Problem 6 - Basic Analysis: Introduction to Real Analysis

Question

Let \(\left(X, d_{X}\right)\) and \(\left(Y, d_{Y}\right)\) be metric spaces. a) Show that \((X \times Y, d)\) with \(d\left(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)\right):=d_{X}\left(x_{1}, x_{2}\right)+d_{Y}\left(y_{1}, y_{2}\right)\) is a metric space. b) Show that \((X \times Y, d)\) with \(d\left(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)\right):=\max \left\\{d_{X}\left(x_{1}, x_{2}\right), d_{Y}\left(y_{1}, y_{2}\right)\right\\}\) is a metric space.

Step-by-Step Solution

Answer

Both definitions satisfy the metric properties, confirming that they are metrics.

Step 1: Understanding Metrics

A metric space is a set equipped with a function (distance) defined between any pair of elements, satisfying positivity, symmetry, and the triangle inequality. Let's verify these properties for the given functions.

Step 2: Part a: Define Positivity for Sum Metric

We are given that \(d(x_1,y_1)(x_2,y_2) = d_X(x_1,x_2) + d_Y(y_1,y_2)\). To check positivity, note that since \(d_X\) and \(d_Y\) are metrics, \(d_X(x_1,x_2) \geq 0\) and \(d_Y(y_1,y_2) \geq 0\), therefore \(d(x_1,y_1)(x_2,y_2) \geq 0\). If \(d(x_1,y_1)(x_2,y_2) = 0\), then both \(d_X(x_1,x_2) = 0\) and \(d_Y(y_1,y_2) = 0\), implying that \(x_1 = x_2\) and \(y_1 = y_2\).

Step 3: Part a: Check Symmetry for Sum Metric

For symmetry, \(d_X(x_1,x_2) = d_X(x_2,x_1)\) and \(d_Y(y_1,y_2) = d_Y(y_2,y_1)\); hence \(d((x_1,y_1),(x_2,y_2)) = d((x_2,y_2),(x_1,y_1))\).

Step 4: Part a: Verify Triangle Inequality for Sum Metric

For the triangle inequality, \(d_X(x_1,x_2) \leq d_X(x_1,x_3) + d_X(x_3,x_2)\) and \(d_Y(y_1,y_2) \leq d_Y(y_1,y_3) + d_Y(y_3,y_2)\) hold. Sum these to get: \(d((x_1,y_1),(x_2,y_2)) = d_X(x_1,x_2) + d_Y(y_1,y_2) \leq (d_X(x_1,x_3) + d_X(x_3,x_2)) + (d_Y(y_1,y_3) + d_Y(y_3,y_2)) = d((x_1,y_1),(x_3,y_3)) + d((x_3,y_3),(x_2,y_2))\).

Step 5: Conclusion for Part a

The function \(d((x_1,y_1),(x_2,y_2)) = d_X(x_1,x_2) + d_Y(y_1,y_2)\) satisfies all the metric space properties, hence is a metric.

Step 6: Part b: Define Positivity for Max Metric

The function is \(d((x_1,y_1),(x_2,y_2)) = \max\{d_X(x_1,x_2), d_Y(y_1,y_2)\}\). Since \(d_X\) and \(d_Y\) are metrics, both are non-negative, so the function is non-negative. If \(d((x_1,y_1),(x_2,y_2)) = 0\), both \(d_X(x_1,x_2) = 0\) and \(d_Y(y_1,y_2) = 0\), thus \(x_1 = x_2\) and \(y_1 = y_2\).

Step 7: Part b: Check Symmetry for Max Metric

Given that \(d_X(x_1,x_2) = d_X(x_2,x_1)\) and \(d_Y(y_1,y_2) = d_Y(y_2,y_1)\), the symmetry holds for the maximum function as well.

