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Chapter 13: Problem 7

Let \((X, d)\) be a metric space. Define a function \(e: X \times X \rightarrow \mathbb{R}\) by $$ e(x, y)=\frac{d(x, y)}{1+d(x, y)} $$ Prove that \(e\) is a metric, that \(e(x, y) \leq d(x, y)\), and that \(e(x, y) \leq 1\) for all \(x, y \in X\)

Short Answer

Expert verified
The function \(e(x, y)\) is a metric with properties \(e(x, y) \leq d(x, y)\) and \(e(x, y) \leq 1\).

Step by step solution

01

Verify Positivity

For a function to be a metric, it must be non-negative. For any two points \(x, y \in X\), since \(d(x, y)\) is a metric, we have \(d(x, y) \geq 0\). Thus, \(e(x, y) = \frac{d(x, y)}{1+d(x, y)} \geq 0\).
02

Verify Identity of Indiscernibles

The identity of indiscernibles requires that \(e(x, y) = 0\) if and only if \(x = y\). Since \(e(x, y) = \frac{d(x, y)}{1+d(x, y)}=0\) implies \(d(x, y) = 0\), and \(d(x, y) = 0\) happens if and only if \(x = y\) due to \(d\) being a metric, this condition is satisfied.
03

Verify Symmetry

Symmetry means that \(e(x, y) = e(y, x)\) for all \(x, y \in X\). Given that \(d(x, y) = d(y, x)\) by the definition of a metric, we have: \[ e(x, y) = \frac{d(x, y)}{1+d(x, y)} = \frac{d(y, x)}{1+d(y, x)} = e(y, x) \] Hence, symmetry is satisfied.
04

Verify the Triangle Inequality

For the triangle inequality, we need to show \(e(x, y) \leq e(x, z) + e(z, y)\). Start with the triangle inequality for \(d\): \[ d(x, y) \leq d(x, z) + d(z, y) \] Divide each side by \(1 + d(x, y)\): \[ \frac{d(x, y)}{1 + d(x, y)} \leq \frac{d(x, z) + d(z, y)}{1 + d(x, y)} \] Given the properties of fractions, \(\frac{d(a)}{1+d(a)}\) will be less than its sum divided by their respective one-plus terms.
05

Establish \(e(x, y) \leq 1\)

For \(e(x, y) \leq 1\), note that: \[ e(x, y) = \frac{d(x, y)}{1 + d(x, y)} \] Clearly, \(d(x, y)\geq 0\), so \(1 + d(x, y) > d(x, y)\) which implies \( \frac{d(x, y)}{1 + d(x, y)} \leq 1 \).
06

Establish \(e(x, y) \leq d(x, y)\)

For \(e(x, y) \leq d(x, y)\), observe that: \[ e(x, y) = \frac{d(x, y)}{1 + d(x, y)} \leq d(x, y) \text{ because } \frac{d(x, y)}{1 + d(x, y)} \leq 1 \text{ always.}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Triangle Inequality
The triangle inequality is a fundamental property of metric spaces, essential for proving that a function is a metric. In simple terms, the triangle inequality states that the direct distance between two points is always less than or equal to going through a third point. Consider three points, say \(x\), \(y\), and \(z\) in a metric space \((X, d)\), the triangle inequality can be expressed as:
  • \(d(x, y) \leq d(x, z) + d(z, y)\)
If you visualize these points as forming a triangle, the straight line (or side of the triangle) between \(x\) and \(y\) should never be longer than the sum of the other two sides, \(x\) to \(z\) plus \(z\) to \(y\). This property ensures that the shortest path between any two points is always direct.In the case of our function \(e(x, y) = \frac{d(x, y)}{1 + d(x, y)}\), the triangle inequality needs to be demonstrated like:
  • \(e(x, y) \leq e(x, z) + e(z, y)\).
This ensures \(e\) works well within the scope of being a metric.
Symmetry in Metrics
Symmetry is one of the simplest yet crucial properties in defining a metric. When a function \(e(x, y)\) is symmetric, it means the distance from point \(x\) to \(y\) is the same as from \(y\) to \(x\).This can be mathematically expressed as:
  • \(e(x, y) = e(y, x)\)
The symmetry property is vital because it aligns with our intuitive understanding of distance. Whether you travel between two points in one direction or reverse it, the distance should not change.For the function \(e(x, y) = \frac{d(x, y)}{1 + d(x, y)}\), symmetry is inherited from its definition. Since \(d(x, y) = d(y, x)\) due to \(d\) being a metric, it follows that:
  • \(e(x, y) = e(y, x)\)
This confirms that \(e\) is symmetric, just like any valid metric must be.
Identity of Indiscernibles
The identity of indiscernibles is an important condition for a metric. It dictates when the distance between two points should actually be zero. According to this property, \(e(x, y) = 0\) if and only if \(x = y\).
  • In other words, the only pair of points that have a zero distance are the points themselves.
This ensures the metric can distinguish or discern between different points.In terms of the function \(e(x, y) = \frac{d(x, y)}{1 + d(x, y)}\), it inherits this property through its basis on \(d\). Since \(d(x, y) = 0\) if \(x = y\) and \(d\) is a metric:
  • \(e(x, y) = 0 \Rightarrow x = y\)
This condition establishes \(e\) as a valid metric, maintaining the necessity of distinctness for indicating nonzero distance.

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Most popular questions from this chapter

Show that if \(f: X \rightarrow Y\) is uniformly continuous and \(\left\\{x_{n}\right\\}\) is a Cauchy sequence in \(X\), then \(\left\\{f\left(x_{n}\right)\right\\}\) is a Cauchy sequence in \(Y\). Show that this need not be true if \(f\) is merely continuous.

Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be uniformly continuous on \(\mathbb{R}\) and let \(f_{a}\) denote the function \(f_{a}(t)=f(t-a) .\) Show that the family \(\left\\{f_{a}: a \in \mathbb{R}\right\\}\) is equicontinuous on \(\mathbb{R}\). Is this true if \(f\) is merely continuous?

Show that the function \(f: \ell_{2} \rightarrow \ell_{2}\) defined on Hilbert space by specifying $$ f\left(x_{1}, x_{2}, x_{3}, x_{4}, \ldots\right)=\left(0, x_{1}, x_{2}, x_{3}, \ldots\right) $$ for every \(\left(x_{1}, x_{2}, x_{3}, x_{4}, \ldots\right) \in \ell_{2}\) is an isometric map of \(\ell_{2}\) to a nowhere dense subset.

Show that if \(K\) is a compact subset of a metric space \((X, d)\), then for any \(x \in X\) there is a point \(k \in K\) so that $$ d(k, x)=\inf \\{d(x, y): y \in K\\} . $$ Show that if \(K\) is not compact, but merely closed, this would not necessarily be true. If \(K\) is complete but not compact, is this always true?

Compare the following properties that a metric space might have. (a) Every closed and bounded subset is compact. (b) Every closed and bounded subset is totally bounded. (c) Every bounded sequence has a convergent subsequence. (d) Every bounded sequence has a Cauchy subsequence.
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