Mixing
The space of sequence is infinite dimensional yet it seems Bolzano-Weiestrass applies
I got the following exercise :
Let $f_n$ be a sequence of functions defined on a countable set $A$ and such that :
$$forall x in A, exists M_x > 0, forall n, mid f_n(x) mid leq M_x$$
Then prove that it’s possible to find an extraction $phi$ such that forall $x$, $f_{phi(n)}(x)$ converges for all $x$ in $A$.
It’s not difficult to construct such an extraction.
Moreover in a recent question I’ve asked it says that Bolzano-Weiestrass is true only in finite dimensional vector spaces.
Yet, let’s say I take the space of sequence $u : mathbb{N} to mathbb{R}$ then this is an infinite vector space.
But every sequence of this vector space : $a_n$ can be represented as follow :
$$a_1 = (f_1(x_1), f_1(x_2),...), a_2=(f_2(x_1), ...), a_n = (f_n(x_1),....,)$$
Hence every sequence can be represented as a sequence of functions $f_n$ such that $f_i : A to mathbb{R}$. Hence, by the property above this infinite vector space have the Bolzano-Weirstrass property since (thank’s to the exercises Îve cited above) we can find an extraction such that the sequence $a_{phi(n)}$ converges.
So what’s the mistake I am doing here ?
real-analysis general-topology vector-spaces
asked Nov 21 '18 at 10:13
Interesting problems
13310
|
show 3 more comments
I got the following exercise :
Let $f_n$ be a sequence of functions defined on a countable set $A$ and such that :
$$forall x in A, exists M_x > 0, forall n, mid f_n(x) mid leq M_x$$
Then prove that it’s possible to find an extraction $phi$ such that forall $x$, $f_{phi(n)}(x)$ converges for all $x$ in $A$.
It’s not difficult to construct such an extraction.
Moreover in a recent question I’ve asked it says that Bolzano-Weiestrass is true only in finite dimensional vector spaces.
Yet, let’s say I take the space of sequence $u : mathbb{N} to mathbb{R}$ then this is an infinite vector space.
But every sequence of this vector space : $a_n$ can be represented as follow :
$$a_1 = (f_1(x_1), f_1(x_2),...), a_2=(f_2(x_1), ...), a_n = (f_n(x_1),....,)$$
Hence every sequence can be represented as a sequence of functions $f_n$ such that $f_i : A to mathbb{R}$. Hence, by the property above this infinite vector space have the Bolzano-Weirstrass property since (thank’s to the exercises Îve cited above) we can find an extraction such that the sequence $a_{phi(n)}$ converges.
So what’s the mistake I am doing here ?
real-analysis general-topology vector-spaces
asked Nov 21 '18 at 10:13
Interesting problems
13310
I'm guessing "extraction" = subsequence...is this correct?
– DonAntonio
Nov 21 '18 at 10:16
Yes your are right !
– Interesting problems
Nov 21 '18 at 10:17
What is the notion of convergence i this space. You need a topology to talk about Bolzano -Weirstarss property.
– Kavi Rama Murthy
Nov 21 '18 at 10:17
@Ravi Rama Murthy, in this space a sequence converges if : $a_1 = (v_{1,1}, v_{1,2},...), a_n = (v_{n,1}, ...)$ converges iff $forall i, lim_{n to infty} v_{n,i} in mathbb{R}$, so that each component of the vector converges component wise
– Interesting problems
Nov 21 '18 at 10:21
1
@Interestingproblems There is no such thing as Euclidean norm on the space of infinite bounded sequences.
– Kavi Rama Murthy
Nov 21 '18 at 11:45
|
show 3 more comments
I got the following exercise :
Let $f_n$ be a sequence of functions defined on a countable set $A$ and such that :
$$forall x in A, exists M_x > 0, forall n, mid f_n(x) mid leq M_x$$
Then prove that it’s possible to find an extraction $phi$ such that forall $x$, $f_{phi(n)}(x)$ converges for all $x$ in $A$.
It’s not difficult to construct such an extraction.
Moreover in a recent question I’ve asked it says that Bolzano-Weiestrass is true only in finite dimensional vector spaces.
