Math Problem Statement

Let's tackle this

Let (X,d) be a metric space. Show that |d(x, z) − d(y, z)| ≤ d(x, y) for all x, y, z ∈ X. Let a function f : X → R+ be defined by f(x) = d(x0, x) for all x ∈ X where x0 is fixed. (i) Show that f is uniformly continuous. (ii) Show that if (xn) is a sequence in X such that lim f(xn) = 0 then xn converges to x0.

Solution

Let's break down and solve the given problems one by one:

Problem (b): Metric Space Property

Let (X,d)(X, d) be a metric space. The problem asks to show the inequality: d(x,z)d(y,z)d(x,y),x,y,zX.|d(x, z) - d(y, z)| \leq d(x, y), \quad \forall x, y, z \in X.

This is an application of the triangle inequality of metric spaces. To prove this:

  1. Using the triangle inequality: By definition of a metric space, for all x,y,zXx, y, z \in X, the following holds: d(x,z)d(x,y)+d(y,z)d(x, z) \leq d(x, y) + d(y, z) Rearranging, we get: d(x,z)d(y,z)d(x,y)d(x, z) - d(y, z) \leq d(x, y)

  2. Switching the roles: Now, switch xx and yy: d(y,z)d(y,x)+d(x,z)which givesd(y,z)d(x,z)d(x,y)d(y, z) \leq d(y, x) + d(x, z) \quad \text{which gives} \quad d(y, z) - d(x, z) \leq d(x, y)

  3. Combining both inequalities: We now have the following two inequalities: d(x,z)d(y,z)d(x,y)andd(y,z)d(x,z)d(x,y)d(x, z) - d(y, z) \leq d(x, y) \quad \text{and} \quad d(y, z) - d(x, z) \leq d(x, y) Thus, the absolute value inequality holds: d(x,z)d(y,z)d(x,y).|d(x, z) - d(y, z)| \leq d(x, y).

This proves the required result.

Problem (c): Function defined by distance

Given the function f:XR+f : X \to \mathbb{R}^+ defined by: f(x)=d(x0,x)xX,f(x) = d(x_0, x) \quad \forall x \in X, where x0Xx_0 \in X is fixed, we need to show two things:

Part (i): Prove that ff is uniformly continuous

To show ff is uniformly continuous, we need to show that for every ϵ>0\epsilon > 0, there exists a δ>0\delta > 0 such that for all x,yXx, y \in X, if d(x,y)<δd(x, y) < \delta, then f(x)f(y)<ϵ|f(x) - f(y)| < \epsilon.

  1. Start by expressing f(x)f(y)|f(x) - f(y)|: f(x)f(y)=d(x0,x)d(x0,y).|f(x) - f(y)| = |d(x_0, x) - d(x_0, y)|.

  2. Using the result from part (b), we know that: d(x0,x)d(x0,y)d(x,y).|d(x_0, x) - d(x_0, y)| \leq d(x, y).

  3. Therefore, we have: f(x)f(y)d(x,y).|f(x) - f(y)| \leq d(x, y).

  4. To satisfy f(x)f(y)<ϵ|f(x) - f(y)| < \epsilon, it is enough to choose δ=ϵ\delta = \epsilon. That is, if d(x,y)<δd(x, y) < \delta, then: f(x)f(y)d(x,y)<ϵ.|f(x) - f(y)| \leq d(x, y) < \epsilon.

Thus, ff is uniformly continuous.

Part (ii): Convergence of a sequence

We are given a sequence (xn)(x_n) in XX such that: limnf(xn)=0,\lim_{n \to \infty} f(x_n) = 0, and need to show that xnx_n converges to x0x_0.

  1. From the definition of f(x)=d(x0,x)f(x) = d(x_0, x), the condition limnf(xn)=0\lim_{n \to \infty} f(x_n) = 0 means: limnd(x0,xn)=0.\lim_{n \to \infty} d(x_0, x_n) = 0.

  2. This implies that the distance between xnx_n and x0x_0 tends to 0, which by the definition of convergence in a metric space means that: xnx0asn.x_n \to x_0 \quad \text{as} \quad n \to \infty.

Thus, the sequence xnx_n converges to x0x_0.


Do you want further details or have any questions on the steps?

Here are 5 related questions to deepen understanding:

  1. What is the triangle inequality in a metric space, and why is it essential for proving part (b)?
  2. How can we generalize the proof of uniform continuity for other types of metric spaces?
  3. Why does the limit limnd(x0,xn)=0\lim_{n \to \infty} d(x_0, x_n) = 0 imply that xnx_n converges to x0x_0?
  4. Can the function f(x)=d(x0,x)f(x) = d(x_0, x) be Lipschitz continuous, and how would you show that?
  5. What is the difference between pointwise and uniform continuity in metric spaces?

Tip: When working with sequences in metric spaces, always think of convergence in terms of distances shrinking to zero between sequence points and the limit.

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Math Problem Analysis

Mathematical Concepts

Metric Space
Uniform Continuity
Sequence Convergence
Triangle Inequality

Formulas

|d(x, z) − d(y, z)| ≤ d(x, y)
f(x) = d(x0, x)

Theorems

Triangle Inequality

Suitable Grade Level

Undergraduate - Advanced Mathematics