Mixing
On the closure of the convex hull of a sequence in normed spaces
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Let $E$ be a infinite dimensional normed spaces and $(x_n)_{n=1}^infty$ be a sequence in $E$ converging to zero. Is it true that the closure of the convex hull of the set ${x_n: nin mathbb{N}}cup{0}$ in $E$ is equal to the closure of the convex hull of the set ${x_n: nin mathbb{N}}$ in $E$, that is, is it true that $overline{co({x_n: nin mathbb{N}}cup{0})}=overline{co({x_n: nin mathbb{N}})}$ ? (Here we assume that for all $nin mathbb{N}$, $x_nneq 0$.)
However, in may books, instead of writing $overline{co({x_n: nin mathbb{N}})}$, I see that it is written $overline{co({x_n: nin mathbb{N}}cup{0})}$. But, I think that this equality should be true, I mean, these two closures should be the same, which can be seen when we consider the well-known description of the convex hull of a subset $Asubset E$, namely, $co(A)={sum_{n=1}^Nlambda_nx_n: x_nin A, lambda_ngeq0, sum_{n=1}^Nlambda_n=1, Ninmathbb{N}}$.
Am I right?
Thanks for any comment/answer.
functional-analysis normed-spaces
asked 6 hours ago
serenus
1786
add a comment |
up vote
0
down vote
favorite
Let $E$ be a infinite dimensional normed spaces and $(x_n)_{n=1}^infty$ be a sequence in $E$ converging to zero. Is it true that the closure of the convex hull of the set ${x_n: nin mathbb{N}}cup{0}$ in $E$ is equal to the closure of the convex hull of the set ${x_n: nin mathbb{N}}$ in $E$, that is, is it true that $overline{co({x_n: nin mathbb{N}}cup{0})}=overline{co({x_n: nin mathbb{N}})}$ ? (Here we assume that for all $nin mathbb{N}$, $x_nneq 0$.)
However, in may books, instead of writing $overline{co({x_n: nin mathbb{N}})}$, I see that it is written $overline{co({x_n: nin mathbb{N}}cup{0})}$. But, I think that this equality should be true, I mean, these two closures should be the same, which can be seen when we consider the well-known description of the convex hull of a subset $Asubset E$, namely, $co(A)={sum_{n=1}^Nlambda_nx_n: x_nin A, lambda_ngeq0, sum_{n=1}^Nlambda_n=1, Ninmathbb{N}}$.
Am I right?
Thanks for any comment/answer.
functional-analysis normed-spaces
asked 6 hours ago
serenus
1786
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $E$ be a infinite dimensional normed spaces and $(x_n)_{n=1}^infty$ be a sequence in $E$ converging to zero. Is it true that the closure of the convex hull of the set ${x_n: nin mathbb{N}}cup{0}$ in $E$ is equal to the closure of the convex hull of the set ${x_n: nin mathbb{N}}$ in $E$, that is, is it true that $overline{co({x_n: nin mathbb{N}}cup{0})}=overline{co({x_n: nin mathbb{N}})}$ ? (Here we assume that for all $nin mathbb{N}$, $x_nneq 0$.)
However, in may books, instead of writing $overline{co({x_n: nin mathbb{N}})}$, I see that it is written $overline{co({x_n: nin mathbb{N}}cup{0})}$. But, I think that this equality should be true, I mean, these two closures should be the same, which can be seen when we consider the well-known description of the convex hull of a subset $Asubset E$, namely, $co(A)={sum_{n=1}^Nlambda_nx_n: x_nin A, lambda_ngeq0, sum_{n=1}^Nlambda_n=1, Ninmathbb{N}}$.
Am I right?
Thanks for any comment/answer.
functional-analysis normed-spaces
asked 6 hours ago
serenus
1786
Let $E$ be a infinite dimensional normed spaces and $(x_n)_{n=1}^infty$ be a sequence in $E$ converging to zero. Is it true that the closure of the convex hull of the set ${x_n: nin mathbb{N}}cup{0}$ in $E$ is equal to the closure of the convex hull of the set ${x_n: nin mathbb{N}}$ in $E$, that is, is it true that $overline{co({x_n: nin mathbb{N}}cup{0})}=overline{co({x_n: nin mathbb{N}})}$ ? (Here we assume that for all $nin mathbb{N}$, $x_nneq 0$.)
