Mixing
Closure of ball in normed space
$begingroup$
I found this question in a text book and wanted to ask for some help. I am not quite sure where to start with this.
Prove that in a normed space
$overline{B^o(x, r)} = B(x, r) forall x ∈ X text{ and } r > 0$.
Hint 1: You need to show that every point $y ∈ X$ such that $||y − x|| = r$ is a limit point of $B^o(x, r)$. To do so, use the sequence $y_n = x + (y − x)(1 − frac{1}{n})$ and show that $y_n ∈ B^o(x, r)$ $forall n text{ and }y_n → y$.
Hint 2: You also need to show that every point $y ∈ X$ such that $||y − x|| > r$ is not a limit point. To do so, use the observation that $Xsetminus B(x, r)$ is open.
I guess I can see how to start with Hint 1. $vert vert y_n-x vert vert=vert vert (y-x)frac{1}{n} vert vert to 0$ ... but then Im not sure
Hint 2 I cant approach as I do not fully understand Hint 1.
real-analysis limits metric-spaces normed-spaces
asked Jan 20 at 17:21
ʎpoqouʎpoqou
3561211
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add a comment |
$begingroup$
I found this question in a text book and wanted to ask for some help. I am not quite sure where to start with this.
Prove that in a normed space
$overline{B^o(x, r)} = B(x, r) forall x ∈ X text{ and } r > 0$.
Hint 1: You need to show that every point $y ∈ X$ such that $||y − x|| = r$ is a limit point of $B^o(x, r)$. To do so, use the sequence $y_n = x + (y − x)(1 − frac{1}{n})$ and show that $y_n ∈ B^o(x, r)$ $forall n text{ and }y_n → y$.
Hint 2: You also need to show that every point $y ∈ X$ such that $||y − x|| > r$ is not a limit point. To do so, use the observation that $Xsetminus B(x, r)$ is open.
I guess I can see how to start with Hint 1. $vert vert y_n-x vert vert=vert vert (y-x)frac{1}{n} vert vert to 0$ ... but then Im not sure
Hint 2 I cant approach as I do not fully understand Hint 1.
real-analysis limits metric-spaces normed-spaces
asked Jan 20 at 17:21
ʎpoqouʎpoqou
3561211
$endgroup$
3
$begingroup$
What is $B^{circ}(x,r) $? Is it interior of $B (x,r) $?
$endgroup$
– Thomas Shelby
Jan 20 at 17:24
add a comment |
$begingroup$
I found this question in a text book and wanted to ask for some help. I am not quite sure where to start with this.
Prove that in a normed space
$overline{B^o(x, r)} = B(x, r) forall x ∈ X text{ and } r > 0$.
Hint 1: You need to show that every point $y ∈ X$ such that $||y − x|| = r$ is a limit point of $B^o(x, r)$. To do so, use the sequence $y_n = x + (y − x)(1 − frac{1}{n})$ and show that $y_n ∈ B^o(x, r)$ $forall n text{ and }y_n → y$.
Hint 2: You also need to show that every point $y ∈ X$ such that $||y − x|| > r$ is not a limit point. To do so, use the observation that $Xsetminus B(x, r)$ is open.
I guess I can see how to start with Hint 1. $vert vert y_n-x vert vert=vert vert (y-x)frac{1}{n} vert vert to 0$ ... but then Im not sure
Hint 2 I cant approach as I do not fully understand Hint 1.
real-analysis limits metric-spaces normed-spaces
asked Jan 20 at 17:21
ʎpoqouʎpoqou
3561211
$endgroup$
I found this question in a text book and wanted to ask for some help. I am not quite sure where to start with this.
Prove that in a normed space
$overline{B^o(x, r)} = B(x, r) forall x ∈ X text{ and } r > 0$.
Hint 1: You need to show that every point $y ∈ X$ such that $||y − x|| = r$ is a limit point of $B^o(x, r)$. To do so, use the sequence $y_n = x + (y − x)(1 − frac{1}{n})$ and show that $y_n ∈ B^o(x, r)$ $forall n text{ and }y_n → y$.
Hint 2: You also need to show that every point $y ∈ X$ such that $||y − x|| > r$ is not a limit point. To do so, use the observation that $Xsetminus B(x, r)$ is open.
I guess I can see how to start with Hint 1. $vert vert y_n-x vert vert=vert vert (y-x)frac{1}{n} vert vert to 0$ ... but then Im not sure
Hint 2 I cant approach as I do not fully understand Hint 1.
real-analysis limits metric-spaces normed-spaces
real-analysis limits metric-spaces normed-spaces
asked Jan 20 at 17:21
ʎpoqouʎpoqou
3561211
asked Jan 20 at 17:21
ʎpoqouʎpoqou
3561211
asked Jan 20 at 17:21
ʎpoqouʎpoqou
3561211
asked Jan 20 at 17:21
ʎpoqouʎpoqou
3561211
asked Jan 20 at 17:21
ʎpoqouʎpoqou
3561211
3561211
3
$begingroup$
What is $B^{circ}(x,r) $? Is it interior of $B (x,r) $?
$endgroup$
– Thomas Shelby
Jan 20 at 17:24
add a comment |
3
$begingroup$
What is $B^{circ}(x,r) $? Is it interior of $B (x,r) $?
$endgroup$
– Thomas Shelby
Jan 20 at 17:24
3
3
$begingroup$
What is $B^{circ}(x,r) $? Is it interior of $B (x,r) $?