Yet, let’s say I take the space of sequence $u : mathbb{N} to mathbb{R}$ then this is an infinite vector space.
But every sequence of this vector space : $a_n$ can be represented as follow :
$$a_1 = (f_1(x_1), f_1(x_2),...), a_2=(f_2(x_1), ...), a_n = (f_n(x_1),....,)$$
Hence every sequence can be represented as a sequence of functions $f_n$ such that $f_i : A to mathbb{R}$. Hence, by the property above this infinite vector space have the Bolzano-Weirstrass property since (thank’s to the exercises Îve cited above) we can find an extraction such that the sequence $a_{phi(n)}$ converges.
So what’s the mistake I am doing here ?
real-analysis general-topology vector-spaces
asked Nov 21 '18 at 10:13
Interesting problems
13310
I got the following exercise :
Let $f_n$ be a sequence of functions defined on a countable set $A$ and such that :
$$forall x in A, exists M_x > 0, forall n, mid f_n(x) mid leq M_x$$
Then prove that it’s possible to find an extraction $phi$ such that forall $x$, $f_{phi(n)}(x)$ converges for all $x$ in $A$.
It’s not difficult to construct such an extraction.
Moreover in a recent question I’ve asked it says that Bolzano-Weiestrass is true only in finite dimensional vector spaces.
Yet, let’s say I take the space of sequence $u : mathbb{N} to mathbb{R}$ then this is an infinite vector space.
But every sequence of this vector space : $a_n$ can be represented as follow :
$$a_1 = (f_1(x_1), f_1(x_2),...), a_2=(f_2(x_1), ...), a_n = (f_n(x_1),....,)$$
Hence every sequence can be represented as a sequence of functions $f_n$ such that $f_i : A to mathbb{R}$. Hence, by the property above this infinite vector space have the Bolzano-Weirstrass property since (thank’s to the exercises Îve cited above) we can find an extraction such that the sequence $a_{phi(n)}$ converges.
So what’s the mistake I am doing here ?
real-analysis general-topology vector-spaces
real-analysis general-topology vector-spaces
asked Nov 21 '18 at 10:13
Interesting problems
13310
asked Nov 21 '18 at 10:13
Interesting problems
13310
asked Nov 21 '18 at 10:13
Interesting problems
13310
asked Nov 21 '18 at 10:13
Interesting problems
13310
asked Nov 21 '18 at 10:13
Interesting problems
13310
13310
I'm guessing "extraction" = subsequence...is this correct?
– DonAntonio
Nov 21 '18 at 10:16
Yes your are right !
– Interesting problems
Nov 21 '18 at 10:17
What is the notion of convergence i this space. You need a topology to talk about Bolzano -Weirstarss property.
– Kavi Rama Murthy
Nov 21 '18 at 10:17
@Ravi Rama Murthy, in this space a sequence converges if : $a_1 = (v_{1,1}, v_{1,2},...), a_n = (v_{n,1}, ...)$ converges iff $forall i, lim_{n to infty} v_{n,i} in mathbb{R}$, so that each component of the vector converges component wise
– Interesting problems
Nov 21 '18 at 10:21
1
@Interestingproblems There is no such thing as Euclidean norm on the space of infinite bounded sequences.
– Kavi Rama Murthy
Nov 21 '18 at 11:45
|
show 3 more comments
I'm guessing "extraction" = subsequence...is this correct?
– DonAntonio
Nov 21 '18 at 10:16
Yes your are right !
– Interesting problems
Nov 21 '18 at 10:17
What is the notion of convergence i this space. You need a topology to talk about Bolzano -Weirstarss property.
– Kavi Rama Murthy
Nov 21 '18 at 10:17
@Ravi Rama Murthy, in this space a sequence converges if : $a_1 = (v_{1,1}, v_{1,2},...), a_n = (v_{n,1}, ...)$ converges iff $forall i, lim_{n to infty} v_{n,i} in mathbb{R}$, so that each component of the vector converges component wise
– Interesting problems
Nov 21 '18 at 10:21
1
@Interestingproblems There is no such thing as Euclidean norm on the space of infinite bounded sequences.
