Mixing
If $(N,|.|)$ is a non trivial vector space with norm, then the diameter of the unit sphere is $ge2$
$begingroup$
It is easy to see that $delta(B_1(0))le 2$ because $$|x-y|le|x|+|-y|=|x|+|-1||y|=|x|+|y|<1+1=2$$
so $sup{|x-y|:x,yin B_1(0)}le 2$
But to show that $delta(B_1(0))ge 2$ we chose $ain B_1(0)backslash{0}$ and $sin[0,1)$ and observe that $|sfrac{a}{|a|}|=s$ $implies sfrac{a}{|a|},-sfrac{a}{|a|}in B_1(0)$ and $|sfrac{a}{|a|}-(-sfrac{a}{|a|})|=2s$
But why does that imply that $delta(B_1(0))ge2s$?
This is how I think I understand it: the diameter of $B_1(0)$ is greater than that of $B_s(0)$ for all $sin[0,1)$ because we it contains it. And the diameter of $B_s(0)$ is $2s$ for each $s$, which is true because there exists a sequence $(epsilon_nfrac{a}{|a|},-epsilon_nfrac{a}{|a|})$ of pairs of points, with $0leepsilon_n<s~forall n$ and $limlimits_{nrightarrowinfty}epsilon_n=s$ such that it is always contained in $B_s(0)$ and $limlimits_{nrightarrowinfty}d(epsilon_nfrac{a}{|a|},-epsilon_nfrac{a}{|a|})=2s$
I'm not sure if I got it right...
sequences-and-series vector-spaces metric-spaces
asked Jan 17 at 15:39
John CataldoJohn Cataldo
1,1881316
$endgroup$
add a comment |
$begingroup$
It is easy to see that $delta(B_1(0))le 2$ because $$|x-y|le|x|+|-y|=|x|+|-1||y|=|x|+|y|<1+1=2$$
so $sup{|x-y|:x,yin B_1(0)}le 2$
But to show that $delta(B_1(0))ge 2$ we chose $ain B_1(0)backslash{0}$ and $sin[0,1)$ and observe that $|sfrac{a}{|a|}|=s$ $implies sfrac{a}{|a|},-sfrac{a}{|a|}in B_1(0)$ and $|sfrac{a}{|a|}-(-sfrac{a}{|a|})|=2s$
But why does that imply that $delta(B_1(0))ge2s$?
This is how I think I understand it: the diameter of $B_1(0)$ is greater than that of $B_s(0)$ for all $sin[0,1)$ because we it contains it. And the diameter of $B_s(0)$ is $2s$ for each $s$, which is true because there exists a sequence $(epsilon_nfrac{a}{|a|},-epsilon_nfrac{a}{|a|})$ of pairs of points, with $0leepsilon_n<s~forall n$ and $limlimits_{nrightarrowinfty}epsilon_n=s$ such that it is always contained in $B_s(0)$ and $limlimits_{nrightarrowinfty}d(epsilon_nfrac{a}{|a|},-epsilon_nfrac{a}{|a|})=2s$
I'm not sure if I got it right...
sequences-and-series vector-spaces metric-spaces
asked Jan 17 at 15:39
John CataldoJohn Cataldo
1,1881316
$endgroup$
$begingroup$
"But why does that imply that $delta(B_1(0))geq 2s$?" - because it has two things in it that are $2s$ apart. That's it.
