Question

Prove that if \(\rho\) is a metric for \(S\), then another metric \(\rho^{\prime}\) for \(S\) is given by $$ \begin{array}{l} \text { (i) } \rho^{\prime}(x, y)=\min \\{1, \rho(x, y)\\} \\ \text { (ii) } \rho^{\prime}(x, y)=\frac{\rho(x, y)}{1+\rho(x, y)} \end{array} $$ In case (i), show that globes \(G_{p}(\varepsilon)\) of radius \(\varepsilon \leq 1\) are the same under \(\rho\) and \(\rho^{\prime} .\) In case (ii), prove that any \(G_{p}(\varepsilon)\) in \((S, \rho)\) is also a globe \(G_{p}\left(\varepsilon^{\prime}\right)\) in \(\left(S, \rho^{\prime}\right)\) of radius $$ \varepsilon^{\prime}=\frac{\varepsilon}{1+\varepsilon}, $$ and any globe of radius \(\varepsilon^{\prime}<1\) in \(\left(S, \rho^{\prime}\right)\) is also a globe in \((S, \rho)\). (Find the converse formula for \(\varepsilon\) as well! \()\)

Short answer

Both definitions of \(\rho'\) are metrics, and globes transform between \(\rho\) and \(\rho'\) per given conditions.

Step-by-step solution

Understanding the Metric Properties

A function \( \rho \) is a metric on a set \( S \) if it satisfies the following properties for all points \( x, y, z \in S \). 1) Non-negativity: \( \rho(x, y) \geq 0 \), and \( \rho(x, y) = 0 \) if and only if \( x = y \). 2) Symmetry: \( \rho(x, y) = \rho(y, x) \). 3) Triangle Inequality: \( \rho(x, z) \leq \rho(x, y) + \rho(y, z) \). You will verify that both definitions of \( \rho' \) also satisfy these properties, ensuring that \( \rho' \) is a metric.

Prove \( \rho'(x, y) = \min(1, \rho(x, y)) \) is a Metric

For the metric \( \rho' \), check the conditions: 1) Non-negativity: Clearly \( \min(1, \rho(x, y)) \geq 0 \).2) Symmetry: \( \min(1, \rho(x, y)) = \min(1, \rho(y, x)) \) by symmetry of \( \rho \).3) Triangle Inequality: \( \rho'(x, z) = \min(1, \rho(x, z)) \leq \min(1, \rho(x, y)) + \min(1, \rho(y, z)) = \rho'(x, y) + \rho'(y, z) \). Thus, \( \rho' \) satisfies the metric properties.

Prove \( \rho'(x, y) = \frac{\rho(x, y)}{1+\rho(x, y)} \) is a Metric

For this case, check:1) Non-negativity: \( \frac{\rho(x, y)}{1+\rho(x, y)} \geq 0 \) since \( \rho(x, y) \geq 0 \).2) Symmetry: Similar to symmetry of \( \rho \), \( \rho'(x, y) = \rho'(y, x) \).3) Triangle Inequality: By manipulating the triangle inequality of \( \rho \), note \( \rho'(x, z) \leq \frac{\rho(x, y) + \rho(y, z)}{1 + \rho(x, y) + \rho(y, z)} = \rho'(x, y) + \rho'(y, z) \). Thus, \( \rho' \) is a metric.

Compare Globes in Case (i) with \(\rho' = \min(1, \rho(x, y))\)

A globe \( G_p(\varepsilon) \) in \( (S, \rho) \) consists of all points \( y \) such that \( \rho(p, y) < \varepsilon \). For \( \varepsilon \leq 1 \), \( \rho'(x, y) = \rho(x, y) \) since it's just the intersection with the sphere of radius 1. Hence, the globes under \( \rho \) and \( \rho' \) are identical.

Convert Radius in Case (ii) for \(\rho' = \frac{\rho(x, y)}{1+\rho(x, y)}\)

For \( (S, \rho') \), the globe \( G_p(\varepsilon') \) consists of points \( y \) such that \( \frac{\rho(p, y)}{1+\rho(p, y)} < \varepsilon' \), implying \( \rho(p, y) < \frac{\varepsilon'}{1-\varepsilon'} \). By substituting \( \varepsilon' = \frac{\varepsilon}{1+\varepsilon} \), this matches the form \( \rho(p, y) < \varepsilon \).

