Question

Let \(M\) be the set of all positive integers together with the "point" \(\infty\). Metrize \(M\) by setting \(\rho(m, n)=\left|\frac{1}{m}-\frac{1}{n}\right|,\) with the convention that \(\frac{1}{\infty}=0\). Verify the metric axioms. Describe \(G_{\infty}\left(\frac{1}{2}\right), S_{\infty}\left(\frac{1}{2}\right),\) and \(G_{1}(1) .\)

Short answer

The metric \(\rho\) satisfies all axioms. \(G_{\infty}(\frac{1}{2}) = \{n > 2\}\), \(S_{\infty}(\frac{1}{2}) = \{n \geq 2\}\), \(G_1(1)\) is all of \(M\).

Step-by-step solution

Verify Non-negativity and Identity

To verify non-negativity and identity, recall a metric \( \rho(a, b) \) must satisfy \( \rho(m, n) \geq 0 \) and \( \rho(m, n) = 0 \) if and only if \( m = n \). Consider the formula \( \rho(m, n) = \left| \frac{1}{m} - \frac{1}{n} \right| \). Since it's the absolute value, it's clearly non-negative. Also, if \( \rho(m, n) = 0 \), then \( \frac{1}{m} - \frac{1}{n} = 0 \), implying that \( \frac{1}{m} = \frac{1}{n} \), which occurs if and only if \( m = n \). Hence, the identity axiom is verified.

Verify Symmetry

A metric \( \rho(a, b) \) must satisfy symmetry: \( \rho(m, n) = \rho(n, m) \). In our case, \( \rho(m, n) = \left| \frac{1}{m} - \frac{1}{n} \right| \). Notice that swapping \( m \) and \( n \) yields \( \left| \frac{1}{n} - \frac{1}{m} \right| \), which is equal to the original expression \( \left| \frac{1}{m} - \frac{1}{n} \right| \) because the absolute value of a difference is the same regardless of order.

Verify Triangle Inequality

The triangle inequality requires that \( \rho(m, n) \leq \rho(m, p) + \rho(p, n) \) for any \( m, n, p \in M \). Consider \[ \rho(m, n) = \left| \frac{1}{m} - \frac{1}{n} \right| \leq \left| \frac{1}{m} - \frac{1}{p} \right| + \left| \frac{1}{p} - \frac{1}{n} \right| = \rho(m, p) + \rho(p, n). \] Use the properties of absolute values \( |a-b| \leq |a-c| + |c-b| \) to justify this. Since it's true for real numbers, it holds here as well.

Describe the Open Ball \\( G_\infty(\frac{1}{2})\\)

An open ball \( G_\infty(\frac{1}{2}) \) is defined by \( \{ n \in M \, | \, \rho(\infty, n) < \frac{1}{2} \} \). Calculate \( \rho(\infty, n) = \left| \frac{1}{\infty} - \frac{1}{n} \right| = \frac{1}{n} \). Thus, we need \( \frac{1}{n} < \frac{1}{2} \), which means \( n > 2 \). Therefore, this set contains all integers greater than 2.

Describe the Closed Ball \\( S_\infty(\frac{1}{2})\\)

A closed ball \( S_\infty(\frac{1}{2}) \) includes \( n \in M \) such that \( \rho(\infty, n) \leq \frac{1}{2} \). This translates to \( \frac{1}{n} \leq \frac{1}{2} \), giving \( n \geq 2 \). Hence, \( S_\infty(\frac{1}{2}) \) includes all integers \( n \geq 2 \).

Describe the Open Ball \\( G_1(1)\\)

For \( G_1(1) \), find the set \( \{ n \in M \, | \, \rho(1, n) < 1 \} \). Here, \( \rho(1, n) = \left| \frac{1}{1} - \frac{1}{n} \right| = \left| 1 - \frac{1}{n} \right| < 1 \), leading to \(-1 < 1 - \frac{1}{n} < 1\). Solving \( -1 < 1 - \frac{1}{n} \) gives \( \frac{1}{n} < 2 \) or \( n > \frac{1}{2} \), which includes all positive integers and \( \infty \). Thus, \( \{ n \geq 1 \} \).