$endgroup$
– user3482749
Jan 17 at 15:41
$begingroup$
Well if $d(x,y)=2$ then that would imply $|x|=|y=-x|=1not<1implies xnotin B_1(0)$, but I'm not sure how to prove that $|pmfrac{a}{|a|}|=1$ so $pmfrac{a}{|a|}notin B_1(0)$
$endgroup$
– John Cataldo
Jan 17 at 15:43
$begingroup$
I am sorry, I realize you are speaking of the open ball. In that case, my argument was wrong. However, you are right : for every $s in [0,1)$, we can find two elements in $B_1(0)$ which have distance $2s$ from each other. This shows that the diameter of $B_1(0)$ is at least $2s$ for every $s in [0,1)$ , and hence at least $2$. Note that no two points inside $B_1(0)$ have distance two between each other, but then the diameter does not require this to occur : if the diameter of a set $S$ is $d$, then it means that there are two points in $S$ separated by $d-epsilon$ for all $epsilon > 0$,
$endgroup$
– астон вілла олоф мэллбэрг
Jan 17 at 15:55
add a comment |
$begingroup$
It is easy to see that $delta(B_1(0))le 2$ because $$|x-y|le|x|+|-y|=|x|+|-1||y|=|x|+|y|<1+1=2$$
so $sup{|x-y|:x,yin B_1(0)}le 2$
But to show that $delta(B_1(0))ge 2$ we chose $ain B_1(0)backslash{0}$ and $sin[0,1)$ and observe that $|sfrac{a}{|a|}|=s$ $implies sfrac{a}{|a|},-sfrac{a}{|a|}in B_1(0)$ and $|sfrac{a}{|a|}-(-sfrac{a}{|a|})|=2s$
But why does that imply that $delta(B_1(0))ge2s$?
This is how I think I understand it: the diameter of $B_1(0)$ is greater than that of $B_s(0)$ for all $sin[0,1)$ because we it contains it. And the diameter of $B_s(0)$ is $2s$ for each $s$, which is true because there exists a sequence $(epsilon_nfrac{a}{|a|},-epsilon_nfrac{a}{|a|})$ of pairs of points, with $0leepsilon_n<s~forall n$ and $limlimits_{nrightarrowinfty}epsilon_n=s$ such that it is always contained in $B_s(0)$ and $limlimits_{nrightarrowinfty}d(epsilon_nfrac{a}{|a|},-epsilon_nfrac{a}{|a|})=2s$
I'm not sure if I got it right...
sequences-and-series vector-spaces metric-spaces
asked Jan 17 at 15:39
John CataldoJohn Cataldo
1,1881316
$endgroup$
It is easy to see that $delta(B_1(0))le 2$ because $$|x-y|le|x|+|-y|=|x|+|-1||y|=|x|+|y|<1+1=2$$
so $sup{|x-y|:x,yin B_1(0)}le 2$
But to show that $delta(B_1(0))ge 2$ we chose $ain B_1(0)backslash{0}$ and $sin[0,1)$ and observe that $|sfrac{a}{|a|}|=s$ $implies sfrac{a}{|a|},-sfrac{a}{|a|}in B_1(0)$ and $|sfrac{a}{|a|}-(-sfrac{a}{|a|})|=2s$
But why does that imply that $delta(B_1(0))ge2s$?
This is how I think I understand it: the diameter of $B_1(0)$ is greater than that of $B_s(0)$ for all $sin[0,1)$ because we it contains it. And the diameter of $B_s(0)$ is $2s$ for each $s$, which is true because there exists a sequence $(epsilon_nfrac{a}{|a|},-epsilon_nfrac{a}{|a|})$ of pairs of points, with $0leepsilon_n<s~forall n$ and $limlimits_{nrightarrowinfty}epsilon_n=s$ such that it is always contained in $B_s(0)$ and $limlimits_{nrightarrowinfty}d(epsilon_nfrac{a}{|a|},-epsilon_nfrac{a}{|a|})=2s$
I'm not sure if I got it right...
sequences-and-series vector-spaces metric-spaces
sequences-and-series vector-spaces metric-spaces
asked Jan 17 at 15:39
John CataldoJohn Cataldo
1,1881316
asked Jan 17 at 15:39
John CataldoJohn Cataldo
1,1881316
asked Jan 17 at 15:39
John CataldoJohn Cataldo
1,1881316
asked Jan 17 at 15:39
John CataldoJohn Cataldo
1,1881316
asked Jan 17 at 15:39
John CataldoJohn Cataldo
1,1881316
1,1881316
$begingroup$
"But why does that imply that $delta(B_1(0))geq 2s$?" - because it has two things in it that are $2s$ apart. That's it.
