Question

Exercise 1. Let $(M, d)$ be a metric space. Show that $d_1$ given by $d_1(x, y) = \frac{d(x, y)}{1 + d(x, y)}$, for $x, y \in M$ is a metric. The diameter of a subset $A$ of $M$ is defined as $\delta(A) = \sup_{x, y \in A} d(x, y) \leq \infty)$. Show that $\delta(A) = 0$ iff $A$ contains only one point. Show that $\delta_1(A) = \sup_{x, y \in A} d_1(x, y) \leq 1$ for all $A \subset M$. Is it possible to find a subset $A$ with $\delta_1(A) = 1$?

          Exercise 1. Let $(M, d)$ be a metric space. Show that $d_1$ given by

$d_1(x, y) = \frac{d(x, y)}{1 + d(x, y)}$, for $x, y \in M$
is a metric.
The diameter of a subset $A$ of $M$ is defined as
$\delta(A) = \sup_{x, y \in A} d(x, y) \leq \infty)$.
Show that $\delta(A) = 0$ iff $A$ contains only one point.
Show that $\delta_1(A) = \sup_{x, y \in A} d_1(x, y) \leq 1$ for all $A \subset M$. Is it possible to find a
subset $A$ with $\delta_1(A) = 1$?

Exercise 1. Let (M, d) be a metric space. Show that d1 given by

d1(x, y) = (d(x, y))/(1 + d(x, y)), for x, y ∈ M
is a metric.
The diameter of a subset A of M is defined as
δ(A) = supx, y ∈ A d(x, y) ≤∞).
Show that δ(A) = 0 iff A contains only one point.
Show that δ1(A) = supx, y ∈ A d1(x, y) ≤ 1 for all A ⊂ M. Is it possible to find a
subset A with δ1(A) = 1?

Added by Nicholas G.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Madhur L

06/03/2022

Step 1

Step 1: To show that d₁ is a metric, we need to show that it satisfies the three properties of a metric: non-negativity, identity of indiscernibles, and the triangle inequality. Show more…

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Exercise 1. Let (M, d) be a metric space. Show that d₁ given by d(x, y) = 1+ d(x, y) is a metric. The diameter of a subset A of M is defined as sup d(x,y) (≤ ∞) for x,y ∈ A. Show that sup d(A) = 0 iff A contains only one point. Show that sup d₁(x, y) ≤ 1 for all x,y ∈ M. Is it possible to find a subset A with sup d(A) = 1?