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does this function fail to be a metric because the triangle inequality does not hold?
Clash Royale CLAN TAG#URR8PPP
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given the function
$d:0,1^Bbb N times 0,1^Bbb N rightarrow Bbb R_0^+ $
$d(a,b):= beginBmatrix 0 &if&a=b\frac1mina_kneq b_k & if & ane b endBmatrix$
Now we are given in the definition of d that d(a,b)=0 iff a=b and we also know that $frac1mina_k neq b_k >0$. So condition one is satisfied.
next if a=b then d(a,b)=0=d(b,a)
and if $aneq b$ then $d(a,b)=frac1min a_k neq b_k=frac1min b_k neq a_k=d(b,a)$
But finally I don't think it satisfies the triangle inequality, because
$d(a,c)leq d(a,b)+d(b,c)$ implies that $frac1minkin Bbb N leqfrac1minqin Bbb N + frac1minpin Bbb N $
but if the minimum k was 1 , and the minimum p and q were say 4 then we would have $1 leq 0.5$ which is not true ....is this correct ?
metric-spaces
asked Aug 6 at 14:03
exodius
772315
add a comment |Â
up vote
2
down vote
favorite
given the function
$d:0,1^Bbb N times 0,1^Bbb N rightarrow Bbb R_0^+ $
$d(a,b):= beginBmatrix 0 &if&a=b\frac1mina_kneq b_k & if & ane b endBmatrix$
Now we are given in the definition of d that d(a,b)=0 iff a=b and we also know that $frac1mina_k neq b_k >0$. So condition one is satisfied.
next if a=b then d(a,b)=0=d(b,a)
and if $aneq b$ then $d(a,b)=frac1min a_k neq b_k=frac1min b_k neq a_k=d(b,a)$
But finally I don't think it satisfies the triangle inequality, because
$d(a,c)leq d(a,b)+d(b,c)$ implies that $frac1minkin Bbb N leqfrac1minqin Bbb N + frac1minpin Bbb N $
but if the minimum k was 1 , and the minimum p and q were say 4 then we would have $1 leq 0.5$ which is not true ....is this correct ?
metric-spaces
asked Aug 6 at 14:03
exodius
772315
The minimum $p$ and $q$ cannot both be 4 when the minimum $k$ is 1. Do you see why? If not, try to give explicit examples of $a, b, c$ to see where your idea goes wrong.
– Mees de Vries
Aug 6 at 14:09
@MeesdeVries we could say that a=n , b=1/n and c=1+n , then k =1 ,q= 2 and p=1. so then $d(a,c)leq d(a,b)+d(b,c)$ I'm confused however as to how to show this generally ?
– exodius
Aug 6 at 14:18
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
given the function
$d:0,1^Bbb N times 0,1^Bbb N rightarrow Bbb R_0^+ $
$d(a,b):= beginBmatrix 0 &if&a=b\frac1mina_kneq b_k & if & ane b endBmatrix$
Now we are given in the definition of d that d(a,b)=0 iff a=b and we also know that $frac1mina_k neq b_k >0$. So condition one is satisfied.
next if a=b then d(a,b)=0=d(b,a)
and if $aneq b$ then $d(a,b)=frac1min a_k neq b_k=frac1min b_k neq a_k=d(b,a)$
But finally I don't think it satisfies the triangle inequality, because
$d(a,c)leq d(a,b)+d(b,c)$ implies that $frac1minkin Bbb N leqfrac1minqin Bbb N + frac1minpin Bbb N $
but if the minimum k was 1 , and the minimum p and q were say 4 then we would have $1 leq 0.5$ which is not true ....is this correct ?
metric-spaces
asked Aug 6 at 14:03
exodius
772315
given the function
$d:0,1^Bbb N times 0,1^Bbb N rightarrow Bbb R_0^+ $
$d(a,b):= beginBmatrix 0 &if&a=b\frac1mina_kneq b_k & if & ane b endBmatrix$
Now we are given in the definition of d that d(a,b)=0 iff a=b and we also know that $frac1mina_k neq b_k >0$. So condition one is satisfied.
