why a subspace is closed?
$begingroup$
Let $E$ be $K-$vector space with a norm $|cdot|$, and $F$ a subspace with dimension $n$. Show that $F$ is a closed set .
I am trying to show that any convergent sequence of elements $(x_n)_{n in N}$ of $F$ converge to an element $x in F$.
This means I need to show that $||x_n-x|| rightarrow 0$
I see that like a distance which is attended by the fuction $y rightarrow ||x-y||$ .
How can I show that $ x in F$?
I have been reading a solution which I don't really understand the intuition behind it :
let $x$ be in $E$ and let $B={ y in F ||y-x|| leq ||x||}$
after showing that its a compact, they tried to show that $inf_{y in F}(||x-y||)= lambda $ exists.
by definition of $lambda$ it is the distance between $x$ and $F$
topological-vector-spaces
$endgroup$
|
show 3 more comments
$begingroup$
Let $E$ be $K-$vector space with a norm $|cdot|$, and $F$ a subspace with dimension $n$. Show that $F$ is a closed set .
I am trying to show that any convergent sequence of elements $(x_n)_{n in N}$ of $F$ converge to an element $x in F$.
This means I need to show that $||x_n-x|| rightarrow 0$
I see that like a distance which is attended by the fuction $y rightarrow ||x-y||$ .
How can I show that $ x in F$?
I have been reading a solution which I don't really understand the intuition behind it :
let $x$ be in $E$ and let $B={ y in F ||y-x|| leq ||x||}$
after showing that its a compact, they tried to show that $inf_{y in F}(||x-y||)= lambda $ exists.
by definition of $lambda$ it is the distance between $x$ and $F$
topological-vector-spaces
$endgroup$
1
$begingroup$
I am voting to close, because it is not at all clear to me what you are asking. Are you asking for the intuition behind closed sets, or are you asking how to show that a finite dimensional subspace is closed?
$endgroup$
– K.Power
5 hours ago
$begingroup$
@K.Power actually I am looking for both
$endgroup$
– user515918
5 hours ago
$begingroup$
If it is the latter then you will find such a question has been asked before on this site. I should also point out that you need to show that any convergent sequence in $F$ has its limit in $F$.
$endgroup$
– K.Power
5 hours ago
$begingroup$
@K.Power I've been searching but I dont have any idea about the methode they using
$endgroup$
– user515918
5 hours ago
$begingroup$
As these are very distinct questions you should definitely ask these in two separate posts.
$endgroup$
– K.Power
5 hours ago
|
show 3 more comments
$begingroup$
Let $E$ be $K-$vector space with a norm $|cdot|$, and $F$ a subspace with dimension $n$. Show that $F$ is a closed set .
I am trying to show that any convergent sequence of elements $(x_n)_{n in N}$ of $F$ converge to an element $x in F$.
This means I need to show that $||x_n-x|| rightarrow 0$
I see that like a distance which is attended by the fuction $y rightarrow ||x-y||$ .
How can I show that $ x in F$?
I have been reading a solution which I don't really understand the intuition behind it :
let $x$ be in $E$ and let $B={ y in F ||y-x|| leq ||x||}$
after showing that its a compact, they tried to show that $inf_{y in F}(||x-y||)= lambda $ exists.
by definition of $lambda$ it is the distance between $x$ and $F$
topological-vector-spaces
$endgroup$
Let $E$ be $K-$vector space with a norm $|cdot|$, and $F$ a subspace with dimension $n$. Show that $F$ is a closed set .
I am trying to show that any convergent sequence of elements $(x_n)_{n in N}$ of $F$ converge to an element $x in F$.
This means I need to show that $||x_n-x|| rightarrow 0$
I see that like a distance which is attended by the fuction $y rightarrow ||x-y||$ .
How can I show that $ x in F$?
I have been reading a solution which I don't really understand the intuition behind it :
let $x$ be in $E$ and let $B={ y in F ||y-x|| leq ||x||}$
after showing that its a compact, they tried to show that $inf_{y in F}(||x-y||)= lambda $ exists.
by definition of $lambda$ it is the distance between $x$ and $F$
topological-vector-spaces
topological-vector-spaces
edited 5 hours ago
user515918
asked 5 hours ago
user515918user515918
565
565
1
$begingroup$
I am voting to close, because it is not at all clear to me what you are asking. Are you asking for the intuition behind closed sets, or are you asking how to show that a finite dimensional subspace is closed?
