Closure of finite span has non-empty interior
$begingroup$
Be $X$ a Banach space with countable basis. Suppose $(x_n)_{n in mathbb{N}} $ is a sequence in $X$ (allowing repetitions), such that
$$(forall x in X)(exists M_x subseteq mathbb{N})left(|M_x| in mathbb{N} ,,wedge left(forall m in M_x)(exists lambda_m in mathbb{F} right) left(x=sum_{m in M_x} lambda_m x_m right)right)
$$
For each $k in mathbb{N}$, define $L_k :=$ sp${x_1, x_2, ldots, x_k}$.
It is supposed to be the case that there must exists at least one $k_0 in mathbb{N}$ such that the closure of $L_{k_0}$ has non-empty interior: $left(bar{L_{k_0}}right)^{circ} neq emptyset$.
I find the question oddly phrased, since each $L_k$ is a finite-dimensional subspace, and is therefore closed, so why speak of its closure? Is there something about the interior of the closure of a set that's useful here?
Having found such a $k_0$ then, how would this imply that $X = L_{k_0}$?
Suggestion
Let $x in X setminus {0}$. Note that since $L_{k_0}$ contains an open ball, say, $B_{varepsilon}(a)$, $z := frac{varepsilon}{2}frac{x}{lVert x rVert} + a in B_{varepsilon}(a) subseteq L_{k_0}$, so $z in L_{k_0}$. But then
$x = frac{2lVert xrVert}{varepsilon}(z-a) in L_{k_0} oplus text{sp}(a)$. This basically brings us back to what we want to prove, since we now need to have that $a in L_{k_0}$...
Is there a way to show that not only $L_{k_0}$ contains some open ball, but an open ball around 0?
Oops, of course $a in L_{k_0}$, since $B_{varepsilon}(a) subseteq L_{k_0}$!
functional-analysis metric-spaces banach-spaces
asked Jan 21 at 17:06
Jos van NieuwmanJos van Nieuwman
519
$endgroup$
add a comment |
$begingroup$
Be $X$ a Banach space with countable basis. Suppose $(x_n)_{n in mathbb{N}} $ is a sequence in $X$ (allowing repetitions), such that
$$(forall x in X)(exists M_x subseteq mathbb{N})left(|M_x| in mathbb{N} ,,wedge left(forall m in M_x)(exists lambda_m in mathbb{F} right) left(x=sum_{m in M_x} lambda_m x_m right)right)
$$
For each $k in mathbb{N}$, define $L_k :=$ sp${x_1, x_2, ldots, x_k}$.
It is supposed to be the case that there must exists at least one $k_0 in mathbb{N}$ such that the closure of $L_{k_0}$ has non-empty interior: $left(bar{L_{k_0}}right)^{circ} neq emptyset$.
I find the question oddly phrased, since each $L_k$ is a finite-dimensional subspace, and is therefore closed, so why speak of its closure? Is there something about the interior of the closure of a set that's useful here?
Having found such a $k_0$ then, how would this imply that $X = L_{k_0}$?
Suggestion
Let $x in X setminus {0}$. Note that since $L_{k_0}$ contains an open ball, say, $B_{varepsilon}(a)$, $z := frac{varepsilon}{2}frac{x}{lVert x rVert} + a in B_{varepsilon}(a) subseteq L_{k_0}$, so $z in L_{k_0}$. But then
$x = frac{2lVert xrVert}{varepsilon}(z-a) in L_{k_0} oplus text{sp}(a)$. This basically brings us back to what we want to prove, since we now need to have that $a in L_{k_0}$...
Is there a way to show that not only $L_{k_0}$ contains some open ball, but an open ball around 0?
Oops, of course $a in L_{k_0}$, since $B_{varepsilon}(a) subseteq L_{k_0}$!
functional-analysis metric-spaces banach-spaces
asked Jan 21 at 17:06
Jos van NieuwmanJos van Nieuwman
519
$endgroup$
3
$begingroup$
The closure is indeed redundant, but only because the $L_k$ are finite-dimensional. You do need closed sets for the Baire Category Theorem.
$endgroup$
– Mindlack
Jan 21 at 17:09
$begingroup$
Aaah yes! I would never have thought of that, thanks!
$endgroup$
– Jos van Nieuwman
Jan 21 at 18:08
add a comment |
$begingroup$
Be $X$ a Banach space with countable basis. Suppose $(x_n)_{n in mathbb{N}} $ is a sequence in $X$ (allowing repetitions), such that
$$(forall x in X)(exists M_x subseteq mathbb{N})left(|M_x| in mathbb{N} ,,wedge left(forall m in M_x)(exists lambda_m in mathbb{F} right) left(x=sum_{m in M_x} lambda_m x_m right)right)
$$
For each $k in mathbb{N}$, define $L_k :=$ sp${x_1, x_2, ldots, x_k}$.
It is supposed to be the case that there must exists at least one $k_0 in mathbb{N}$ such that the closure of $L_{k_0}$ has non-empty interior: $left(bar{L_{k_0}}right)^{circ} neq emptyset$.
I find the question oddly phrased, since each $L_k$ is a finite-dimensional subspace, and is therefore closed, so why speak of its closure? Is there something about the interior of the closure of a set that's useful here?