next if a=b then d(a,b)=0=d(b,a)
and if $aneq b$ then $d(a,b)=frac1min a_k neq b_k=frac1min b_k neq a_k=d(b,a)$
But finally I don't think it satisfies the triangle inequality, because
$d(a,c)leq d(a,b)+d(b,c)$ implies that $frac1minkin Bbb N leqfrac1minqin Bbb N + frac1minpin Bbb N $
but if the minimum k was 1 , and the minimum p and q were say 4 then we would have $1 leq 0.5$ which is not true ....is this correct ?
metric-spaces
asked Aug 6 at 14:03
exodius
772315
asked Aug 6 at 14:03
exodius
772315
asked Aug 6 at 14:03
exodius
772315
asked Aug 6 at 14:03
exodius
772315
772315
The minimum $p$ and $q$ cannot both be 4 when the minimum $k$ is 1. Do you see why? If not, try to give explicit examples of $a, b, c$ to see where your idea goes wrong.
– Mees de Vries
Aug 6 at 14:09
@MeesdeVries we could say that a=n , b=1/n and c=1+n , then k =1 ,q= 2 and p=1. so then $d(a,c)leq d(a,b)+d(b,c)$ I'm confused however as to how to show this generally ?
– exodius
Aug 6 at 14:18
add a comment |Â
The minimum $p$ and $q$ cannot both be 4 when the minimum $k$ is 1. Do you see why? If not, try to give explicit examples of $a, b, c$ to see where your idea goes wrong.
– Mees de Vries
Aug 6 at 14:09
@MeesdeVries we could say that a=n , b=1/n and c=1+n , then k =1 ,q= 2 and p=1. so then $d(a,c)leq d(a,b)+d(b,c)$ I'm confused however as to how to show this generally ?
– exodius
Aug 6 at 14:18
The minimum $p$ and $q$ cannot both be 4 when the minimum $k$ is 1. Do you see why? If not, try to give explicit examples of $a, b, c$ to see where your idea goes wrong.
– Mees de Vries
Aug 6 at 14:09
The minimum $p$ and $q$ cannot both be 4 when the minimum $k$ is 1. Do you see why? If not, try to give explicit examples of $a, b, c$ to see where your idea goes wrong.
– Mees de Vries
Aug 6 at 14:09
@MeesdeVries we could say that a=n , b=1/n and c=1+n , then k =1 ,q= 2 and p=1. so then $d(a,c)leq d(a,b)+d(b,c)$ I'm confused however as to how to show this generally ?
– exodius
Aug 6 at 14:18
@MeesdeVries we could say that a=n , b=1/n and c=1+n , then k =1 ,q= 2 and p=1. so then $d(a,c)leq d(a,b)+d(b,c)$ I'm confused however as to how to show this generally ?
– exodius
Aug 6 at 14:18
add a comment |Â
2 Answers
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3
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Let me prove the triangle inequality just in the special case that $a,b,c$ are all different from each other (the case where two of them are equal, or where all three are equal, is easy).
First let's evaluate the left hand side of the triangle inequality: let $K = mink in mathbb N mid a_k ne c_k$, so $d(a,c)=frac1K$.
Notice that $a_K ne c_K$. As a consequence of this, it is not possible that both of the equations $a_K = b_K$ and $b_K = c_K$ are true: at least one of those two equations must be an inequality, $a_K ne b_K$ or $b_K ne c_K$. Therefore, at least one of the two numbers $Q = minq in mathbb N mid a_q ne b_q$ or $P = minp in mathbb N mid b_p = c_p$ must be less than or equal to $K$.
It follows that $d(a,c)=frac1K$ is less than or equal to at least one of the two numbers $frac1Q$ or $frac1P$, so $d(a,c)$ must be less than or equal to their sum $frac1Q + frac1P$, which proves that
$$d(a,c) le d(a,b) + d(b,c)
$$
answered Aug 6 at 14:37
Lee Mosher
45.7k33478
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0
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Let $a,c in 0,1^mathbb N$. Suppose $a_i = c_i$ for all $i leq l-1$ and $a_l neq c_l$. Note that the inequality is obvious if $a=c$, since $d$ is non-negative.
