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 $?

Question

Let $ (\mathbb{R}^2, d_1) $ and $ (\mathbb{R}^2, d_2) $ be two metric spaces with \[ d_1\left( (x_1, x_2), (y_1, y_2) \right) = |x_1 - y_1| + |x_2 - y_2| \] \[ {and} \quad d_2\left( (x_1, x_2), (y_1, y_2) \right) = \frac{d_1\left( (x_1, x_2), (y_1, y_2) \right)}{1 + d_1\left( (x_1, x_2), (y_1, y_2) \right)}. \] If the open ball centered at $ (0,0) $ with radius $ \frac{1}{7} $ in $ (\mathbb{R}^2, d_1) $ is equal to the open ball centered at $ (0,0) $ with radius $ \frac{1}{\alpha} $ in $ (\mathbb{R}^2, d_2) $, then the value of $ \alpha $ is (in integer).

Answer 1

There is a general pattern worth knowing here: if $d$ is a metric and $c > 0$ is a constant, then $\min(d, c)$ is always again a metric, since it just caps the distance at $c$, but $\max(d, c)$ is never a metric, because it forces every point to be at distance at least $c$ from itself. Let's use this idea to spot the wrong option quickly, then verify the rest.

(A) min(d1(x,y), 3): At $x = y$, this gives $\min(0,3) = 0$, so the identity property survives. Symmetry is inherited directly from $d_1$. For the triangle inequality, if $a = d_1(x,z) \le b + e = d_1(x,y) + d_1(y,z)$, then capping each side at $3$ still respects the inequality, checking the cases where $b,e$ are above or below $3$ shows $\min(a,3) \le \min(b,3)+\min(e,3)$ always. So this is a genuine metric.
(B) max(d2(x,y), 3): Set $x = y$. Since $d_2(x,x) = 0$, as $d_2$ is a metric, we get $d(x,x) = \max(0,3) = 3 \neq 0$. Every point ends up at distance $3$ from itself, which directly breaks the requirement that $d(x,x) = 0$. This fails to be a metric, and the failure does not depend on which particular metric $d_2$ we started with; it happens purely because of the $\max(\cdot, 3)$ construction.
(C) 10 d3(x,y) / (1 + d3(x,y)): This is the classic bounded-metric transform. It is $0$ exactly when $d_3(x,y) = 0$, that is when $x = y$, it is symmetric since $d_3$ is, and because $t \mapsto \frac{10t}{1+t}$ is increasing and subadditive on $t \ge 0$, applying it to the triangle inequality for $d_3$ preserves the triangle inequality for $d$. So this is a genuine metric.
(D) (1/3)(d1(x,y)+d2(x,y)+d3(x,y)): A positive-weighted sum of metrics is again a metric: it vanishes only when all three vanish, which happens only at $x=y$ since each $d_i$ is already a metric, it is symmetric, and the triangle inequality follows by adding the three individual triangle inequalities and dividing by $3$. So this is a genuine metric.

Only option (B) breaks down, and it breaks down specifically at the diagonal case $x = y$, giving a nonzero self-distance of $3$.

Let's summarize:

Capping a metric from above with $\min(d,c)$ always keeps it a metric.
Flooring a metric from below with $\max(d,c)$ for $c>0$ always destroys the metric property, since it makes $d(x,x) = c \neq 0$.
The bounded transform $\frac{10d_3}{1+d_3}$ and the average $\frac{1}{3}(d_1+d_2+d_3)$ both preserve every metric axiom.

\[ \boxed{\text{(B) } \max(d_2(x,y), 3) \text{ is NOT a metric}} \]

Answer 2

There is a general pattern worth knowing here: if $d$ is a metric and $c > 0$ is a constant, then $\min(d, c)$ is always again a metric, since it just caps the distance at $c$, but $\max(d, c)$ is never a metric, because it forces every point to be at distance at least $c$ from itself. Let's use this idea to spot the wrong option quickly, then verify the rest.

(A) min(d1(x,y), 3): At $x = y$, this gives $\min(0,3) = 0$, so the identity property survives. Symmetry is inherited directly from $d_1$. For the triangle inequality, if $a = d_1(x,z) \le b + e = d_1(x,y) + d_1(y,z)$, then capping each side at $3$ still respects the inequality, checking the cases where $b,e$ are above or below $3$ shows $\min(a,3) \le \min(b,3)+\min(e,3)$ always. So this is a genuine metric.
(B) max(d2(x,y), 3): Set $x = y$. Since $d_2(x,x) = 0$, as $d_2$ is a metric, we get $d(x,x) = \max(0,3) = 3 \neq 0$. Every point ends up at distance $3$ from itself, which directly breaks the requirement that $d(x,x) = 0$. This fails to be a metric, and the failure does not depend on which particular metric $d_2$ we started with; it happens purely because of the $\max(\cdot, 3)$ construction.
(C) 10 d3(x,y) / (1 + d3(x,y)): This is the classic bounded-metric transform. It is $0$ exactly when $d_3(x,y) = 0$, that is when $x = y$, it is symmetric since $d_3$ is, and because $t \mapsto \frac{10t}{1+t}$ is increasing and subadditive on $t \ge 0$, applying it to the triangle inequality for $d_3$ preserves the triangle inequality for $d$. So this is a genuine metric.
(D) (1/3)(d1(x,y)+d2(x,y)+d3(x,y)): A positive-weighted sum of metrics is again a metric: it vanishes only when all three vanish, which happens only at $x=y$ since each $d_i$ is already a metric, it is symmetric, and the triangle inequality follows by adding the three individual triangle inequalities and dividing by $3$. So this is a genuine metric.

Only option (B) breaks down, and it breaks down specifically at the diagonal case $x = y$, giving a nonzero self-distance of $3$.

Let's summarize:

Capping a metric from above with $\min(d,c)$ always keeps it a metric.
Flooring a metric from below with $\max(d,c)$ for $c>0$ always destroys the metric property, since it makes $d(x,x) = c \neq 0$.
The bounded transform $\frac{10d_3}{1+d_3}$ and the average $\frac{1}{3}(d_1+d_2+d_3)$ both preserve every metric axiom.

\[ \boxed{\text{(B) } \max(d_2(x,y), 3) \text{ is NOT a metric}} \]

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 \( X \)?

Show Hint

The Correct Option is B

Solution and Explanation

Top Questions on Metric Spaces

Questions Asked in GATE MA exam

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 \( X \)?