:21
freakish
11.5k1629
answered Nov 21 '18 at 9:21
freakish
11.5k1629
answered Nov 21 '18 at 9:21
freakish
11.5k1629
11.5k1629
Thank you for your answer. It’s quite clear, yet I’ve asked an other question because it seems like I misunderstood something in the case where the vector space is infinite dimensional.
– Interesting problems
Nov 21 '18 at 10:14
add a comment |
Thank you for your answer. It’s quite clear, yet I’ve asked an other question because it seems like I misunderstood something in the case where the vector space is infinite dimensional.
– Interesting problems
Nov 21 '18 at 10:14
Thank you for your answer. It’s quite clear, yet I’ve asked an other question because it seems like I misunderstood something in the case where the vector space is infinite dimensional.
– Interesting problems
Nov 21 '18 at 10:14
Thank you for your answer. It’s quite clear, yet I’ve asked an other question because it seems like I misunderstood something in the case where the vector space is infinite dimensional.
– Interesting problems
Nov 21 '18 at 10:14
add a comment |