However, in may books, instead of writing $overline{co({x_n: nin mathbb{N}})}$, I see that it is written $overline{co({x_n: nin mathbb{N}}cup{0})}$. But, I think that this equality should be true, I mean, these two closures should be the same, which can be seen when we consider the well-known description of the convex hull of a subset $Asubset E$, namely, $co(A)={sum_{n=1}^Nlambda_nx_n: x_nin A, lambda_ngeq0, sum_{n=1}^Nlambda_n=1, Ninmathbb{N}}$.
Am I right?
Thanks for any comment/answer.
functional-analysis normed-spaces
functional-analysis normed-spaces
asked 6 hours ago
serenus
1786
asked 6 hours ago
serenus
1786
asked 6 hours ago
serenus
1786
asked 6 hours ago
serenus
1786
asked 6 hours ago
serenus
1786
1786
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
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Just verify that each side is contained in the other. Obviously, RHS is contained in LHS. For the other way it is enough to show that ${x_n:n in mathbb N} cup {0} $ is contained in RHS (because RHS is closed and convex). This is true because RHS contains the closure of ${x_n:n in mathbb N}$.
answered 6 hours ago
Kavi Rama Murthy
39.9k31750
Kavi Rama Murthy: I agree with you. But, since I saw the LHS in many books, I thought that "Am I missing something ?". So, the equality is indeed the case. Thanks for the explanation.
– serenus
5 hours ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Just verify that each side is contained in the other. Obviously, RHS is contained in LHS. For the other way it is enough to show that ${x_n:n in mathbb N} cup {0} $ is contained in RHS (because RHS is closed and convex). This is true because RHS contains the closure of ${x_n:n in mathbb N}$.
answered 6 hours ago
Kavi Rama Murthy
39.9k31750
Kavi Rama Murthy: I agree with you. But, since I saw the LHS in many books, I thought that "Am I missing something ?". So, the equality is indeed the case. Thanks for the explanation.
– serenus
5 hours ago
add a comment |
up vote
0
down vote
Just verify that each side is contained in the other. Obviously, RHS is contained in LHS. For the other way it is enough to show that ${x_n:n in mathbb N} cup {0} $ is contained in RHS (because RHS is closed and convex). This is true because RHS contains the closure of ${x_n:n in mathbb N}$.
answered 6 hours ago
Kavi Rama Murthy
39.9k31750
Kavi Rama Murthy: I agree with you. But, since I saw the LHS in many books, I thought that "Am I missing something ?". So, the equality is indeed the case. Thanks for the explanation.
– serenus
5 hours ago
add a comment |
up vote
0
down vote
up vote
0
down vote
Just verify that each side is contained in the other. Obviously, RHS is contained in LHS. For the other way it is enough to show that ${x_n:n in mathbb N} cup {0} $ is contained in RHS (because RHS is closed and convex). This is true because RHS contains the closure of ${x_n:n in mathbb N}$.
answered 6 hours ago
Kavi Rama Murthy
39.9k31750
Just verify that each side is contained in the other. Obviously, RHS is contained in LHS. For the other way it is enough to show that ${x_n:n in mathbb N} cup {0} $ is contained in RHS (because RHS is closed and convex). This is true because RHS contains the closure of ${x_n:n in mathbb N}$.
answered 6 hours ago
Kavi Rama Murthy
39.9k31750
answered 6 hours ago
Kavi Rama Murthy
39.9k31750
answered 6 hours ago
Kavi Rama Murthy
39.9k31750
answered 6 hours ago
Kavi Rama Murthy
39.9k31750
39.9k31750
Kavi Rama Murthy: I agree with you. But, since I saw the LHS in many books, I thought that "Am I missing something ?". So, the equality is indeed the case. Thanks for the explanation.
– serenus
5 hours ago
add a comment |
Kavi Rama Murthy: I agree with you. But, since I saw the LHS in many books, I thought that "Am I missing something ?". So, the equality is indeed the case. Thanks for the explanation.
– serenus
5 hours ago
Kavi Rama Murthy: I agree with you. But, since I saw the LHS in many books, I thought that "Am I missing something ?". So, the equality is indeed the case. Thanks for the explanation.
– serenus
5 hours ago
Kavi Rama Murthy: I agree with you. But, since I saw the LHS in many books, I thought that "Am I missing something ?". So, the equality is indeed the case. Thanks for the explanation.
– serenus
5 hours ago
add a comment |
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