Problem 11 - Advanced Calculus

Question

Let \(f\) be continuous on the interval \([a, b] .\) Given \(\varepsilon>0\) and a point \(t\) in the interval, choose \(\rho=\rho(t)\) so that, if \(|x-t|<\rho\), then \(|f(x)-f(t)|<\varepsilon\). Let \(U_{t}\) be the symmetric interval centered on \(t\) of radius \(\frac{1}{2} \rho(t) .\) Show that there are points \(t_{1}, t_{2}, \cdots, t_{m}\) such that the sets \(U_{t,}\) together cover the interval \([a, b]\).

Step-by-Step Solution

Answer

For a continuous function on a closed interval, it is always possible to find a finite number of “special” intervals \(U_{t}\) such that they collectively cover the original interval.

Step 1: Defining the Intervals

Since \(f\) is continuous on \([a, b]\), for every \(t\) in \([a, b]\), we can find a corresponding \(\rho (t)>0\) such that, if \(|x-t|<\rho(t)\), then \(|f(x)-f(t)|<\varepsilon\). Now, for each \(t\) in \([a, b]\), form an interval \(U_t\) with radius equal to \(1/2\rho(t)\), meaning \(U_t=(t-\frac{1}{2}\rho(t),t+\frac{1}{2}\rho(t))\).

Step 2: Construction of the Covering Set

Start at the left endpoint \(a\) of the interval \([a, b]\). There is an interval \(U_{t_1}\) containing \(a\). Let \(t_1\) be the right endpoint of this interval. If \(t_1

Step 3: Covering the Interval

By construction, each point in the interval \([a, b]\) lies in at least one of the intervals \(U_{t_i}\). Thus \([a, b]\) is the union of these \(U_{t_i}\), meaning the sets \(U_{t_i}\) successfully cover the interval \([a, b]\).

Key Concepts

Open Cover

An open cover is a way to cover an entire space using open sets. In our exercise, we're looking at the interval \( [a, b] \), covered by open intervals \( U_t \). Each \( U_t \) is centered on a point \( t \) and extends to a certain radius, defined as \( \frac{1}{2}\rho(t) \). This radius ensures no gaps occur in the coverage.

Here's how it works:

• Given any point \( t \) in \( [a, b] \), you get a small interval around it.
• These intervals, when arranged correctly, form an open cover that spans the whole main interval.
• The objective is to create a chain so each part of the interval \( [a, b] \) is inside at least one \( U_t \).

By carefully selecting points and constructing subsequent intervals, we ensure each part of \( [a, b] \) is covered neatly.Open covers are fundamental in topology and real analysis, helping us understand properties like compactness and continuity.

Uniform Continuity

Uniform continuity strengthens the idea of continuity by requiring a more consistent behavior over its entire domain. For a function \( f \) on an interval \( [a, b] \), uniform continuity means that the distance between the function's outputs is small whenever the inputs are sufficiently close, no matter where you are in the interval.

What's special about uniform continuity?

• It doesn't rely on varying the distance based on points, unlike normal continuity, which can change with each point.
• For any given \( \varepsilon > 0 \), you can choose a single \( \delta \) that works for the entire interval.

In the given exercise, the process involves determining \( \rho(t) \) for each point such that \( |x-t| < \rho(t) \) ensures \( |f(x) - f(t)| < \varepsilon \). This speaks to the fact that \( f \) behaves consistently over the entire interval.

Uniform continuity is particularly valuable when working with closed and bounded intervals and assures us about the function's limits and bounds.

Archimedean Property

The Archimedean Property is an essential concept in real analysis that helps us understand how numbers work. It states that given any two positive numbers, there exists an integer that can be multiplied with one of the numbers to exceed the other. This common sense truth means no gap is infinitely large compared to our measuring stick of 1.

How does this apply to our exercise?

• As we select points \( t_1, t_2, \ldots \) along the interval, the Archimedean Property ensures the process doesn't continue infinitely without covering \( [a, b] \).
• It guarantees that a finite number of intervals \( U_{t_i} \) can cover the whole space from \( a \) to \( b \), thanks to the bounded nature of the interval.

The Archimedean Property reassures us that by building one interval at a time, we will eventually complete our coverage task, filling the entire interval \( [a, b] \) efficiently. This concept provides a bridge between infinite processes and finite results.

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Problem 10

Next

Problem 12

Other exercises in this chapter

Problem 10

Let \(p_{1}, p_{2}, p_{3}\) be the vertices of a triangle \(D\) in the plane. (a) Show that any point \(p\) in \(D\) can be expressed as $$ p=\alpha_{1} p_{1}+\

View solution

Problem 10

Give a mathematical argument to show that a heated wire in the shape of a circle must always have two diametrically opposite points with the same temperature.

View solution

Problem 12

How are \(f^{-1}(A \cap B)\) and \(f^{-1}(A \cup B)\) related to \(f^{-1}(A)\) and \(f^{-1}(B)\) ?

View solution

Problem 13

Let \(f: X \rightarrow Y\) be a function, and \(S\) and \(T\) arbitrary sets. Show that: (a) \(f f^{-1}(S) \subset S\) (b) \(T \subset f^{-1} f(T)\)

View solution

Key Concepts

Open Cover

An open cover is a way to cover an entire space using open sets. In our exercise, we're looking at the interval \( [a, b] \), covered by open intervals \( U_t \). Each \( U_t \) is centered on a point \( t \) and extends to a certain radius, defined as \( \frac{1}{2}\rho(t) \). This radius ensures no gaps occur in the coverage.

Here's how it works:

• Given any point \( t \) in \( [a, b] \), you get a small interval around it.
• These intervals, when arranged correctly, form an open cover that spans the whole main interval.
• The objective is to create a chain so each part of the interval \( [a, b] \) is inside at least one \( U_t \).

By carefully selecting points and constructing subsequent intervals, we ensure each part of \( [a, b] \) is covered neatly.Open covers are fundamental in topology and real analysis, helping us understand properties like compactness and continuity.

Uniform Continuity

Uniform continuity strengthens the idea of continuity by requiring a more consistent behavior over its entire domain. For a function \( f \) on an interval \( [a, b] \), uniform continuity means that the distance between the function's outputs is small whenever the inputs are sufficiently close, no matter where you are in the interval.

What's special about uniform continuity?

• It doesn't rely on varying the distance based on points, unlike normal continuity, which can change with each point.
• For any given \( \varepsilon > 0 \), you can choose a single \( \delta \) that works for the entire interval.

In the given exercise, the process involves determining \( \rho(t) \) for each point such that \( |x-t| < \rho(t) \) ensures \( |f(x) - f(t)| < \varepsilon \). This speaks to the fact that \( f \) behaves consistently over the entire interval.

Uniform continuity is particularly valuable when working with closed and bounded intervals and assures us about the function's limits and bounds.

Archimedean Property

The Archimedean Property is an essential concept in real analysis that helps us understand how numbers work. It states that given any two positive numbers, there exists an integer that can be multiplied with one of the numbers to exceed the other. This common sense truth means no gap is infinitely large compared to our measuring stick of 1.

How does this apply to our exercise?

• As we select points \( t_1, t_2, \ldots \) along the interval, the Archimedean Property ensures the process doesn't continue infinitely without covering \( [a, b] \).
• It guarantees that a finite number of intervals \( U_{t_i} \) can cover the whole space from \( a \) to \( b \), thanks to the bounded nature of the interval.

The Archimedean Property reassures us that by building one interval at a time, we will eventually complete our coverage task, filling the entire interval \( [a, b] \) efficiently. This concept provides a bridge between infinite processes and finite results.