$endgroup$
– K.Power
5 hours ago
$begingroup$
@K.Power actually I am looking for both
$endgroup$
– user515918
5 hours ago
$begingroup$
If it is the latter then you will find such a question has been asked before on this site. I should also point out that you need to show that any convergent sequence in $F$ has its limit in $F$.
$endgroup$
– K.Power
5 hours ago
$begingroup$
@K.Power I've been searching but I dont have any idea about the methode they using
$endgroup$
– user515918
5 hours ago
$begingroup$
As these are very distinct questions you should definitely ask these in two separate posts.
$endgroup$
– K.Power
5 hours ago
|
show 3 more comments
1
$begingroup$
I am voting to close, because it is not at all clear to me what you are asking. Are you asking for the intuition behind closed sets, or are you asking how to show that a finite dimensional subspace is closed?
$endgroup$
– K.Power
5 hours ago
$begingroup$
@K.Power actually I am looking for both
$endgroup$
– user515918
5 hours ago
$begingroup$
If it is the latter then you will find such a question has been asked before on this site. I should also point out that you need to show that any convergent sequence in $F$ has its limit in $F$.
$endgroup$
– K.Power
5 hours ago
$begingroup$
@K.Power I've been searching but I dont have any idea about the methode they using
$endgroup$
– user515918
5 hours ago
$begingroup$
As these are very distinct questions you should definitely ask these in two separate posts.
$endgroup$
– K.Power
5 hours ago
1
1
$begingroup$
I am voting to close, because it is not at all clear to me what you are asking. Are you asking for the intuition behind closed sets, or are you asking how to show that a finite dimensional subspace is closed?
$endgroup$
– K.Power
5 hours ago
$begingroup$
I am voting to close, because it is not at all clear to me what you are asking. Are you asking for the intuition behind closed sets, or are you asking how to show that a finite dimensional subspace is closed?
$endgroup$
– K.Power
5 hours ago
$begingroup$
@K.Power actually I am looking for both
$endgroup$
– user515918
5 hours ago
$begingroup$
@K.Power actually I am looking for both
$endgroup$
– user515918
5 hours ago
$begingroup$
If it is the latter then you will find such a question has been asked before on this site. I should also point out that you need to show that any convergent sequence in $F$ has its limit in $F$.
$endgroup$
– K.Power
5 hours ago
$begingroup$
If it is the latter then you will find such a question has been asked before on this site. I should also point out that you need to show that any convergent sequence in $F$ has its limit in $F$.
$endgroup$
– K.Power
5 hours ago
$begingroup$
@K.Power I've been searching but I dont have any idea about the methode they using
$endgroup$
– user515918
5 hours ago
$begingroup$
@K.Power I've been searching but I dont have any idea about the methode they using
$endgroup$
– user515918
5 hours ago
$begingroup$
As these are very distinct questions you should definitely ask these in two separate posts.
$endgroup$
– K.Power
5 hours ago
$begingroup$
As these are very distinct questions you should definitely ask these in two separate posts.
$endgroup$
– K.Power
5 hours ago
|
show 3 more comments
5 Answers
5
active
oldest
votes
$begingroup$
Another approach will be to show that a finite-dimensional normed space (over $mathbb R$ or $mathbb C$ or indeed over any complete normed field) is complete. Then conclude that it is closed in a larger normed space.
$endgroup$
add a comment |
$begingroup$
Hint:
Use that all norms on $F$ are equivalent to show that all bounded subsets of $F$ which are closed in $F$ are compact.
Show that your sequence is contained in a bounded subset of $F$ which is closed in $F$.
Find a subsequence which has a limit in $F$. What is the limit?
$endgroup$
add a comment |
$begingroup$
To elaborate a bit on GEdgar's suggestion: We know that $F$ has a basis ${e_1,dots,e_n}$. Using Bolzano-Weierstrass you can show that we have the existence of some $r>0$ such that for any choice of scalars $c_1,dots,c_nin K$ we have
$$|sum_{i=1}^nc_ie_i|geq rsum_{i=1}^n|c_i|.$$
Using this you can show that $F$ is complete. Consider a cauchy sequence $(x_m)subset F$. We know that each $x_m=sum_{i=1}^nc_i^{(m)}e_i$. Using the inequality above and the cauchyness of $(x_m)$ try and show that each sequence of scalars $(c_i^{(m)})$ is also Cauchy. As $K$ is complete we know that these sequences each have a limit in $K$. Show that $(x_m)$ converges to the element of $F$ defined by these scalar limits.
