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Now,
D(x,y) + D(y,z) = σ(d(x,y)) + σ(d(y,z))
≥ max d(x,y)+d(y,z)
u=|d(x,y)−d(y,z)| {σ(u)} [From Condition(3)]
≥ D(x,z), ∀x,y ∈ A.
Hence D is a metric on A.
Next we prove the necessity by contradiction. Let σ be an MPT that
does not satisfy one or more of the conditions (1), (2), and (3). Immediately
the violation of conditions (1) and (2) are meaningless and impossible for an
MPT. So, we assume that ∃x,y ∈ R such that ∃z,|x−y| ≤ z ≤ x + y for
which σ(x) + σ(y) < σ(z). Let A = R 2 , d = E 2 , and D = D E 2 = σ◦E 2 . Now
it is always possible to find u,v,w ∈ R 2 such that E 2 (u,v) = x, E 2 (v,w) = y
and E 2 (u,w) = z, so that σ(E 2 (u,v)) + σ(E 2 (v,w)) < σ(E 2 (u,w)) [see Fig.
2.2.4]. Or, D E 2 (u,v) + D E 2 (v,w) < D E 2 (u,w). Hence, D E 2 is not triangular.
But D E 2 should be triangular since σ is metricity preserving, which yields the
contradiction.
Reprinted from Pattern Recognition Letters 10(1989), P. P. Das, Metricity preserving transforms, 73-76, Copyright
(1989), with permission from Elsevier.
FIGURE 2.5: Construction of counter-example for Theorem 2.1.
Corollary 2.1. If σ is monotonically increasing, that is, if x > y implies
σ(x) ≥ σ(y), ∀x,y ∈ R + ∪{0}, then Condition (3) of MPT simplifies to
σ(x) + σ(y) ≥ σ(x + y).
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