Question

Let \( X \) be a set with at least two elements. Let \( d_1, d_2 \) and \( d_3 \) be metrics on \( X \).
Which one of the following is NOT a metric on \( X \)?

• A. \( d(x,y) := \min(d_1(x,y), 3) \) for all \( x, y \in X \).
• B. \( d(x,y) := \max(d_2(x,y), 3) \) for all \( x, y \in X \).
• C. \( d(x,y) := \dfrac{10\, d_3(x,y)}{1+d_3(x,y)} \) for all \( x, y \in X \).
• D. \( d(x,y) := \dfrac{1}{3}(d_1(x,y) + d_2(x,y) + d_3(x,y)) \) for all \( x, y \in X \).

Hint:

Test each candidate distance function at x = y first and see whether it still comes out to zero.

Answer 1

The Correct Option is B

Step 1: Recall the axioms a metric must satisfy.
A function \( d : X \times X \to [0, \infty) \) is a metric if, for all \( x, y, z \in X \):
(i) \( d(x,y) = 0 \) if and only if \( x = y \);
(ii) \( d(x,y) = d(y,x) \), symmetry;
(iii) \( d(x,z) \leq d(x,y) + d(y,z) \), the triangle inequality.
We are told \( d_1, d_2, d_3 \) are already metrics on \( X \), and we must find which combination built from them fails one of these three properties.

Step 2: Check option (B) at x = y first, since that is the fastest way to break it.
Option (B) defines \( d(x,y) = \max(d_2(x,y), 3) \). Take \( x = y \); this is allowed since the definition is stated "for all \( x, y \in X \)", which includes \( x = y \).
Since \( d_2 \) is a metric, \( d_2(x,x) = 0 \). So
\[ d(x,x) = \max(d_2(x,x), 3) = \max(0, 3) = 3 \]
But a metric must give \( d(x,x) = 0 \), and here we get \( 3 \) instead. So option (B) violates axiom (i) for every single point \( x \in X \), and is therefore not a metric.

Step 3: Confirm the other three options do satisfy all the axioms.
For option (A), \( d(x,y) = \min(d_1(x,y), 3) \): at \( x=y \), \( \min(0,3) = 0 \), so axiom (i) holds there; symmetry is immediate since \( d_1 \) is symmetric; and it is a standard fact that capping a metric with \( \min(d, c) \) for a constant \( c > 0 \) keeps the triangle inequality, since \( \min(a,c) \le \min(b,c) + \min(e,c) \) whenever \( a \le b + e \), which can be checked by splitting into cases on whether \( b, e \) exceed \( c \). So (A) is a genuine metric.
For option (C), \( d(x,y) = \dfrac{10\, d_3(x,y)}{1 + d_3(x,y)} \): this is the standard bounded-metric construction. Since \( t \mapsto \dfrac{10t}{1+t} \) is increasing and satisfies \( \dfrac{10(a+b)}{1+a+b} \le \dfrac{10a}{1+a} + \dfrac{10b}{1+b} \) for \( a, b \ge 0 \), a routine algebraic check, applying it to \( d_3 \), which already satisfies the triangle inequality, keeps the triangle inequality; it clearly vanishes only when \( d_3(x,y) = 0 \), meaning \( x=y \); so (C) is a genuine metric.
For option (D), \( d(x,y) = \tfrac13(d_1(x,y)+d_2(x,y)+d_3(x,y)) \): a positive combination, an average, of metrics is again a metric, since each axiom, being zero exactly on the diagonal, symmetry, and the triangle inequality, is preserved under addition and positive scaling. So (D) is a genuine metric.

Final Answer:
Option (B) is the only one that fails, because forcing a floor of \( 3 \) with \( \max(\cdot, 3) \) makes \( d(x,x) = 3 \) instead of \( 0 \). \[ \boxed{\text{(B) } d(x,y) := \max(d_2(x,y), 3) \text{ is NOT a metric.}} \]