$endgroup$
add a comment |
$begingroup$
I think K.Power's answer is the best elementary proof but if you don't know the inequality used in his/her answer, you can argue successfully without it:
Let $F=text{span}{e^1,cdots, e^n},$ where the $|e^i|=1; 1le ile n.$
Now, suppose $Fsupset (x_n)to xin E$. Then, $x_i=a^i_1e^1+cdots a^i_ne^n; 1le ile n.$
Of course, $(x_n)$ is a Cauchy sequence in $E$, so since
$|x_m-x_k|=|(a^m_1-a^k_1)e^1+cdots (a^m_n-a^k_n)e^n|$
the triangle inequality shows that each $(a^m_i); 1le ile n$ is a Cauchy sequence in $K$.
Assuming now that $K$ is complete, we have $a^m_ito a_iin K; 1le ile n$.
To finish, consider the vector $x=a_1e^1+cdots +a_ne^n$.
Clearly, $xin F$ and now, another application of the triangle inequality shows that
$|x-x_n|to 0$, from which it follows that $x_nto xin F$
$endgroup$
add a comment |
$begingroup$
You haven't correctly stated what you need to prove -- you need to show that any sequence of elements of $F$ that converges in $E$ in fact has its limit in $F$.
But I think it may be easier to show that the complement of F is open. Consider an element $x in E setminus F$. It has a distance $gamma gt 0$ from $F$, given by $||z||$, where $x = y+z$ with $y in F$ and $z perp F$. Then $B(x, gamma /2) subseteq E setminus F$ so $x$ is in the interior of $E setminus F$. We chose $x$ arbitrarily, so this proves $E setminus F$ is open so that $F$ is closed.
$endgroup$
$begingroup$
how can we write x=y+z ? E is not an euclidien space to talk about the orthogonal of a subspace !
$endgroup$
– user515918
5 hours ago
$begingroup$
Aren't you assuming that the OP knows that finite dimensional subspaces are complemented? This implies they already know that they are closed.
$endgroup$
– K.Power
4 hours ago
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Another approach will be to show that a finite-dimensional normed space (over $mathbb R$ or $mathbb C$ or indeed over any complete normed field) is complete. Then conclude that it is closed in a larger normed space.
$endgroup$
add a comment |
$begingroup$
Another approach will be to show that a finite-dimensional normed space (over $mathbb R$ or $mathbb C$ or indeed over any complete normed field) is complete. Then conclude that it is closed in a larger normed space.
$endgroup$
add a comment |
$begingroup$
Another approach will be to show that a finite-dimensional normed space (over $mathbb R$ or $mathbb C$ or indeed over any complete normed field) is complete. Then conclude that it is closed in a larger normed space.
$endgroup$
Another approach will be to show that a finite-dimensional normed space (over $mathbb R$ or $mathbb C$ or indeed over any complete normed field) is complete. Then conclude that it is closed in a larger normed space.
answered 5 hours ago
GEdgarGEdgar
62.7k267171
62.7k267171
add a comment |
add a comment |
$begingroup$
Hint:
Use that all norms on $F$ are equivalent to show that all bounded subsets of $F$ which are closed in $F$ are compact.
Show that your sequence is contained in a bounded subset of $F$ which is closed in $F$.
Find a subsequence which has a limit in $F$. What is the limit?
$endgroup$
add a comment |
$begingroup$
Hint:
Use that all norms on $F$ are equivalent to show that all bounded subsets of $F$ which are closed in $F$ are compact.
Show that your sequence is contained in a bounded subset of $F$ which is closed in $F$.
Find a subsequence which has a limit in $F$. What is the limit?
$endgroup$
add a comment |
$begingroup$
Hint:
Use that all norms on $F$ are equivalent to show that all bounded subsets of $F$ which are closed in $F$ are compact.
Show that your sequence is contained in a bounded subset of $F$ which is closed in $F$.
Find a subsequence which has a limit in $F$. What is the limit?
$endgroup$
Hint:
Use that all norms on $F$ are equivalent to show that all bounded subsets of $F$ which are closed in $F$ are compact.