– Kavi Rama Murthy
Nov 21 '18 at 11:45
I'm guessing "extraction" = subsequence...is this correct?
– DonAntonio
Nov 21 '18 at 10:16
I'm guessing "extraction" = subsequence...is this correct?
– DonAntonio
Nov 21 '18 at 10:16
Yes your are right !
– Interesting problems
Nov 21 '18 at 10:17
Yes your are right !
– Interesting problems
Nov 21 '18 at 10:17
What is the notion of convergence i this space. You need a topology to talk about Bolzano -Weirstarss property.
– Kavi Rama Murthy
Nov 21 '18 at 10:17
What is the notion of convergence i this space. You need a topology to talk about Bolzano -Weirstarss property.
– Kavi Rama Murthy
Nov 21 '18 at 10:17
@Ravi Rama Murthy, in this space a sequence converges if : $a_1 = (v_{1,1}, v_{1,2},...), a_n = (v_{n,1}, ...)$ converges iff $forall i, lim_{n to infty} v_{n,i} in mathbb{R}$, so that each component of the vector converges component wise
– Interesting problems
Nov 21 '18 at 10:21
@Ravi Rama Murthy, in this space a sequence converges if : $a_1 = (v_{1,1}, v_{1,2},...), a_n = (v_{n,1}, ...)$ converges iff $forall i, lim_{n to infty} v_{n,i} in mathbb{R}$, so that each component of the vector converges component wise
– Interesting problems
Nov 21 '18 at 10:21
1
1
@Interestingproblems There is no such thing as Euclidean norm on the space of infinite bounded sequences.
– Kavi Rama Murthy
Nov 21 '18 at 11:45
@Interestingproblems There is no such thing as Euclidean norm on the space of infinite bounded sequences.
– Kavi Rama Murthy
Nov 21 '18 at 11:45
|
show 3 more comments
1 Answer
1
active
oldest
votes
Your are working in a compact subset of an infinite-dimensional vectorspace. The (general) Bolzano-Weierstrass theorem says that every sequence in a compact metrizable space has a convergent subsequence.
You work in the product $prod_{xin A}[-M_x,M_x]$, which is compact and metrizable.
answered Nov 21 '18 at 10:19
hartkp
1,29965
Thank you, si it means that B-W applies in this infinite vector space ?
– Interesting problems
Nov 21 '18 at 11:44
The set you describe is not a vector space; it neither closed under addition nor under scalar multiplication.
– hartkp
Nov 21 '18 at 16:52
The set of sequence is clearly closed under scalar multiplication and addition.
– Interesting problems
Nov 21 '18 at 17:30
No, the function $f$ with values $f(x)=M_x$ is in the set but $2f$ is not.
– hartkp
Nov 21 '18 at 20:58
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Your are working in a compact subset of an infinite-dimensional vectorspace. The (general) Bolzano-Weierstrass theorem says that every sequence in a compact metrizable space has a convergent subsequence.
You work in the product $prod_{xin A}[-M_x,M_x]$, which is compact and metrizable.
answered Nov 21 '18 at 10:19
hartkp
1,29965
Thank you, si it means that B-W applies in this infinite vector space ?
– Interesting problems
Nov 21 '18 at 11:44
The set you describe is not a vector space; it neither closed under addition nor under scalar multiplication.
– hartkp
Nov 21 '18 at 16:52
The set of sequence is clearly closed under scalar multiplication and addition.
– Interesting problems
Nov 21 '18 at 17:30
No, the function $f$ with values $f(x)=M_x$ is in the set but $2f$ is not.
– hartkp
Nov 21 '18 at 20:58
add a comment |
Your are working in a compact subset of an infinite-dimensional vectorspace. The (general) Bolzano-Weierstrass theorem says that every sequence in a compact metrizable space has a convergent subsequence.
You work in the product $prod_{xin A}[-M_x,M_x]$, which is compact and metrizable.
answered Nov 21 '18 at 10:19
hartkp
1,29965
Thank you, si it means that B-W applies in this infinite vector space ?
– Interesting problems
Nov 21 '18 at 11:44
The set you describe is not a vector space; it neither closed under addition nor under scalar multiplication.