Summary

M is a metric space with metric \(\rho(m, n)\). The sets are: \(G_{\infty}(\frac{1}{2})\) is \(\{n > 2\}\), \(S_{\infty}(\frac{1}{2})\) is \(\{n \geq 2\}\), and \(G_1(1)\) is the entire space \(M\).

Key concepts

Metric Axioms

A metric space is a set paired with a metric, which is a function defining the distances between elements of the set. For a function \( \rho(a, b) \) to be considered a metric, it must satisfy three key axioms: non-negativity, identity of indiscernibles, and symmetry.

• Non-negativity: This means \( \rho(m, n) \geq 0 \) for all elements \( m, n \). Our given metric \( \rho(m, n) = \left| \frac{1}{m} - \frac{1}{n} \right| \) is an absolute value, which ensures it's never negative.


• Identity of Indiscernibles: This states that \( \rho(m, n) = 0 \) if and only if \( m = n \). With the given metric, \( \left| \frac{1}{m} - \frac{1}{n} \right| = 0 \) implies \( \frac{1}{m} = \frac{1}{n} \), which happens only if \( m = n \).


• Symmetry: This requires \( \rho(m, n) = \rho(n, m) \). For our metric, swapping \( m \) and \( n \) leaves the expression unchanged, as the absolute value is indifferent to order.

Together, these properties establish a function as a metric, laying the foundation for exploring more complex properties and structures within metric spaces.

Open Ball

An open ball in a metric space is a set of points that are within a certain distance from a specific point, yet not including the boundary.
In our context, consider the open ball \( G_\infty(\frac{1}{2}) \), which is defined as the set \( \{ n \in M \mid \rho(\infty, n) < \frac{1}{2} \} \).
Since \( \rho(\infty, n) = \left| \frac{1}{\infty} - \frac{1}{n} \right| = \frac{1}{n} \), the condition \( \frac{1}{n} < \frac{1}{2} \) translates to \( n > 2 \).
Thus, the open ball \( G_\infty(\frac{1}{2}) \) includes all positive integers greater than 2. This concept highlights the idea of points being close to \( \infty \), excluding any boundary points like 2.

Closed Ball

A closed ball in a metric space similarly gathers points around a central point, but includes points exactly at the boundary.
In this situation, the closed ball \( S_\infty(\frac{1}{2}) \) is defined as \( \{ n \in M \mid \rho(\infty, n) \leq \frac{1}{2} \} \).
This requirement translates into \( \frac{1}{n} \leq \frac{1}{2} \), which means \( n \geq 2 \).
Consequently, the closed ball \( S_\infty(\frac{1}{2}) \) includes every positive integer starting from 2 and exceeding it. This inclusion of the boundary point 2 differentiates it from the open ball, offering an expansive view of the neighborhood around \( \infty \).

Triangle Inequality

The triangle inequality is a critical property of metric spaces, reflecting how distances relate when a third point is introduced.
It mandates that for any points \( m, n, \) and \( p \), the metric satisfies \( \rho(m, n) \leq \rho(m, p) + \rho(p, n) \). This principle ensures that, intuitively, going from \( m \) to \( n \) directly is at most as long as going through an intermediate \( p \).
For the metric \( \rho(m, n) = \left| \frac{1}{m} - \frac{1}{n} \right| \), the inequality \( \left| \frac{1}{m} - \frac{1}{n} \right| \leq \left| \frac{1}{m} - \frac{1}{p} \right| + \left| \frac{1}{p} - \frac{1}{n} \right| \) holds due to the nature of absolute values, echoing the triangle inequality theorem \( |a-b| \leq |a-c| + |c-b| \) for real numbers.
This overarching concept enforces a sturdy framework within which complex spatial relationships can be consistently understood and applied in metric spaces.