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fucoat
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1. The problem
(this is from Rudin's Principles of Mathematical Analysis, p. 43 #11)
Determine if the following is a metric
d(x,y) = abs(x-y)/(1 + abs(x-y))
where the set in question is the real number line (R1)
the definition of a metric on a set requires that it be:
(1) positive definite
(2) symmetric
(3) satisfies the triangle inequality, that is d(x,y) ≤ d(x,r) + d(r,y) for all x, y, r
the positive definiteness of d(•,•) is apparent since both the numerator and denominator are positive for all x, y such that x≠y. and 0 when x=y.
The symmetry likewise follows quickly from the symmetry of the absolute value function.
The triangle inequality is more difficult:
the problem is given x,y, z in R1 is the following satisfied:
abs(x-y)/(1+ abs(x-y)) ≤ abs(x-r)/(1+ abs(x-r)) + abs(r-y)/(1+abs(r-y).
If we look only at the numerator we see that this is the ordinary triangle inequality for the standard metric on R1. I started with the simple case with y=-x and r=0
which gives:
2|x|/(1+|2x|) for the left hand side and:
|x|/(1+|x|) +|x|/(1+|x|) for the right hand side which reduces to:
2|x|/(1+|x|) which is obviously greater than the left (since the denominator for the LHS is strictly greater than that for the RHS)
when I branch out from this very simple case I can't either compare the quantities or they conform to the triangle inequality (I can post the other values I used if they help at all).
(this is from Rudin's Principles of Mathematical Analysis, p. 43 #11)
Determine if the following is a metric
d(x,y) = abs(x-y)/(1 + abs(x-y))
where the set in question is the real number line (R1)
Homework Equations
the definition of a metric on a set requires that it be:
(1) positive definite
(2) symmetric
(3) satisfies the triangle inequality, that is d(x,y) ≤ d(x,r) + d(r,y) for all x, y, r
The Attempt at a Solution
the positive definiteness of d(•,•) is apparent since both the numerator and denominator are positive for all x, y such that x≠y. and 0 when x=y.
The symmetry likewise follows quickly from the symmetry of the absolute value function.
The triangle inequality is more difficult:
the problem is given x,y, z in R1 is the following satisfied:
abs(x-y)/(1+ abs(x-y)) ≤ abs(x-r)/(1+ abs(x-r)) + abs(r-y)/(1+abs(r-y).
If we look only at the numerator we see that this is the ordinary triangle inequality for the standard metric on R1. I started with the simple case with y=-x and r=0
which gives:
2|x|/(1+|2x|) for the left hand side and:
|x|/(1+|x|) +|x|/(1+|x|) for the right hand side which reduces to:
2|x|/(1+|x|) which is obviously greater than the left (since the denominator for the LHS is strictly greater than that for the RHS)
when I branch out from this very simple case I can't either compare the quantities or they conform to the triangle inequality (I can post the other values I used if they help at all).