$endgroup$
– Thomas Shelby
Jan 20 at 17:24
$begingroup$
What is $B^{circ}(x,r) $? Is it interior of $B (x,r) $?
$endgroup$
– Thomas Shelby
Jan 20 at 17:24
add a comment |
1 Answer
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When using Hint 1, you made an error in "$|y_n-x|=|(y-x)frac{1}{n}|to0$". Note that
$$y_n-x=left[x +(y−x)left(1−frac{1}{n}right)right]-x=(y−x)left(1−frac{1}{n}right),$$
which is not what you said. Instead, from the correct expression above, we can find the norm of $y_n-x$. You should work it out and see that this norm is less than $r$, which proves the first part of Hint 1, that all $y_nin B^o(x,r)$.
For the second part of Hint 1, set up and simplify $y_n-y$. Here it is true that $y_n-y=(y-x)frac{1}{n}$, and so $|y_n-y|=|(y-x)frac{1}{n}|to0$, which is exactly what we want, as it shows that $y_nto y$, i.e. that $yinoverline{B^o(x,r)}$, as desired.
answered Jan 20 at 18:13
zipirovichzipirovich
11.3k11731
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add a comment |
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$begingroup$
When using Hint 1, you made an error in "$|y_n-x|=|(y-x)frac{1}{n}|to0$". Note that
$$y_n-x=left[x +(y−x)left(1−frac{1}{n}right)right]-x=(y−x)left(1−frac{1}{n}right),$$
which is not what you said. Instead, from the correct expression above, we can find the norm of $y_n-x$. You should work it out and see that this norm is less than $r$, which proves the first part of Hint 1, that all $y_nin B^o(x,r)$.
For the second part of Hint 1, set up and simplify $y_n-y$. Here it is true that $y_n-y=(y-x)frac{1}{n}$, and so $|y_n-y|=|(y-x)frac{1}{n}|to0$, which is exactly what we want, as it shows that $y_nto y$, i.e. that $yinoverline{B^o(x,r)}$, as desired.
answered Jan 20 at 18:13
zipirovichzipirovich
11.3k11731
$endgroup$
add a comment |
$begingroup$
When using Hint 1, you made an error in "$|y_n-x|=|(y-x)frac{1}{n}|to0$". Note that
$$y_n-x=left[x +(y−x)left(1−frac{1}{n}right)right]-x=(y−x)left(1−frac{1}{n}right),$$
which is not what you said. Instead, from the correct expression above, we can find the norm of $y_n-x$. You should work it out and see that this norm is less than $r$, which proves the first part of Hint 1, that all $y_nin B^o(x,r)$.
For the second part of Hint 1, set up and simplify $y_n-y$. Here it is true that $y_n-y=(y-x)frac{1}{n}$, and so $|y_n-y|=|(y-x)frac{1}{n}|to0$, which is exactly what we want, as it shows that $y_nto y$, i.e. that $yinoverline{B^o(x,r)}$, as desired.
answered Jan 20 at 18:13
zipirovichzipirovich
11.3k11731
$endgroup$
add a comment |
$begingroup$
When using Hint 1, you made an error in "$|y_n-x|=|(y-x)frac{1}{n}|to0$". Note that
$$y_n-x=left[x +(y−x)left(1−frac{1}{n}right)right]-x=(y−x)left(1−frac{1}{n}right),$$
which is not what you said. Instead, from the correct expression above, we can find the norm of $y_n-x$. You should work it out and see that this norm is less than $r$, which proves the first part of Hint 1, that all $y_nin B^o(x,r)$.
For the second part of Hint 1, set up and simplify $y_n-y$. Here it is true that $y_n-y=(y-x)frac{1}{n}$, and so $|y_n-y|=|(y-x)frac{1}{n}|to0$, which is exactly what we want, as it shows that $y_nto y$, i.e. that $yinoverline{B^o(x,r)}$, as desired.
answered Jan 20 at 18:13
zipirovichzipirovich
11.3k11731
$endgroup$
When using Hint 1, you made an error in "$|y_n-x|=|(y-x)frac{1}{n}|to0$". Note that
$$y_n-x=left[x +(y−x)left(1−frac{1}{n}right)right]-x=(y−x)left(1−frac{1}{n}right),$$
which is not what you said. Instead, from the correct expression above, we can find the norm of $y_n-x$. You should work it out and see that this norm is less than $r$, which proves the first part of Hint 1, that all $y_nin B^o(x,r)$.
For the second part of Hint 1, set up and simplify $y_n-y$. Here it is true that $y_n-y=(y-x)frac{1}{n}$, and so $|y_n-y|=|(y-x)frac{1}{n}|to0$, which is exactly what we want, as it shows that $y_nto y$, i.e. that $yinoverline{B^o(x,r)}$, as desired.
answered Jan 20 at 18:13
zipirovichzipirovich
11.3k11731
answered Jan 20 at 18:13
zipirovichzipirovich
11.3k11731
answered Jan 20 at 18:13
zipirovichzipirovich
11.3k11731
answered Jan 20 at 18:13
zipirovichzipirovich
11.3k11731
11.3k11731
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$begingroup$
What is $B^{circ}(x,r) $? Is it interior of $B (x,r) $?
$endgroup$
– Thomas Shelby
Jan 20 at 17:24