Having found such a $k_0$ then, how would this imply that $X = L_{k_0}$?
Suggestion
Let $x in X setminus {0}$. Note that since $L_{k_0}$ contains an open ball, say, $B_{varepsilon}(a)$, $z := frac{varepsilon}{2}frac{x}{lVert x rVert} + a in B_{varepsilon}(a) subseteq L_{k_0}$, so $z in L_{k_0}$. But then
$x = frac{2lVert xrVert}{varepsilon}(z-a) in L_{k_0} oplus text{sp}(a)$. This basically brings us back to what we want to prove, since we now need to have that $a in L_{k_0}$...
Is there a way to show that not only $L_{k_0}$ contains some open ball, but an open ball around 0?
Oops, of course $a in L_{k_0}$, since $B_{varepsilon}(a) subseteq L_{k_0}$!
functional-analysis metric-spaces banach-spaces
asked Jan 21 at 17:06
Jos van NieuwmanJos van Nieuwman
519
$endgroup$
Be $X$ a Banach space with countable basis. Suppose $(x_n)_{n in mathbb{N}} $ is a sequence in $X$ (allowing repetitions), such that
$$(forall x in X)(exists M_x subseteq mathbb{N})left(|M_x| in mathbb{N} ,,wedge left(forall m in M_x)(exists lambda_m in mathbb{F} right) left(x=sum_{m in M_x} lambda_m x_m right)right)
$$
For each $k in mathbb{N}$, define $L_k :=$ sp${x_1, x_2, ldots, x_k}$.
It is supposed to be the case that there must exists at least one $k_0 in mathbb{N}$ such that the closure of $L_{k_0}$ has non-empty interior: $left(bar{L_{k_0}}right)^{circ} neq emptyset$.
I find the question oddly phrased, since each $L_k$ is a finite-dimensional subspace, and is therefore closed, so why speak of its closure? Is there something about the interior of the closure of a set that's useful here?
Having found such a $k_0$ then, how would this imply that $X = L_{k_0}$?
Suggestion
Let $x in X setminus {0}$. Note that since $L_{k_0}$ contains an open ball, say, $B_{varepsilon}(a)$, $z := frac{varepsilon}{2}frac{x}{lVert x rVert} + a in B_{varepsilon}(a) subseteq L_{k_0}$, so $z in L_{k_0}$. But then
$x = frac{2lVert xrVert}{varepsilon}(z-a) in L_{k_0} oplus text{sp}(a)$. This basically brings us back to what we want to prove, since we now need to have that $a in L_{k_0}$...
Is there a way to show that not only $L_{k_0}$ contains some open ball, but an open ball around 0?
Oops, of course $a in L_{k_0}$, since $B_{varepsilon}(a) subseteq L_{k_0}$!
functional-analysis metric-spaces banach-spaces
functional-analysis metric-spaces banach-spaces
asked Jan 21 at 17:06
Jos van NieuwmanJos van Nieuwman
519
asked Jan 21 at 17:06
Jos van NieuwmanJos van Nieuwman
519
edited Jan 21 at 19:47
Jos van Nieuwman
asked Jan 21 at 17:06
Jos van NieuwmanJos van Nieuwman
519
asked Jan 21 at 17:06
Jos van NieuwmanJos van Nieuwman
519
asked Jan 21 at 17:06
Jos van NieuwmanJos van Nieuwman
519
519
3
$begingroup$
The closure is indeed redundant, but only because the $L_k$ are finite-dimensional. You do need closed sets for the Baire Category Theorem.
$endgroup$
– Mindlack
Jan 21 at 17:09
$begingroup$
Aaah yes! I would never have thought of that, thanks!
$endgroup$
– Jos van Nieuwman
Jan 21 at 18:08
add a comment |
3
$begingroup$
The closure is indeed redundant, but only because the $L_k$ are finite-dimensional. You do need closed sets for the Baire Category Theorem.
$endgroup$
– Mindlack
Jan 21 at 17:09
$begingroup$
Aaah yes! I would never have thought of that, thanks!
$endgroup$
– Jos van Nieuwman
Jan 21 at 18:08
3
3
$begingroup$
The closure is indeed redundant, but only because the $L_k$ are finite-dimensional. You do need closed sets for the Baire Category Theorem.
$endgroup$
– Mindlack
Jan 21 at 17:09
$begingroup$
The closure is indeed redundant, but only because the $L_k$ are finite-dimensional. You do need closed sets for the Baire Category Theorem.
$endgroup$
– Mindlack
Jan 21 at 17:09
$begingroup$
Aaah yes! I would never have thought of that, thanks!
$endgroup$
– Jos van Nieuwman
Jan 21 at 18:08
$begingroup$
Aaah yes! I would never have thought of that, thanks!
$endgroup$
– Jos van Nieuwman
Jan 21 at 18:08
add a comment |
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$begingroup$
The closure is indeed redundant, but only because the $L_k$ are finite-dimensional. You do need closed sets for the Baire Category Theorem.
$endgroup$
– Mindlack
Jan 21 at 17:09
$begingroup$
Aaah yes! I would never have thought of that, thanks!
$endgroup$
– Jos van Nieuwman
Jan 21 at 18:08