Now let $b$ be any other sequence in $0,1^mathbb N$. Then, since $a_l neq c_l$, we see that either $b_l neq a_l$ or $b_l neq c_l$ must happen. Consequently, $d(a,b) geq frac 1l$ or $d(a,c) geq frac 1l$ must happen. By the fact that $d$ is non-negative, we get $d(a,b)+ d(b,c) geq frac 1l = d(a,c)$.
answered Aug 6 at 14:43
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
32k22463
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Let me prove the triangle inequality just in the special case that $a,b,c$ are all different from each other (the case where two of them are equal, or where all three are equal, is easy).
First let's evaluate the left hand side of the triangle inequality: let $K = mink in mathbb N mid a_k ne c_k$, so $d(a,c)=frac1K$.
Notice that $a_K ne c_K$. As a consequence of this, it is not possible that both of the equations $a_K = b_K$ and $b_K = c_K$ are true: at least one of those two equations must be an inequality, $a_K ne b_K$ or $b_K ne c_K$. Therefore, at least one of the two numbers $Q = minq in mathbb N mid a_q ne b_q$ or $P = minp in mathbb N mid b_p = c_p$ must be less than or equal to $K$.
It follows that $d(a,c)=frac1K$ is less than or equal to at least one of the two numbers $frac1Q$ or $frac1P$, so $d(a,c)$ must be less than or equal to their sum $frac1Q + frac1P$, which proves that
$$d(a,c) le d(a,b) + d(b,c)
$$
answered Aug 6 at 14:37
Lee Mosher
45.7k33478
add a comment |Â
up vote
3
down vote
accepted
Let me prove the triangle inequality just in the special case that $a,b,c$ are all different from each other (the case where two of them are equal, or where all three are equal, is easy).
First let's evaluate the left hand side of the triangle inequality: let $K = mink in mathbb N mid a_k ne c_k$, so $d(a,c)=frac1K$.
Notice that $a_K ne c_K$. As a consequence of this, it is not possible that both of the equations $a_K = b_K$ and $b_K = c_K$ are true: at least one of those two equations must be an inequality, $a_K ne b_K$ or $b_K ne c_K$. Therefore, at least one of the two numbers $Q = minq in mathbb N mid a_q ne b_q$ or $P = minp in mathbb N mid b_p = c_p$ must be less than or equal to $K$.
It follows that $d(a,c)=frac1K$ is less than or equal to at least one of the two numbers $frac1Q$ or $frac1P$, so $d(a,c)$ must be less than or equal to their sum $frac1Q + frac1P$, which proves that
$$d(a,c) le d(a,b) + d(b,c)
$$
answered Aug 6 at 14:37
Lee Mosher
45.7k33478
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Let me prove the triangle inequality just in the special case that $a,b,c$ are all different from each other (the case where two of them are equal, or where all three are equal, is easy).
First let's evaluate the left hand side of the triangle inequality: let $K = mink in mathbb N mid a_k ne c_k$, so $d(a,c)=frac1K$.
Notice that $a_K ne c_K$. As a consequence of this, it is not possible that both of the equations $a_K = b_K$ and $b_K = c_K$ are true: at least one of those two equations must be an inequality, $a_K ne b_K$ or $b_K ne c_K$. Therefore, at least one of the two numbers $Q = minq in mathbb N mid a_q ne b_q$ or $P = minp in mathbb N mid b_p = c_p$ must be less than or equal to $K$.
It follows that $d(a,c)=frac1K$ is less than or equal to at least one of the two numbers $frac1Q$ or $frac1P$, so $d(a,c)$ must be less than or equal to their sum $frac1Q + frac1P$, which proves that
$$d(a,c) le d(a,b) + d(b,c)
$$
answered Aug 6 at 14:37
Lee Mosher
45.7k33478
Let me prove the triangle inequality just in the special case that $a,b,c$ are all different from each other (the case where two of them are equal, or where all three are equal, is easy).
First let's evaluate the left hand side of the triangle inequality: let $K = mink in mathbb N mid a_k ne c_k$, so $d(a,c)=frac1K$.
Notice that $a_K ne c_K$. As a consequence of this, it is not possible that both of the equations $a_K = b_K$ and $b_K = c_K$ are true: at least one of those two equations must be an inequality, $a_K ne b_K$ or $b_K ne c_K$. Therefore, at least one of the two numbers $Q = minq in mathbb N mid a_q ne b_q$ or $P = minp in mathbb N mid b_p = c_p$ must be less than or equal to $K$.