Step 8: Part b: Verify Triangle Inequality for Max Metric

For \(\max\), we need \(\max\{d_X(x_1,x_2), d_Y(y_1,y_2)\}\) to satisfy the triangle inequality. Since each component individually satisfies it, \(d_X(x_1,x_2) \leq d_X(x_1,x_3) + d_X(x_3,x_2)\) and \(d_Y(y_1,y_2) \leq d_Y(y_1,y_3) + d_Y(y_3,y_2)\), then \(\max\{d_X(x_1,x_2), d_Y(y_1,y_2)\}\) \(\leq \max\{d_X(x_1,x_3) + d_X(x_3,x_2), d_Y(y_1,y_3) + d_Y(y_3,y_2)\}\).

Step 9: Conclusion for Part b

The function \(d((x_1,y_1),(x_2,y_2)) = \max\{d_X(x_1,x_2), d_Y(y_1,y_2)\}\) is verified to satisfy all metric space properties and thus qualifies as a metric.

Key Concepts

Positivity in Metric Spaces

In the realm of metric spaces, positivity is an essential property of distance functions. A metric, which represents the distance between two points, must always be non-negative. This means that for any two points, the metric, often denoted as "d", must satisfy the condition \(d(x_1, x_2) \geq 0\). This requirement ensures that distances are meaningful and represent the actual "gap" or "span" between points, both physically and conceptually.

Consider the metric defined by the sum: \(d((x_1, y_1), (x_2, y_2)) = d_X(x_1, x_2) + d_Y(y_1, y_2)\). Because both \(d_X\) and \(d_Y\) individually satisfy the positivity condition (i.e., they are non-negative), their sum also remains non-negative. Thus, the overall distance function retains the positivity property.

Another example is the maximum metric: \(d((x_1, y_1), (x_2, y_2)) = \max\{d_X(x_1, x_2), d_Y(y_1, y_2)\}\). The maximum of two non-negative numbers is also non-negative, fulfilling the positivity condition for this metric space as well. This ensures that such metrics are valid distances.

Symmetry in Metric Spaces

A core characteristic of metric spaces is symmetry, which means the distance between two points is the same, regardless of the order of those points. Mathematically, this property is expressed as \(d(x_1, x_2) = d(x_2, x_1)\).

For the sum metric, \(d((x_1, y_1), (x_2, y_2)) = d_X(x_1, x_2) + d_Y(y_1, y_2)\), symmetry is easily verified. Since both \(d_X\) and \(d_Y\) are symmetric functions in themselves \(d_X(x_1, x_2) = d_X(x_2, x_1)\) and \(d_Y(y_1, y_2) = d_Y(y_2, y_1)\), their sum retains the symmetry. Thus, the sum metric respects the symmetry property required of all metric spaces.

Symmetry also holds for the max metric \(d((x_1, y_1), (x_2, y_2)) = \max\{d_X(x_1, x_2), d_Y(y_1, y_2)\}\). The maximum function doesn't alter the order when selecting between two values, so symmetry is preserved, highlighting that both formulations abide by this essential requirement.

Triangle Inequality

One of the pivotal properties of metric spaces is the triangle inequality. This rule asserts that the direct distance between two points should always be less than or equal to the distance when travelling through a third point. In a formula, this states \(d(x_1, x_2) \leq d(x_1, x_3) + d(x_3, x_2)\).

For the sum metric form \(d((x_1, y_1), (x_2, y_2)) = d_X(x_1, x_2) + d_Y(y_1, y_2)\), the triangle inequality holds due to the inequalities in each component: \(d_X(x_1, x_2) \leq d_X(x_1, x_3) + d_X(x_3, x_2)\) and \(d_Y(y_1, y_2) \leq d_Y(y_1, y_3) + d_Y(y_3, y_2)\). Adding these together confirms the overall inequality.

The max metric \(d((x_1, y_1), (x_2, y_2)) = \max\{d_X(x_1, x_2), d_Y(y_1, y_2)\}\) also satisfies the triangle inequality because each component individually adheres to it. The max is simply the larger of the two distances, ensuring that it does not exceed the sum of the third party distances. These constructs repeatedly affirm the dependency of metric spaces upon adhering to the triangle inequality.