Show that $d(x, y) = |f (x) − f (y)|$ is a metric on $mathbb R$
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0
down vote
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Prove that for any 1-1 function $f : mathbb R to mathbb R$, the function $d : mathbb R to mathbb R$ defined by
$$ d(x, y) = |f (x) − f (y)| $$
is a metric on $mathbb R$.
I need to prove these properties:
$d(x,y)ge0$
$d(x,y)=0$ iff $x=y$
$d(x,y)=d(y,x)$
Triangle inequality: $d(x, y) + d(y, z) ge d(x, z)$
I was able to prove first three but couldn't prove the last one.
real-analysis functions metric-spaces
asked Nov 23 at 10:57
Pumpkin
4841417
add a comment |
up vote
0
down vote
favorite
Prove that for any 1-1 function $f : mathbb R to mathbb R$, the function $d : mathbb R to mathbb R$ defined by
$$ d(x, y) = |f (x) − f (y)| $$
is a metric on $mathbb R$.
I need to prove these properties:
$d(x,y)ge0$
$d(x,y)=0$ iff $x=y$
$d(x,y)=d(y,x)$
Triangle inequality: $d(x, y) + d(y, z) ge d(x, z)$
I was able to prove first three but couldn't prove the last one.
real-analysis functions metric-spaces
asked Nov 23 at 10:57
Pumpkin
4841417
2
Can you prove $|x-y|+|y-z|geq|x-z|$? If so then you replace $x,y,z$ by $f(x),f(y),f(z)$ and you are ready. For this $f$ does not have to be 1-1. That property is only needed for $d(x,y)=0implies x=y$.
– drhab
Nov 23 at 11:02
Hint : $|f(x)-f(y)|=|f(x)-f(z)+f(z)-f(y)|$
– nicomezi
Nov 23 at 11:02
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Prove that for any 1-1 function $f : mathbb R to mathbb R$, the function $d : mathbb R to mathbb R$ defined by
$$ d(x, y) = |f (x) − f (y)| $$
is a metric on $mathbb R$.
I need to prove these properties:
$d(x,y)ge0$
$d(x,y)=0$ iff $x=y$
$d(x,y)=d(y,x)$
Triangle inequality: $d(x, y) + d(y, z) ge d(x, z)$
I was able to prove first three but couldn't prove the last one.
real-analysis functions metric-spaces
asked Nov 23 at 10:57
Pumpkin
4841417
Prove that for any 1-1 function $f : mathbb R to mathbb R$, the function $d : mathbb R to mathbb R$ defined by
$$ d(x, y) = |f (x) − f (y)| $$
is a metric on $mathbb R$.
I need to prove these properties:
$d(x,y)ge0$
$d(x,y)=0$ iff $x=y$
$d(x,y)=d(y,x)$
Triangle inequality: $d(x, y) + d(y, z) ge d(x, z)$
I was able to prove first three but couldn't prove the last one.
real-analysis functions metric-spaces
real-analysis functions metric-spaces
asked Nov 23 at 10:57
Pumpkin
4841417
asked Nov 23 at 10:57
Pumpkin
4841417
edited Nov 23 at 11:06
asked Nov 23 at 10:57
Pumpkin
4841417
asked Nov 23 at 10:57
Pumpkin
4841417
asked Nov 23 at 10:57
Pumpkin
4841417
4841417
2
Can you prove $|x-y|+|y-z|geq|x-z|$? If so then you replace $x,y,z$ by $f(x),f(y),f(z)$ and you are ready. For this $f$ does not have to be 1-1. That property is only needed for $d(x,y)=0implies x=y$.
– drhab
Nov 23 at 11:02
Hint : $|f(x)-f(y)|=|f(x)-f(z)+f(z)-f(y)|$
– nicomezi
Nov 23 at 11:02
add a comment |
2
Can you prove $|x-y|+|y-z|geq|x-z|$? If so then you replace $x,y,z$ by $f(x),f(y),f(z)$ and you are ready. For this $f$ does not have to be 1-1. That property is only needed for $d(x,y)=0implies x=y$.
– drhab
Nov 23 at 11:02
Hint : $|f(x)-f(y)|=|f(x)-f(z)+f(z)-f(y)|$
– nicomezi
Nov 23 at 11:02
2
2
Can you prove $|x-y|+|y-z|geq|x-z|$? If so then you replace $x,y,z$ by $f(x),f(y),f(z)$ and you are ready. For this $f$ does not have to be 1-1. That property is only needed for $d(x,y)=0implies x=y$.
– drhab
Nov 23 at 11:02
Can you prove $|x-y|+|y-z|geq|x-z|$? If so then you replace $x,y,z$ by $f(x),f(y),f(z)$ and you are ready. For this $f$ does not have to be 1-1. That property is only needed for $d(x,y)=0implies x=y$.
– drhab
Nov 23 at 11:02
Hint : $|f(x)-f(y)|=|f(x)-f(z)+f(z)-f(y)|$
– nicomezi
Nov 23 at 11:02
Hint : $|f(x)-f(y)|=|f(x)-f(z)+f(z)-f(y)|$
– nicomezi
Nov 23 at 11:02
add a comment |
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
If you know that it's true for the identity function, then:
$$|f(x)-f(z)|=|f(x)-f(y)+f(y)-f(z)|=|(f(x)-f(y))+(f(y)-f(z))|leq |f(x)-f(y)|+|f(y)-f(z)|$$
answered Nov 23 at 11:02
Botond
5,2012732
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
If you know that it's true for the identity function, then:
$$|f(x)-f(z)|=|f(x)-f(y)+f(y)-f(z)|=|(f(x)-f(y))+(f(y)-f(z))|leq |f(x)-f(y)|+|f(y)-f(z)|$$
answered Nov 23 at 11:02
Botond
5,2012732
add a comment |
up vote
4
down vote
accepted
If you know that it's true for the identity function, then:
$$|f(x)-f(z)|=|f(x)-f(y)+f(y)-f(z)|=|(f(x)-f(y))+(f(y)-f(z))|leq |f(x)-f(y)|+|f(y)-f(z)|$$
answered Nov 23 at 11:02
Botond
5,2012732
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
If you know that it's true for the identity function, then:
$$|f(x)-f(z)|=|f(x)-f(y)+f(y)-f(z)|=|(f(x)-f(y))+(f(y)-f(z))|leq |f(x)-f(y)|+|f(y)-f(z)|$$
answered Nov 23 at 11:02
Botond
5,2012732
If you know that it's true for the identity function, then:
$$|f(x)-f(z)|=|f(x)-f(y)+f(y)-f(z)|=|(f(x)-f(y))+(f(y)-f(z))|leq |f(x)-f(y)|+|f(y)-f(z)|$$
answered Nov 23 at 11:02
Botond
5,2012732
edited Nov 23 at 11:10
answered Nov 23 at 11:02
Botond
5,2012732
answered Nov 23 at 11:02
Botond
5,2012732
answered Nov 23 at 11:02
Botond
5,2012732
5,2012732
add a comment |
add a comment |
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2
Can you prove $|x-y|+|y-z|geq|x-z|$? If so then you replace $x,y,z$ by $f(x),f(y),f(z)$ and you are ready. For this $f$ does not have to be 1-1. That property is only needed for $d(x,y)=0implies x=y$.
– drhab
Nov 23 at 11:02
Hint : $|f(x)-f(y)|=|f(x)-f(z)+f(z)-f(y)|$
– nicomezi
Nov 23 at 11:02