It follows that $d(a,c)=frac1K$ is less than or equal to at least one of the two numbers $frac1Q$ or $frac1P$, so $d(a,c)$ must be less than or equal to their sum $frac1Q + frac1P$, which proves that
$$d(a,c) le d(a,b) + d(b,c)
$$
answered Aug 6 at 14:37
Lee Mosher
45.7k33478
answered Aug 6 at 14:37
Lee Mosher
45.7k33478
answered Aug 6 at 14:37
Lee Mosher
45.7k33478
answered Aug 6 at 14:37
Lee Mosher
45.7k33478
45.7k33478
add a comment |Â
add a comment |Â
up vote
0
down vote
Let $a,c in 0,1^mathbb N$. Suppose $a_i = c_i$ for all $i leq l-1$ and $a_l neq c_l$. Note that the inequality is obvious if $a=c$, since $d$ is non-negative.
Now let $b$ be any other sequence in $0,1^mathbb N$. Then, since $a_l neq c_l$, we see that either $b_l neq a_l$ or $b_l neq c_l$ must happen. Consequently, $d(a,b) geq frac 1l$ or $d(a,c) geq frac 1l$ must happen. By the fact that $d$ is non-negative, we get $d(a,b)+ d(b,c) geq frac 1l = d(a,c)$.
answered Aug 6 at 14:43
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
32k22463
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up vote
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Let $a,c in 0,1^mathbb N$. Suppose $a_i = c_i$ for all $i leq l-1$ and $a_l neq c_l$. Note that the inequality is obvious if $a=c$, since $d$ is non-negative.
Now let $b$ be any other sequence in $0,1^mathbb N$. Then, since $a_l neq c_l$, we see that either $b_l neq a_l$ or $b_l neq c_l$ must happen. Consequently, $d(a,b) geq frac 1l$ or $d(a,c) geq frac 1l$ must happen. By the fact that $d$ is non-negative, we get $d(a,b)+ d(b,c) geq frac 1l = d(a,c)$.
answered Aug 6 at 14:43
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
32k22463
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let $a,c in 0,1^mathbb N$. Suppose $a_i = c_i$ for all $i leq l-1$ and $a_l neq c_l$. Note that the inequality is obvious if $a=c$, since $d$ is non-negative.
Now let $b$ be any other sequence in $0,1^mathbb N$. Then, since $a_l neq c_l$, we see that either $b_l neq a_l$ or $b_l neq c_l$ must happen. Consequently, $d(a,b) geq frac 1l$ or $d(a,c) geq frac 1l$ must happen. By the fact that $d$ is non-negative, we get $d(a,b)+ d(b,c) geq frac 1l = d(a,c)$.
answered Aug 6 at 14:43
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
32k22463
Let $a,c in 0,1^mathbb N$. Suppose $a_i = c_i$ for all $i leq l-1$ and $a_l neq c_l$. Note that the inequality is obvious if $a=c$, since $d$ is non-negative.
Now let $b$ be any other sequence in $0,1^mathbb N$. Then, since $a_l neq c_l$, we see that either $b_l neq a_l$ or $b_l neq c_l$ must happen. Consequently, $d(a,b) geq frac 1l$ or $d(a,c) geq frac 1l$ must happen. By the fact that $d$ is non-negative, we get $d(a,b)+ d(b,c) geq frac 1l = d(a,c)$.
answered Aug 6 at 14:43
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
32k22463
answered Aug 6 at 14:43
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
32k22463
answered Aug 6 at 14:43
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
32k22463
answered Aug 6 at 14:43
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
32k22463
32k22463
add a comment |Â
add a comment |Â
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The minimum $p$ and $q$ cannot both be 4 when the minimum $k$ is 1. Do you see why? If not, try to give explicit examples of $a, b, c$ to see where your idea goes wrong.
– Mees de Vries
Aug 6 at 14:09
@MeesdeVries we could say that a=n , b=1/n and c=1+n , then k =1 ,q= 2 and p=1. so then $d(a,c)leq d(a,b)+d(b,c)$ I'm confused however as to how to show this generally ?
– exodius
Aug 6 at 14:18