Chapter 1: Problem 1
Show that if \(d\) is a metric on \(X\), then both \(\bar{d}=d /(1+d)\) and \(\bar{d}=\) \(\min (1, d)\) are also metrics and that they are equivalent to \(d\) (i.e., the identity map \(1:(X, d) \rightarrow(X, \bar{d})\) is a homeomorphism).
Short Answer
Expert verified
Both \(\bar{d} = d /(1+d)\) and \(\bar{d} = \min (1, d)\) are metrics and equivalent to \(d\).
Step by step solution
01
Verify \(\bar{d} = d / (1 + d)\) is a metric
To show \(\bar{d}\) is a metric, verify the three metric properties: non-negativity, identity of indiscernibles, symmetry, and triangle inequality. 1. Non-negativity: \(\bar{d}(x, y) = d(x, y) / (1 + d(x, y)) \geq 0\). 2. Identity of indiscernibles: \(\bar{d}(x, y) = 0 \iff d(x, y) = 0\). 3. Symmetry: \(\bar{d}(x, y) = d(x, y) / (1 + d(x, y)) = d(y, x) / (1 + d(y, x)) = \bar{d}(y, x)\). 4. Triangle inequality: \(\bar{d}(x, z) \leq \bar{d}(x, y) + \bar{d}(y, z)\) follows from the fact that \(f(x) = \frac{d(x, y)}{1 + d(x, y)}\) is concave and non-increasing.
02
Verify \(\bar{d} = \min (1, d)\) is a metric
To show \(\bar{d}\) is a metric, again verify the three metric properties: 1. Non-negativity: \(\bar{d}(x, y) = \min (1, d(x, y)) \geq 0\). 2. Identity of indiscernibles: \(\bar{d}(x, y) = 0 \iff d(x, y) = 0\). 3. Symmetry: \(\bar{d}(x, y) = \min (1, d(x, y)) = \min (1, d(y, x)) = \bar{d}(y, x)\). 4. Triangle inequality: \(\bar{d}(x, z) \leq \min (1, d(x, y) + d(y, z)) \leq \min (1, d(x, y)) + \min (1, d(y, z)) = \bar{d}(x, y) + \bar{d}(y, z)\).
03
Show equivalence of \(d\) and \(\bar{d}\)
To confirm equivalence, demonstrate that the identity map is a homeomorphism: \(1:(X, d) \rightarrow(X, \bar{d})\).For \(\bar{d} = \frac{d}{1 + d}\): Consider the open ball \(B_r^d (x)\). Since \(\frac{d}{1 + d}\) is monotonically increasing and continuous, any \(\bar{d}\)-neighborhood around \(x\) corresponds to a \(d\)-neighborhood.For \(\bar{d} = \min (1, d)\): The identity map preserves openness since open sets in \(d\)-metric space translate directly into \(\bar{d}\)-metric space.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Metric Properties
In differential geometry, a metric is a function that defines a distance between elements in a space. For a metric, certain properties must be satisfied:
- Non-negativity: The distance between two points is always non-negative: \( d(x, y) \geq 0 \).
- Identity of indiscernibles: The distance is zero if and only if the two points are the same: \( d(x, y) = 0 \iff x = y \).
- Symmetry: The distance between two points is the same in either direction: \( d(x, y) = d(y, x) \).
- Triangle Inequality: The direct distance between two points is less than or equal to the sum of the distances via a third point: \( d(x, z) \leq d(x, y) + d(y, z) \).
Homeomorphism
A homeomorphism is a special type of function between topological spaces, particularly those with metric properties. It ensures that two spaces are topologically the same, meaning there exists a continuous, bijective function with a continuous inverse function that maps one space onto the other. In other words, a homeomorphism maintains the structure of the space and translates it exactly.
For two metrics to be considered equivalent, the identity map \( 1:(X, d) \rightarrow (X, \bar{d}) \) must be a homeomorphism. This means that every open set in the original metric \( d \) corresponds to an open set in the new metric \( \bar{d} \). For our exercise, confirming this property for both \( \bar{d} = \frac{d}{1 + d} \) and \( \bar{d} = \min(1, d) \) assures us that they are indeed equivalent to the original metric \( d \). Essentially, a homeomorphism allows us to argue that the spaces defined by these metrics are the same in terms of topology.
For two metrics to be considered equivalent, the identity map \( 1:(X, d) \rightarrow (X, \bar{d}) \) must be a homeomorphism. This means that every open set in the original metric \( d \) corresponds to an open set in the new metric \( \bar{d} \). For our exercise, confirming this property for both \( \bar{d} = \frac{d}{1 + d} \) and \( \bar{d} = \min(1, d) \) assures us that they are indeed equivalent to the original metric \( d \). Essentially, a homeomorphism allows us to argue that the spaces defined by these metrics are the same in terms of topology.
Open Sets
In topology, an open set is a set that, informally speaking, does not include its boundary. Within any open set, you can 'move a little bit' without leaving the set. Open sets play a fundamental role in defining topological spaces and understanding how metrics relate to continuity.
To confirm that the identity map between \( d \) and \( \bar{d} \) is a homeomorphism, we need to show that the image of any open set in \( (X, d) \) remains open in \( (X, \bar{d}) \). For the specific cases in our exercise:
To confirm that the identity map between \( d \) and \( \bar{d} \) is a homeomorphism, we need to show that the image of any open set in \( (X, d) \) remains open in \( (X, \bar{d}) \). For the specific cases in our exercise:
- For \( \bar{d} = \frac{d}{1 + d} \), the mapping preserves openness because the function \( \frac{d}{1 + d} \) is monotonically increasing and continuous.
- For \( \bar{d} = \min(1, d) \), any open set in \( d \)-metric space corresponds directly to an open set in the \( \bar{d} \)-metric space since it preserves the bounded distances without altering the openness property.
Triangle Inequality
One key aspect of a metric is the triangle inequality. This property states that for any three points \( x, y, \text{and} z \) in the space, the direct distance from \( x \) to \( z \) is never greater than the sum of the distances from \( x \) to \( y \) and from \( y \) to \( z \): \( d(x, z) \leq d(x, y) + d(y, z) \). This property ensures that the shortest distance between two points is a straight path, aligning with our intuitive understanding of distance.
When examining both \( \bar{d} = \frac{d}{1 + d} \) and \( \bar{d} = \min(1, d) \), we must verify they adhere to this principle:
When examining both \( \bar{d} = \frac{d}{1 + d} \) and \( \bar{d} = \min(1, d) \), we must verify they adhere to this principle:
- For \( \bar{d}(x, z) = \frac{d(x,z)}{1 + d(x, z)} \), we use the fact that \( \frac{d(x, y)}{1 + d(x, y)} \) is a concave, non-increasing function, ensuring that the sum of transformed distances does not exceed \( \bar{d}(x,z) \).
- For \( \bar{d}(x, z) = \min(1, d(x,z)) \), the new metric directly reflects the strong triangle inequality property because \( \min(1, d(x, y) + d(y, z)) \) never exceeds \( \bar{d}(x, y) + \bar{d}(y, z)\).