Show that your sequence is contained in a bounded subset of $F$ which is closed in $F$.
Find a subsequence which has a limit in $F$. What is the limit?
edited 4 hours ago
answered 4 hours ago
triitrii
1515
1515
add a comment |
add a comment |
$begingroup$
To elaborate a bit on GEdgar's suggestion: We know that $F$ has a basis ${e_1,dots,e_n}$. Using Bolzano-Weierstrass you can show that we have the existence of some $r>0$ such that for any choice of scalars $c_1,dots,c_nin K$ we have
$$|sum_{i=1}^nc_ie_i|geq rsum_{i=1}^n|c_i|.$$
Using this you can show that $F$ is complete. Consider a cauchy sequence $(x_m)subset F$. We know that each $x_m=sum_{i=1}^nc_i^{(m)}e_i$. Using the inequality above and the cauchyness of $(x_m)$ try and show that each sequence of scalars $(c_i^{(m)})$ is also Cauchy. As $K$ is complete we know that these sequences each have a limit in $K$. Show that $(x_m)$ converges to the element of $F$ defined by these scalar limits.
$endgroup$
add a comment |
$begingroup$
To elaborate a bit on GEdgar's suggestion: We know that $F$ has a basis ${e_1,dots,e_n}$. Using Bolzano-Weierstrass you can show that we have the existence of some $r>0$ such that for any choice of scalars $c_1,dots,c_nin K$ we have
$$|sum_{i=1}^nc_ie_i|geq rsum_{i=1}^n|c_i|.$$
Using this you can show that $F$ is complete. Consider a cauchy sequence $(x_m)subset F$. We know that each $x_m=sum_{i=1}^nc_i^{(m)}e_i$. Using the inequality above and the cauchyness of $(x_m)$ try and show that each sequence of scalars $(c_i^{(m)})$ is also Cauchy. As $K$ is complete we know that these sequences each have a limit in $K$. Show that $(x_m)$ converges to the element of $F$ defined by these scalar limits.
$endgroup$
add a comment |
$begingroup$
To elaborate a bit on GEdgar's suggestion: We know that $F$ has a basis ${e_1,dots,e_n}$. Using Bolzano-Weierstrass you can show that we have the existence of some $r>0$ such that for any choice of scalars $c_1,dots,c_nin K$ we have
$$|sum_{i=1}^nc_ie_i|geq rsum_{i=1}^n|c_i|.$$
Using this you can show that $F$ is complete. Consider a cauchy sequence $(x_m)subset F$. We know that each $x_m=sum_{i=1}^nc_i^{(m)}e_i$. Using the inequality above and the cauchyness of $(x_m)$ try and show that each sequence of scalars $(c_i^{(m)})$ is also Cauchy. As $K$ is complete we know that these sequences each have a limit in $K$. Show that $(x_m)$ converges to the element of $F$ defined by these scalar limits.
$endgroup$
To elaborate a bit on GEdgar's suggestion: We know that $F$ has a basis ${e_1,dots,e_n}$. Using Bolzano-Weierstrass you can show that we have the existence of some $r>0$ such that for any choice of scalars $c_1,dots,c_nin K$ we have
$$|sum_{i=1}^nc_ie_i|geq rsum_{i=1}^n|c_i|.$$
Using this you can show that $F$ is complete. Consider a cauchy sequence $(x_m)subset F$. We know that each $x_m=sum_{i=1}^nc_i^{(m)}e_i$. Using the inequality above and the cauchyness of $(x_m)$ try and show that each sequence of scalars $(c_i^{(m)})$ is also Cauchy. As $K$ is complete we know that these sequences each have a limit in $K$. Show that $(x_m)$ converges to the element of $F$ defined by these scalar limits.
edited 4 hours ago
answered 4 hours ago
K.PowerK.Power
3,065926
3,065926
add a comment |
add a comment |
$begingroup$
I think K.Power's answer is the best elementary proof but if you don't know the inequality used in his/her answer, you can argue successfully without it:
Let $F=text{span}{e^1,cdots, e^n},$ where the $|e^i|=1; 1le ile n.$
Now, suppose $Fsupset (x_n)to xin E$. Then, $x_i=a^i_1e^1+cdots a^i_ne^n; 1le ile n.$
Of course, $(x_n)$ is a Cauchy sequence in $E$, so since
$|x_m-x_k|=|(a^m_1-a^k_1)e^1+cdots (a^m_n-a^k_n)e^n|$
the triangle inequality shows that each $(a^m_i); 1le ile n$ is a Cauchy sequence in $K$.