$endgroup$
– user3482749
Jan 17 at 15:41
$begingroup$
Well if $d(x,y)=2$ then that would imply $|x|=|y=-x|=1not<1implies xnotin B_1(0)$, but I'm not sure how to prove that $|pmfrac{a}{|a|}|=1$ so $pmfrac{a}{|a|}notin B_1(0)$
$endgroup$
– John Cataldo
Jan 17 at 15:43
$begingroup$
I am sorry, I realize you are speaking of the open ball. In that case, my argument was wrong. However, you are right : for every $s in [0,1)$, we can find two elements in $B_1(0)$ which have distance $2s$ from each other. This shows that the diameter of $B_1(0)$ is at least $2s$ for every $s in [0,1)$ , and hence at least $2$. Note that no two points inside $B_1(0)$ have distance two between each other, but then the diameter does not require this to occur : if the diameter of a set $S$ is $d$, then it means that there are two points in $S$ separated by $d-epsilon$ for all $epsilon > 0$,
$endgroup$
– астон вілла олоф мэллбэрг
Jan 17 at 15:55
add a comment |
$begingroup$
"But why does that imply that $delta(B_1(0))geq 2s$?" - because it has two things in it that are $2s$ apart. That's it.
$endgroup$
– user3482749
Jan 17 at 15:41
$begingroup$
Well if $d(x,y)=2$ then that would imply $|x|=|y=-x|=1not<1implies xnotin B_1(0)$, but I'm not sure how to prove that $|pmfrac{a}{|a|}|=1$ so $pmfrac{a}{|a|}notin B_1(0)$
$endgroup$
– John Cataldo
Jan 17 at 15:43
$begingroup$
I am sorry, I realize you are speaking of the open ball. In that case, my argument was wrong. However, you are right : for every $s in [0,1)$, we can find two elements in $B_1(0)$ which have distance $2s$ from each other. This shows that the diameter of $B_1(0)$ is at least $2s$ for every $s in [0,1)$ , and hence at least $2$. Note that no two points inside $B_1(0)$ have distance two between each other, but then the diameter does not require this to occur : if the diameter of a set $S$ is $d$, then it means that there are two points in $S$ separated by $d-epsilon$ for all $epsilon > 0$,
$endgroup$
– астон вілла олоф мэллбэрг
Jan 17 at 15:55
$begingroup$
"But why does that imply that $delta(B_1(0))geq 2s$?" - because it has two things in it that are $2s$ apart. That's it.
$endgroup$
– user3482749
Jan 17 at 15:41
$begingroup$
"But why does that imply that $delta(B_1(0))geq 2s$?" - because it has two things in it that are $2s$ apart. That's it.
$endgroup$
– user3482749
Jan 17 at 15:41
$begingroup$
Well if $d(x,y)=2$ then that would imply $|x|=|y=-x|=1not<1implies xnotin B_1(0)$, but I'm not sure how to prove that $|pmfrac{a}{|a|}|=1$ so $pmfrac{a}{|a|}notin B_1(0)$
$endgroup$
– John Cataldo
Jan 17 at 15:43
$begingroup$
Well if $d(x,y)=2$ then that would imply $|x|=|y=-x|=1not<1implies xnotin B_1(0)$, but I'm not sure how to prove that $|pmfrac{a}{|a|}|=1$ so $pmfrac{a}{|a|}notin B_1(0)$
$endgroup$
– John Cataldo
Jan 17 at 15:43
$begingroup$
I am sorry, I realize you are speaking of the open ball. In that case, my argument was wrong. However, you are right : for every $s in [0,1)$, we can find two elements in $B_1(0)$ which have distance $2s$ from each other. This shows that the diameter of $B_1(0)$ is at least $2s$ for every $s in [0,1)$ , and hence at least $2$. Note that no two points inside $B_1(0)$ have distance two between each other, but then the diameter does not require this to occur : if the diameter of a set $S$ is $d$, then it means that there are two points in $S$ separated by $d-epsilon$ for all $epsilon > 0$,
$endgroup$
– астон вілла олоф мэллбэрг
Jan 17 at 15:55
$begingroup$
I am sorry, I realize you are speaking of the open ball. In that case, my argument was wrong. However, you are right : for every $s in [0,1)$, we can find two elements in $B_1(0)$ which have distance $2s$ from each other. This shows that the diameter of $B_1(0)$ is at least $2s$ for every $s in [0,1)$ , and hence at least $2$. Note that no two points inside $B_1(0)$ have distance two between each other, but then the diameter does not require this to occur : if the diameter of a set $S$ is $d$, then it means that there are two points in $S$ separated by $d-epsilon$ for all $epsilon > 0$,
$endgroup$
– астон вілла олоф мэллбэрг
Jan 17 at 15:55
add a comment |
1 Answer
1
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oldest
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$begingroup$
You suppose that you can choose $sin [0,1)$ as near to 1 as you want, but it isn't generally true in a $mathbb{K}$-vector space.
