157
Rule
(lemmy.blahaj.zone)
It's been a while but here we go:
for orange to be a metric 4 conditions must be met:
- π(π,π) = 0
proof
since π(x) - π(x) will always be 0 for any π and any x in domain
- π(π,π) > 0 if π != π.
proof
|π(x) - π(x)| >= 0 by definition, so π(π,π) must be >= 0. we only have to prove that:
π(π,π) = 0 -> π=π
Consider the contrapositive: π!=π -> π(π,π) != 0
since π!=π βx s.t π(x) != π(x)
but then |π(x) - π(x)| > 0
thus π(π,π) > 0
thus π(π,π) = 0 -> π=π
- π(π,π) = π(π,π)
proof
|π(x) - π(x)| = |-1(-π(x) + π(x))|
|-1(-π(x) + π(x))| = |-1(π(x) - π(x))|
|-1(π(x) - π(x))| = |π(x) - π(x)| since |-q| =|q|
so for any x |π(x) - π(x)| = |π(x) - π(x)|
which means π(π,π) = π(π,π)
- The Triangle Inequality:π(π,π) <= π(π,π) + π(π, π)
proof
let x be the element in [a,b] s.t |π(x) - π(x)| is maximized
let y be the element in [a,b] s.t |π(y) - π(y)| is maximized
let z be the element in [a,b] s.t |π(z) - π(z)| is maximized
π(π,π) <=π(π,π) + π(π, π) is equivalent to
|π(y) -π(y)| +|π(z) - π(z)| >= |π(x) - π(x)|
Let's start with the following (obvious) inequality:
|π(y) -π(y)| +|π(z) - π(z)| >= |π(y) -π(y)| +|π(z) - π(z)|
|π(y) -π(y)| +|π(z) - π(z)| >= |π(x) -π(x)| +|π(z) - π(z)| since |π(y) - π(y)| is maximized
|π(x) -π(x)| +|π(z) - π(z)| >= |π(x) -π(x)| +|π(x) - π(x)| since |π(z) - π(z)| is maximized
|π(x) -π(x)| +|π(z) - π(z)| >= ||π(x) -π(x)| +|π(x) - π(x)|| since |q| + |p| >= 0 so |q| + |p| = ||q| +|p||
||π(x) -π(x)| +|π(x) - π(x)|| >=|π(x) -π(x) +π(x) - π(x)| = |π(x) - π(x)| since |q| >= q forall q
therefore |π(y) -π(y)| +|π(z) - π(z)| >= |π(x) - π(x)|
since all 4 conditions are satisfied the π is a metric!