Assuming now that $K$ is complete, we have $a^m_ito a_iin K; 1le ile n$.
To finish, consider the vector $x=a_1e^1+cdots +a_ne^n$.
Clearly, $xin F$ and now, another application of the triangle inequality shows that
$|x-x_n|to 0$, from which it follows that $x_nto xin F$
$endgroup$
add a comment |
$begingroup$
I think K.Power's answer is the best elementary proof but if you don't know the inequality used in his/her answer, you can argue successfully without it:
Let $F=text{span}{e^1,cdots, e^n},$ where the $|e^i|=1; 1le ile n.$
Now, suppose $Fsupset (x_n)to xin E$. Then, $x_i=a^i_1e^1+cdots a^i_ne^n; 1le ile n.$
Of course, $(x_n)$ is a Cauchy sequence in $E$, so since
$|x_m-x_k|=|(a^m_1-a^k_1)e^1+cdots (a^m_n-a^k_n)e^n|$
the triangle inequality shows that each $(a^m_i); 1le ile n$ is a Cauchy sequence in $K$.
Assuming now that $K$ is complete, we have $a^m_ito a_iin K; 1le ile n$.
To finish, consider the vector $x=a_1e^1+cdots +a_ne^n$.
Clearly, $xin F$ and now, another application of the triangle inequality shows that
$|x-x_n|to 0$, from which it follows that $x_nto xin F$
$endgroup$
add a comment |
$begingroup$
I think K.Power's answer is the best elementary proof but if you don't know the inequality used in his/her answer, you can argue successfully without it:
Let $F=text{span}{e^1,cdots, e^n},$ where the $|e^i|=1; 1le ile n.$
Now, suppose $Fsupset (x_n)to xin E$. Then, $x_i=a^i_1e^1+cdots a^i_ne^n; 1le ile n.$
Of course, $(x_n)$ is a Cauchy sequence in $E$, so since
$|x_m-x_k|=|(a^m_1-a^k_1)e^1+cdots (a^m_n-a^k_n)e^n|$
the triangle inequality shows that each $(a^m_i); 1le ile n$ is a Cauchy sequence in $K$.
Assuming now that $K$ is complete, we have $a^m_ito a_iin K; 1le ile n$.
To finish, consider the vector $x=a_1e^1+cdots +a_ne^n$.
Clearly, $xin F$ and now, another application of the triangle inequality shows that
$|x-x_n|to 0$, from which it follows that $x_nto xin F$
$endgroup$
I think K.Power's answer is the best elementary proof but if you don't know the inequality used in his/her answer, you can argue successfully without it:
Let $F=text{span}{e^1,cdots, e^n},$ where the $|e^i|=1; 1le ile n.$
Now, suppose $Fsupset (x_n)to xin E$. Then, $x_i=a^i_1e^1+cdots a^i_ne^n; 1le ile n.$
Of course, $(x_n)$ is a Cauchy sequence in $E$, so since
$|x_m-x_k|=|(a^m_1-a^k_1)e^1+cdots (a^m_n-a^k_n)e^n|$
the triangle inequality shows that each $(a^m_i); 1le ile n$ is a Cauchy sequence in $K$.
Assuming now that $K$ is complete, we have $a^m_ito a_iin K; 1le ile n$.
To finish, consider the vector $x=a_1e^1+cdots +a_ne^n$.
Clearly, $xin F$ and now, another application of the triangle inequality shows that
$|x-x_n|to 0$, from which it follows that $x_nto xin F$
edited 3 hours ago
answered 4 hours ago
MatematletaMatematleta
11.4k2920
11.4k2920
add a comment |
add a comment |
$begingroup$
You haven't correctly stated what you need to prove -- you need to show that any sequence of elements of $F$ that converges in $E$ in fact has its limit in $F$.
But I think it may be easier to show that the complement of F is open. Consider an element $x in E setminus F$. It has a distance $gamma gt 0$ from $F$, given by $||z||$, where $x = y+z$ with $y in F$ and $z perp F$. Then $B(x, gamma /2) subseteq E setminus F$ so $x$ is in the interior of $E setminus F$. We chose $x$ arbitrarily, so this proves $E setminus F$ is open so that $F$ is closed.
