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Chapter 3: Problem 32

Define a relation \(R\) on \(\mathbf{R}^{\mathbf{R}},\) the set of functions from \(\mathbf{R}\) to \(\mathbf{R}\), by \(f R g\) if \(f(0)=g(0)\). Prove that \(R\) is an equivalence relation on \(\mathbf{R}^{\mathbf{R}}\). Let \(f(x)=x\) for all \(x \in \mathbf{R}\). Describe \([f]\).

Short Answer

Expert verified
The relation \(R\) is an equivalence relation on \(\mathbb{R}^\mathbb{R}\) because it satisfies reflexivity, symmetry, and transitivity. The equivalence class of \(f\) under \(R\), \([f]\), is the set of all functions \(g(x)\) in \(\mathbb{R}^\mathbb{R}\) such that \(g(0) = 0\).

Step by step solution

01

Proving Reflexivity

Try and prove that the relation \(R\) is reflexive. Remember that a relation is reflexive if, for all elements \(a\) in the set, \(aRa\) holds true. Here, we need to prove that for all functions \(f\) in the set \(\mathbb{R}^\mathbb{R}\), \(fRf\) holds true, which means that \(f(0) = f(0)\). This is clearly true, since any element (in this case any function) is equal to itself. Thus, \(R\) is reflexive.
02

Proving Symmetry

Next, try and prove that the relation \(R\) is symmetric. A relation is symmetric if for any two elements \(a\) and \(b\) in the set, \(aRb\) implies that \(bRa\). In this scenario, we need to prove that if \(fRg\) then \(gRf\) for all functions \(f, g \in \mathbb{R}^\mathbb{R}\). Since \(fRg\) means \(f(0) = g(0)\), the reverse, \(g(0) = f(0)\), is also true. Consequently, \(gRf\) is proven, hence \(R\) is symmetric.
03

Proving Transitivity

Finally, you should aim to prove that the relation \(R\) is transitive. A relation is transitive if, for any three elements \(a\), \(b\), and \(c\) in the set, \(aRb\) and \(bRc\) implies that \(aRc\). In this situation we need to prove that if \(fRg\) and \(gRh\) then \(fRh\) for all functions \(f, g, h \in \mathbb{R}^\mathbb{R}\). If \(fRg\) and \(gRh\), it means that \(f(0) = g(0)\) and \(g(0) = h(0)\). Therefore, \(f(0) = h(0)\), which implies \(fRh\). Accordingly, \(R\) is transitive.
04

Describing [f]

By definition, the equivalence class of \(f\) under the equivalence relation \(R\), denoted \([f]\), is the set of all elements in \(\mathbb{R}^\mathbb{R}\) that are related to \(f\) via \(R\). Since the equivalence relation \(R\) is defined by \(fRg\) if \(f(0) = g(0)\), this means that any function \(g\) that satisfies \(g(0) = f(0)\) belongs to \([f]\). As \(f(x) = x\), it means \(f(0) = 0\). Therefore, \([f]\) is the set of all functions \(g(x)\) in \(\mathbb{R}^\mathbb{R}\) such that \(g(0) = 0\).

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Key Concepts

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

Reflexivity
Reflexivity is a fundamental property of an equivalence relation. To say that a relation is reflexive, it means that every element in the set is related to itself.
In our example, we are considering functions from real numbers to real numbers, described as \(\mathbb{R}^\mathbb{R}\). The relationship \(R\) specifies that two functions \(f\) and \(g\) are related if they equal at the point 0; thus, \(f(0) = g(0)\).
To prove that \(R\) is reflexive, we must show that for every function \(f\) in our set, it holds that \(f(0) = f(0)\). Clearly, this is true because a function evaluated at the same point twice yields the same result. Reflexivity is, therefore, quite straightforward to establish in this case.
As a bullet point summary:
  • Reflexivity implies self-relation for every function \(f(x)\).
  • For relation \(R\), it checks \(f(0) = f(0)\) for all functions.
  • This property assures every function relates to itself at 0.
Symmetry
The property of symmetry in a relation suggests that if one element is related to another, then the reverse must also be true.
Specifically, for the relation \(R\) on functions from \(\mathbb{R}^\mathbb{R}\), symmetry must show that whenever \(fRg\), then \(gRf\) is also valid. In simpler terms, if two functions agree at 0, then changing their order does not affect their relation.
To prove symmetry: Start with \(fRg\), meaning \(f(0) = g(0)\). From this equality, it naturally follows that \(g(0) = f(0)\), which implies \(gRf\). Thus, the relation is symmetric.
Quick symmetry recap:
  • Symmetry checks for reciprocal relationship.
  • For functions, it looks at \(fRg\) and asks if \(gRf\) follows.
  • Being symmetric means relation holds in both directions.
Transitivity
Transitivity is essential for a relation to be classified as an equivalence relation. It demands that if one element relates to a second and that second relates to a third, the first must relate to the third.
For the function relation \(R\), transitivity needs to show: if \(fRg\) and \(gRh\), then \(fRh\) must be true. Suppose we know \(f(0)=g(0)\) and \(g(0)=h(0)\). By substituting, it directly follows that \(f(0)=h(0)\), fulfilling the transitivity requirement.
In sum, transitivity:
  • Ensures that relationships propagate across three or more elements.
  • In our relation, if \(fRg\) and \(gRh\), then \(fRh\) completes the chain.
  • Transitivity confirms consistent chaining of relationships.

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