Consider for example a $mathbb{Z}_{p}$-vector space, than $B_{1}(0)={0}$.
That's why in general the statement is false.
Anyway, i think your proof works in every $mathbb{Q}$, $mathbb{R}$ or $mathbb{C}$-vector space
answered Jan 17 at 22:17
ecrinecrin
3497
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
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oldest
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$begingroup$
You suppose that you can choose $sin [0,1)$ as near to 1 as you want, but it isn't generally true in a $mathbb{K}$-vector space.
Consider for example a $mathbb{Z}_{p}$-vector space, than $B_{1}(0)={0}$.
That's why in general the statement is false.
Anyway, i think your proof works in every $mathbb{Q}$, $mathbb{R}$ or $mathbb{C}$-vector space
answered Jan 17 at 22:17
ecrinecrin
3497
$endgroup$
add a comment |
$begingroup$
You suppose that you can choose $sin [0,1)$ as near to 1 as you want, but it isn't generally true in a $mathbb{K}$-vector space.
Consider for example a $mathbb{Z}_{p}$-vector space, than $B_{1}(0)={0}$.
That's why in general the statement is false.
Anyway, i think your proof works in every $mathbb{Q}$, $mathbb{R}$ or $mathbb{C}$-vector space
answered Jan 17 at 22:17
ecrinecrin
3497
$endgroup$
add a comment |
$begingroup$
You suppose that you can choose $sin [0,1)$ as near to 1 as you want, but it isn't generally true in a $mathbb{K}$-vector space.
Consider for example a $mathbb{Z}_{p}$-vector space, than $B_{1}(0)={0}$.
That's why in general the statement is false.
Anyway, i think your proof works in every $mathbb{Q}$, $mathbb{R}$ or $mathbb{C}$-vector space
answered Jan 17 at 22:17
ecrinecrin
3497
$endgroup$
You suppose that you can choose $sin [0,1)$ as near to 1 as you want, but it isn't generally true in a $mathbb{K}$-vector space.
Consider for example a $mathbb{Z}_{p}$-vector space, than $B_{1}(0)={0}$.
That's why in general the statement is false.
Anyway, i think your proof works in every $mathbb{Q}$, $mathbb{R}$ or $mathbb{C}$-vector space
answered Jan 17 at 22:17
ecrinecrin
3497
answered Jan 17 at 22:17
ecrinecrin
3497
answered Jan 17 at 22:17
ecrinecrin
3497
answered Jan 17 at 22:17
ecrinecrin
3497
3497
add a comment |
add a comment |
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$begingroup$
"But why does that imply that $delta(B_1(0))geq 2s$?" - because it has two things in it that are $2s$ apart. That's it.
$endgroup$
– user3482749
Jan 17 at 15:41
$begingroup$
Well if $d(x,y)=2$ then that would imply $|x|=|y=-x|=1not<1implies xnotin B_1(0)$, but I'm not sure how to prove that $|pmfrac{a}{|a|}|=1$ so $pmfrac{a}{|a|}notin B_1(0)$
$endgroup$
– John Cataldo
Jan 17 at 15:43
$begingroup$
I am sorry, I realize you are speaking of the open ball. In that case, my argument was wrong. However, you are right : for every $s in [0,1)$, we can find two elements in $B_1(0)$ which have distance $2s$ from each other. This shows that the diameter of $B_1(0)$ is at least $2s$ for every $s in [0,1)$ , and hence at least $2$. Note that no two points inside $B_1(0)$ have distance two between each other, but then the diameter does not require this to occur : if the diameter of a set $S$ is $d$, then it means that there are two points in $S$ separated by $d-epsilon$ for all $epsilon > 0$,
$endgroup$
– астон вілла олоф мэллбэрг
Jan 17 at 15:55