$endgroup$
$begingroup$
how can we write x=y+z ? E is not an euclidien space to talk about the orthogonal of a subspace !
$endgroup$
– user515918
5 hours ago
$begingroup$
Aren't you assuming that the OP knows that finite dimensional subspaces are complemented? This implies they already know that they are closed.
$endgroup$
– K.Power
4 hours ago
add a comment |
$begingroup$
You haven't correctly stated what you need to prove -- you need to show that any sequence of elements of $F$ that converges in $E$ in fact has its limit in $F$.
But I think it may be easier to show that the complement of F is open. Consider an element $x in E setminus F$. It has a distance $gamma gt 0$ from $F$, given by $||z||$, where $x = y+z$ with $y in F$ and $z perp F$. Then $B(x, gamma /2) subseteq E setminus F$ so $x$ is in the interior of $E setminus F$. We chose $x$ arbitrarily, so this proves $E setminus F$ is open so that $F$ is closed.
$endgroup$
$begingroup$
how can we write x=y+z ? E is not an euclidien space to talk about the orthogonal of a subspace !
$endgroup$
– user515918
5 hours ago
$begingroup$
Aren't you assuming that the OP knows that finite dimensional subspaces are complemented? This implies they already know that they are closed.
$endgroup$
– K.Power
4 hours ago
add a comment |
$begingroup$
You haven't correctly stated what you need to prove -- you need to show that any sequence of elements of $F$ that converges in $E$ in fact has its limit in $F$.
But I think it may be easier to show that the complement of F is open. Consider an element $x in E setminus F$. It has a distance $gamma gt 0$ from $F$, given by $||z||$, where $x = y+z$ with $y in F$ and $z perp F$. Then $B(x, gamma /2) subseteq E setminus F$ so $x$ is in the interior of $E setminus F$. We chose $x$ arbitrarily, so this proves $E setminus F$ is open so that $F$ is closed.
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You haven't correctly stated what you need to prove -- you need to show that any sequence of elements of $F$ that converges in $E$ in fact has its limit in $F$.
But I think it may be easier to show that the complement of F is open. Consider an element $x in E setminus F$. It has a distance $gamma gt 0$ from $F$, given by $||z||$, where $x = y+z$ with $y in F$ and $z perp F$. Then $B(x, gamma /2) subseteq E setminus F$ so $x$ is in the interior of $E setminus F$. We chose $x$ arbitrarily, so this proves $E setminus F$ is open so that $F$ is closed.
answered 5 hours ago
Robert ShoreRobert Shore
1,719115
1,719115
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how can we write x=y+z ? E is not an euclidien space to talk about the orthogonal of a subspace !
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– user515918
5 hours ago
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Aren't you assuming that the OP knows that finite dimensional subspaces are complemented? This implies they already know that they are closed.
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– K.Power
4 hours ago
add a comment |
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how can we write x=y+z ? E is not an euclidien space to talk about the orthogonal of a subspace !
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– user515918
5 hours ago
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Aren't you assuming that the OP knows that finite dimensional subspaces are complemented? This implies they already know that they are closed.
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– K.Power
4 hours ago
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how can we write x=y+z ? E is not an euclidien space to talk about the orthogonal of a subspace !
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– user515918
5 hours ago
$begingroup$
how can we write x=y+z ? E is not an euclidien space to talk about the orthogonal of a subspace !
$endgroup$
– user515918
5 hours ago
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Aren't you assuming that the OP knows that finite dimensional subspaces are complemented? This implies they already know that they are closed.
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– K.Power
4 hours ago
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Aren't you assuming that the OP knows that finite dimensional subspaces are complemented? This implies they already know that they are closed.
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– K.Power
4 hours ago
add a comment |
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I am voting to close, because it is not at all clear to me what you are asking. Are you asking for the intuition behind closed sets, or are you asking how to show that a finite dimensional subspace is closed?
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– K.Power
5 hours ago
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@K.Power actually I am looking for both
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– user515918
5 hours ago
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If it is the latter then you will find such a question has been asked before on this site. I should also point out that you need to show that any convergent sequence in $F$ has its limit in $F$.
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– K.Power
5 hours ago
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@K.Power I've been searching but I dont have any idea about the methode they using
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– user515918
5 hours ago
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As these are very distinct questions you should definitely ask these in two separate posts.
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– K.Power
5 hours ago