Mixing
Verification of metric axiom 3
$begingroup$
If I have a $3$ element set, say $X={a,b,c}$ where $d$ is a metric on $X$ with $d(a,b)=5$ and $d(b,c)=7$, then do we have $d(a,c)in{[0,12]}$ by the triangle inequality? since the distance is greater than or equal to $0$, but less than or equal to the sum of the other $2$ sides.
$$d(a,c)leq d(a,b)+d(b,c),$$ $$d(a,c)leq 5+7,$$ $$d(a,c)leq 12.$$
proof-verification metric-spaces
$endgroup$
add a comment |
$begingroup$
If I have a $3$ element set, say $X={a,b,c}$ where $d$ is a metric on $X$ with $d(a,b)=5$ and $d(b,c)=7$, then do we have $d(a,c)in{[0,12]}$ by the triangle inequality? since the distance is greater than or equal to $0$, but less than or equal to the sum of the other $2$ sides.
$$d(a,c)leq d(a,b)+d(b,c),$$ $$d(a,c)leq 5+7,$$ $$d(a,c)leq 12.$$
proof-verification metric-spaces
$endgroup$
2
$begingroup$
You also need $d(b,c)le d(b,a)+d(a,c)$, so $d(a,c)ge 2$
$endgroup$
– Hagen von Eitzen
Jan 17 at 20:38
$begingroup$
Can you please explain to me why we also need this?
$endgroup$
– user606466
Jan 17 at 20:41
1
$begingroup$
Because the triangle inequality must be true for any three points in your space. That is, what @HagenvonEitzen has written must also be true, since the distance from $b$ to $c$ must be less than or equal to the distance from $b$ to $a$ plus the distance from $a$ to $c$.
$endgroup$
– Chickenmancer
Jan 17 at 20:42
$begingroup$
Of course, thanks to you both. :)
$endgroup$
– user606466
Jan 17 at 20:50
add a comment |
$begingroup$
If I have a $3$ element set, say $X={a,b,c}$ where $d$ is a metric on $X$ with $d(a,b)=5$ and $d(b,c)=7$, then do we have $d(a,c)in{[0,12]}$ by the triangle inequality? since the distance is greater than or equal to $0$, but less than or equal to the sum of the other $2$ sides.
$$d(a,c)leq d(a,b)+d(b,c),$$ $$d(a,c)leq 5+7,$$ $$d(a,c)leq 12.$$
proof-verification metric-spaces
$endgroup$
If I have a $3$ element set, say $X={a,b,c}$ where $d$ is a metric on $X$ with $d(a,b)=5$ and $d(b,c)=7$, then do we have $d(a,c)in{[0,12]}$ by the triangle inequality? since the distance is greater than or equal to $0$, but less than or equal to the sum of the other $2$ sides.
$$d(a,c)leq d(a,b)+d(b,c),$$ $$d(a,c)leq 5+7,$$ $$d(a,c)leq 12.$$
proof-verification metric-spaces
proof-verification metric-spaces
asked Jan 17 at 20:37
user606466
2
$begingroup$
You also need $d(b,c)le d(b,a)+d(a,c)$, so $d(a,c)ge 2$
$endgroup$
– Hagen von Eitzen
Jan 17 at 20:38
$begingroup$
Can you please explain to me why we also need this?
$endgroup$
– user606466
Jan 17 at 20:41
1
$begingroup$
Because the triangle inequality must be true for any three points in your space. That is, what @HagenvonEitzen has written must also be true, since the distance from $b$ to $c$ must be less than or equal to the distance from $b$ to $a$ plus the distance from $a$ to $c$.
$endgroup$
– Chickenmancer
Jan 17 at 20:42
$begingroup$
Of course, thanks to you both. :)
$endgroup$
– user606466
Jan 17 at 20:50
add a comment |
2
$begingroup$
You also need $d(b,c)le d(b,a)+d(a,c)$, so $d(a,c)ge 2$
$endgroup$
– Hagen von Eitzen
Jan 17 at 20:38
$begingroup$
Can you please explain to me why we also need this?
$endgroup$
– user606466
Jan 17 at 20:41
1
$begingroup$
Because the triangle inequality must be true for any three points in your space. That is, what @HagenvonEitzen has written must also be true, since the distance from $b$ to $c$ must be less than or equal to the distance from $b$ to $a$ plus the distance from $a$ to $c$.
$endgroup$
– Chickenmancer
Jan 17 at 20:42
$begingroup$
Of course, thanks to you both. :)
$endgroup$
– user606466
Jan 17 at 20:50
2
2
$begingroup$
You also need $d(b,c)le d(b,a)+d(a,c)$, so $d(a,c)ge 2$
$endgroup$
– Hagen von Eitzen
Jan 17 at 20:38
$begingroup$
You also need $d(b,c)le d(b,a)+d(a,c)$, so $d(a,c)ge 2$
$endgroup$
– Hagen von Eitzen
Jan 17 at 20:38
$begingroup$
Can you please explain to me why we also need this?
$endgroup$
– user606466
Jan 17 at 20:41
$begingroup$
Can you please explain to me why we also need this?
$endgroup$
– user606466
Jan 17 at 20:41
1
1
$begingroup$
Because the triangle inequality must be true for any three points in your space. That is, what @HagenvonEitzen has written must also be true, since the distance from $b$ to $c$ must be less than or equal to the distance from $b$ to $a$ plus the distance from $a$ to $c$.
$endgroup$
– Chickenmancer
Jan 17 at 20:42
$begingroup$
Because the triangle inequality must be true for any three points in your space. That is, what @HagenvonEitzen has written must also be true, since the distance from $b$ to $c$ must be less than or equal to the distance from $b$ to $a$ plus the distance from $a$ to $c$.
$endgroup$
– Chickenmancer
Jan 17 at 20:42
$begingroup$
Of course, thanks to you both. :)
$endgroup$
– user606466
Jan 17 at 20:50
$begingroup$
Of course, thanks to you both. :)
$endgroup$
– user606466
Jan 17 at 20:50
add a comment |
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2
$begingroup$
You also need $d(b,c)le d(b,a)+d(a,c)$, so $d(a,c)ge 2$
$endgroup$
– Hagen von Eitzen
Jan 17 at 20:38
$begingroup$
Can you please explain to me why we also need this?
$endgroup$
– user606466
Jan 17 at 20:41
1
$begingroup$
Because the triangle inequality must be true for any three points in your space. That is, what @HagenvonEitzen has written must also be true, since the distance from $b$ to $c$ must be less than or equal to the distance from $b$ to $a$ plus the distance from $a$ to $c$.
$endgroup$
– Chickenmancer
Jan 17 at 20:42
$begingroup$
Of course, thanks to you both. :)
$endgroup$
– user606466
Jan 17 at 20:50