– hartkp
Nov 21 '18 at 16:52
The set of sequence is clearly closed under scalar multiplication and addition.
– Interesting problems
Nov 21 '18 at 17:30
No, the function $f$ with values $f(x)=M_x$ is in the set but $2f$ is not.
– hartkp
Nov 21 '18 at 20:58
add a comment |
Your are working in a compact subset of an infinite-dimensional vectorspace. The (general) Bolzano-Weierstrass theorem says that every sequence in a compact metrizable space has a convergent subsequence.
You work in the product $prod_{xin A}[-M_x,M_x]$, which is compact and metrizable.
answered Nov 21 '18 at 10:19
hartkp
1,29965
Your are working in a compact subset of an infinite-dimensional vectorspace. The (general) Bolzano-Weierstrass theorem says that every sequence in a compact metrizable space has a convergent subsequence.
You work in the product $prod_{xin A}[-M_x,M_x]$, which is compact and metrizable.
answered Nov 21 '18 at 10:19
hartkp
1,29965
answered Nov 21 '18 at 10:19
hartkp
1,29965
answered Nov 21 '18 at 10:19
hartkp
1,29965
answered Nov 21 '18 at 10:19
hartkp
1,29965
1,29965
Thank you, si it means that B-W applies in this infinite vector space ?
– Interesting problems
Nov 21 '18 at 11:44
The set you describe is not a vector space; it neither closed under addition nor under scalar multiplication.
– hartkp
Nov 21 '18 at 16:52
The set of sequence is clearly closed under scalar multiplication and addition.
– Interesting problems
Nov 21 '18 at 17:30
No, the function $f$ with values $f(x)=M_x$ is in the set but $2f$ is not.
– hartkp
Nov 21 '18 at 20:58
add a comment |
Thank you, si it means that B-W applies in this infinite vector space ?
– Interesting problems
Nov 21 '18 at 11:44
The set you describe is not a vector space; it neither closed under addition nor under scalar multiplication.
– hartkp
Nov 21 '18 at 16:52
The set of sequence is clearly closed under scalar multiplication and addition.
– Interesting problems
Nov 21 '18 at 17:30
No, the function $f$ with values $f(x)=M_x$ is in the set but $2f$ is not.
– hartkp
Nov 21 '18 at 20:58
Thank you, si it means that B-W applies in this infinite vector space ?
– Interesting problems
Nov 21 '18 at 11:44
Thank you, si it means that B-W applies in this infinite vector space ?
– Interesting problems
Nov 21 '18 at 11:44
The set you describe is not a vector space; it neither closed under addition nor under scalar multiplication.
– hartkp
Nov 21 '18 at 16:52
The set you describe is not a vector space; it neither closed under addition nor under scalar multiplication.
– hartkp
Nov 21 '18 at 16:52
The set of sequence is clearly closed under scalar multiplication and addition.
– Interesting problems
Nov 21 '18 at 17:30
The set of sequence is clearly closed under scalar multiplication and addition.
– Interesting problems
Nov 21 '18 at 17:30
No, the function $f$ with values $f(x)=M_x$ is in the set but $2f$ is not.
– hartkp
Nov 21 '18 at 20:58
No, the function $f$ with values $f(x)=M_x$ is in the set but $2f$ is not.
– hartkp
Nov 21 '18 at 20:58
add a comment |
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I'm guessing "extraction" = subsequence...is this correct?
– DonAntonio
Nov 21 '18 at 10:16
Yes your are right !
– Interesting problems
Nov 21 '18 at 10:17
What is the notion of convergence i this space. You need a topology to talk about Bolzano -Weirstarss property.
– Kavi Rama Murthy
Nov 21 '18 at 10:17
@Ravi Rama Murthy, in this space a sequence converges if : $a_1 = (v_{1,1}, v_{1,2},...), a_n = (v_{n,1}, ...)$ converges iff $forall i, lim_{n to infty} v_{n,i} in mathbb{R}$, so that each component of the vector converges component wise
– Interesting problems
Nov 21 '18 at 10:21
1
@Interestingproblems There is no such thing as Euclidean norm on the space of infinite bounded sequences.
– Kavi Rama Murthy
Nov 21 '18 at 11:45