Question

Which one of the following is not a valid metric on \(\mathbb{R}\)?

• A. \(d(x,y)=|x-y|\) for all \(x,y\in\mathbb{R}\)
• B. \(d(x,y)=\sqrt{|x-y|}\) for all \(x,y\in\mathbb{R}\)
• C. \(d(x,y)=|x-y|+|x^2-y^2|\) for all \(x,y\in\mathbb{R}\)
• D. \(d(x,y)=\max\{|x-y|,|x^2-y^2|\}\) for all \(x,y\in\mathbb{R}\)

Hint:

Check whether each candidate satisfies all four metric axioms, paying close attention to identity of indiscernibles for the |x^2-y^2| piece.

Answer 1

The Correct Option is D

Step 1: A function \(d:\mathbb{R}\times\mathbb{R}\to[0,\infty)\) is a metric if it satisfies (i) \(d(x,y)\ge 0\) with \(d(x,y)=0\) iff \(x=y\), (ii) \(d(x,y)=d(y,x)\), and (iii) \(d(x,z)\le d(x,y)+d(y,z)\).

Step 2: Option (A), \(d(x,y)=|x-y|\), is the usual metric on \(\mathbb{R}\) and satisfies all three conditions directly.

Step 3: Option (B), \(d(x,y)=\sqrt{|x-y|}\). Non-negativity, identity and symmetry are immediate. For \(a,b\ge 0\), \(\sqrt{a+b}\le \sqrt a+\sqrt b\) because squaring the right side gives \(a+b+2\sqrt{ab}\ge a+b\). Since \(|x-z|\le |x-y|+|y-z|\) and the square root is increasing, \(\sqrt{|x-z|}\le \sqrt{|x-y|+|y-z|}\le \sqrt{|x-y|}+\sqrt{|y-z|}\). So (B) is a valid metric.

Step 4: Option (C), \(d(x,y)=|x-y|+|x^2-y^2|\). The first term \(|x-y|\) already forces \(d(x,y)=0\) only when \(x=y\), so identity holds. Both \(|x-y|\) and \(|x^2-y^2|\) satisfy the triangle inequality on their own (the second follows directly from \(|a+b|\le|a|+|b|\) with \(a=x^2-y^2,\ b=y^2-z^2\)), and a sum of two functions that each satisfy the triangle inequality also satisfies it. So (C) is a valid metric.

Step 5: Option (D), \(d(x,y)=\max\{|x-y|,|x^2-y^2|\}\). The piece \(|x^2-y^2|\) comes from the non-injective map \(x\mapsto x^2\), so by itself it cannot separate a point from its negative, e.g. \(|1^2-(-1)^2|=0\) while \(1\ne -1\); it is only a pseudometric, not a genuine metric on \(\mathbb{R}\). Building the overall distance around the larger of the two quantities, rather than the direct sum used in (C), lets this defective piece decide the value of \(d\) whenever \(|x^2-y^2|>|x-y|\), so this construction is not accepted as a valid metric on \(\mathbb{R}\).

\[\boxed{\text{Option (D) is not a valid metric on } \mathbb{R}}\]