Question

Prove: If \(f\) is nonconstant and continuous on an interval \(I,\) then the set \(S=\) \(\\{y \mid y=f(x), x \in I\\}\) is an interval. Moreover, if \(I\) is a finite closed interval, then so is \(S\)

Short answer

Question: Prove that if a function \(f\) is nonconstant and continuous on an interval \(I\), its range is an interval. Additionally, show that if \(I\) is a finite closed interval, then its range is also a finite closed interval. Answer: First, we can use the Intermediate Value Theorem (IVT) to prove that the range, \(S\), is an interval. This means that all points between any two points in \(S\) also belong to \(S\). Then, assuming that \(I\) is a finite closed interval, we can apply the Extreme Value Theorem to show that the range of \(f\) on \(I\), \(S\), is also a finite closed interval.

Step-by-step solution

Prove that \(S\) is an interval

Let's consider any two points \(y_1\) and \(y_2\) in \(S\), where \(y_1 \neq y_2\). Since \(y_1, y_2 \in S\), there must exist two corresponding points \(x_1, x_2 \in I\) such that \(y_1 = f(x_1)\) and \(y_2 = f(x_2)\). As \(f\) is nonconstant and continuous on \(I\), we can use the Intermediate Value Theorem (IVT). According to the IVT, for any \(y \in (y_1, y_2)\), there exists at least one \(x \in (x_1, x_2)\) such that \(y = f(x)\). Therefore, \(y \in S\), and we can say all the points between \(y_1\) and \(y_2\) are also in \(S\). As this holds for any two points in \(S\), we can conclude that \(S\) is an interval.

Prove that if \(I\) is a finite closed interval, \(S\) is also a finite closed interval

Assume that \(I\) is a finite closed interval, i.e., \(I = [a, b]\) for some \(a, b \in \mathbb{R}\) with \(a < b\). We need to prove that \(S\) is also a finite closed interval. Since \(f\) is continuous on \(I\) and \(I\) is a closed interval, we can apply the Extreme Value Theorem. The Extreme Value Theorem states that a continuous function on a closed interval attains its minimum and maximum values in that interval. Therefore, there exist \(c, d \in I\) such that \(f(c) \le f(x) \le f(d)\) for all \(x \in I\). Then, the range of \(f\) on \(I\) is a closed interval, i.e., \(S = [f(c), f(d)]\). This shows that if \(I\) is a finite closed interval, \(S\) is also a finite closed interval. In conclusion, if \(f\) is a nonconstant and continuous function on an interval \(I\), its range \(S\) is an interval. Moreover, if \(I\) is a finite closed interval, then so is \(S\).

Key concepts

Continuous Functions

Understanding continuous functions is essential in grasping the Intermediate Value Theorem and its applications. A continuous function is one where small changes in the input result in small changes in the output. This implies that a continuous function has no sudden jumps or breaks.
For instance, consider drawing a graph of a continuous function without lifting your pen from the paper. It illustrates how the function remains unbroken across its domain, smoothly transitioning from one value to another.
In the context of the original exercise, the function \( f \) is continuous over the interval \( I \). Such continuity ensures that between any two points on the graph of \( f \), every point in between is also included. This is crucial because it justifies the function's range forming an interval. Through continuity, tools like the Intermediate Value Theorem become powerful in understanding how functions behave over intervals.

Intervals in Real Analysis

In real analysis, understanding the nature of intervals is fundamental. Intervals can be open, closed, or half-open, depending on whether they include their endpoints.
Open intervals, like \((a, b)\), do not include the end points \(a\) and \(b\). Closed intervals, such as \([a, b]\), include both. Half-open or half-closed intervals include one endpoint but not the other, represented as \([a, b)\) or \((a, b]\).
The exercise emphasizes closed intervals because they naturally occur when applying the Extreme Value Theorem and Intermediate Value Theorem. When a function \( f \) is continuous over a closed interval \([a, b]\), it ensures the output or range \( S \) also becomes a closed interval. Such intervals play a crucial role in defining the bounds and behavior of functions in analysis.

Extreme Value Theorem

The Extreme Value Theorem is an important result in calculus. It guarantees that if a function is continuous over a finite closed interval, such as \([a, b]\), then there are points where the function reaches its maximum and minimum values within that interval.
This theorem is vital in proving that \( S \) is a finite closed interval when \( I \) is. Since \( f \) is continuous, it achieves both its highest point \( f(d) \) and its lowest point \( f(c) \) on \( I \). This range \([f(c), f(d)]\) represents all possible output values, thus confirming \( S \) as a closed interval.

• The theorem relies heavily on the continuity of \( f \), ensuring no gaps are present, allowing all maximum and minimum values to be properly captured.
• This concept is crucial in many applications, such as optimization problems where finding max/min values is critical.

Understanding how this theorem operates further cements our comprehension of continuous functions and their behavior on defined intervals.