Find the Converse Formula for \(\varepsilon'\)

To convert \( \varepsilon' \) back to \( \varepsilon \), solve \( \varepsilon' = \frac{\varepsilon}{1+\varepsilon} \) for \( \varepsilon \), yielding \( \varepsilon = \frac{\varepsilon'}{1-\varepsilon'} \). Thus, any globe in \( (S, \rho') \) can be mapped back to a globe in \( (S, \rho) \).

Key concepts

Non-negativity in Metric Spaces

In the realm of metric spaces, non-negativity is one of the fundamental properties. When we say that a metric \( \rho \) is non-negative, we mean that the distance between any two points \( x \) and \( y \) in the space \( S \) is always greater than or equal to zero: \( \rho(x, y) \geq 0 \). This seems intuitive, as distances in the physical world are never negative. Moreover, this property includes the condition that \( \rho(x, y) = 0 \) if and only if \( x = y \), capturing the idea that zero distance means the same location.

When defining \( \rho'(x, y) = \min(1, \rho(x, y)) \), non-negativity holds because both \( \rho(x, y) \) and the constant 1 are non-negative, hence their minimum is also non-negative.
In the case of \( \rho'(x, y) = \frac{\rho(x, y)}{1+\rho(x, y)} \), since \( \rho(x, y) \geq 0 \), the fraction remains non-negative. This ensures that both variations of \( \rho' \) inherit the non-negativity of the original metric \( \rho \).
Remember, non-negativity not only reassures that the concept is grounded in real-world representation but also helps maintain the logical structure of the metric properties.

Symmetry in Distance Metrics

The symmetry property in a metric space is quite simple yet crucial. It states that the distance between two points is the same regardless of the order in which you measure it: \( \rho(x, y) = \rho(y, x) \). Think of symmetry as a two-way street — the route from point \( x \) to \( y \) is the same as from \( y \) to \( x \).

When you redefine the metric as \( \rho'(x, y) = \min(1, \rho(x, y)) \), the symmetry naturally carries over because the function \( \min \) doesn't depend on the order of its inputs, particularly when the original \( \rho \) is symmetric.
Likewise, for the metric \( \rho'(x, y) = \frac{\rho(x, y)}{1+\rho(x, y)} \), symmetry is preserved. Mathematically, the equation holds true when swapping \( x \) and \( y \), as both the numerator and the denominator remain unchanged with this swap.
This symmetric property ensures that we can treat any pair of points equally, which is particularly helpful in defining consistent and predictable distances across the metric space.

Understanding the Triangle Inequality

The triangle inequality is a powerful tool in metric space theory. It essentially structures how distances between points relate to one another. Mathematically, it states: \( \rho(x, z) \leq \rho(x, y) + \rho(y, z) \). In simpler terms, the direct route from \( x \) to \( z \) should be shorter or equal to any indirect path, going through another point \( y \).

For the first variation, \( \rho'(x, y) = \min(1, \rho(x, y)) \), the triangle inequality adapts as follows: \( \rho'(x, z) \leq \min(1, \rho(x, z)) \leq \rho'(x, y) + \rho'(y, z) \). The minimum function ensures that even in capped distances, the inequality holds.

In the case of the transformed metric \( \rho'(x, y) = \frac{\rho(x, y)}{1+\rho(x, y)} \), proving the triangle inequality involves some algebraic manipulation, yet follows conceptually similar reasoning. The transformation maintains that any potential shortcuts in the triangle are captured.

• If you consider three points and the possibility of moving between them through another point, the inequality captures the intuitive notion that the detour should not be less attractive than the direct path.
• This ensures that the triangle remains compact and logically structured, aiding practical applications like navigation and clustering within metric spaces.

Triange Inequality Simplified

Understanding the triangle inequality is key to grasping the structure of distance within metric spaces. It states that: \( \rho(x, z) \leq \rho(x, y) + \rho(y, z) \). Consider it like the shortest path guidance: moving from \( x \) to \( z \) should not exceed the detour through any third point \( y \).

In the transformed case \( \rho'(x, y) = \min(1, \rho(x, y)) \), the minimum function holds the triangle inequality by ensuring no distance exceeds 1, logically maintaining the structure.
With \( \rho'(x, y) = \frac{\rho(x, y)}{1+\rho(x, y)} \), maintaining inequality requires algebra, yet intuitively, it holds because the relation to the original \( \rho \) is proportionately maintained.

• This property is like ensuring pathways in a GPS that favor direct routes, avoiding unnecessary deviations.
• It’s crucial for ensuring distances in metric spaces align with real-world or computational pathways.