Problem 8 - Advanced Calculus

Question

Use the example \(f(x)=x^{2} /\left(1+x^{2}\right)\) to show that a continuous function does not always have to map a closed set onto a closed set.

Step-by-Step Solution

Answer

Using the provided example function \(f(x)=x^{2} /\left(1+x^{2}\right)\), it was demonstrated that a continuous function does not always map a closed set [-1, 1] onto a closed set. The image of the function over the interval [-1, 1] is [0,1], which does not include the limit point of 1, thus proving the argument.

Step 1: Understand the function and the concept of a closed set

Firstly, it's important to understand the given function \(f(x)=x^{2} /\left(1+x^{2}\right)\). Another crucial concept to grasp is that of a closed set, which in the simplest terms, is a set that includes all its limit points.

Step 2: Evaluate the function on a closed set

Let's evaluate this function over the closed set [-1, 1]. You can plug these values into the function to see what results it yields.

Step 3: Check the image of the function

Now that values are obtained, we look at the image of function \(f(x)\) over the interval [-1,1], it works out to be [0,1]. Comparatively, [0,1] is a relatively smaller range than [-1, 1].

Step 4: Make the conclusion

Since the image of the function on the interval [-1, 1] does not include the limit point '1', we see that the closed interval [-1, 1] is not mapped to a closed set [0, 1] by function \(f\). Therefore, this example shows that a continuous function does not always have to map a closed set onto a closed set.

Key Concepts

Closed Sets

In mathematics, a closed set is a fundamental concept in topology. A set is termed "closed" if it includes all its limit points. This means, roughly speaking, if you can approach an edge of the set without actually leaving it, that same edge (limit point) must be part of the set. Closed sets in the real number line typically contain their boundary points, such as the interval

• Closed intervals like \([-1, 1]\) always include their endpoints.
• Examples: More common examples include shapes like circles or rectangular regions fully contained within a plane, including their edges.

Understanding closed sets is crucial when dealing with functions and mappings. It helps predict how the behavior of a function will interact with the properties of the set.

Image of a Function

When we talk about the image of a function, we're referring to the set of output values (or range) that the function produces from a given set of input values. For a function like \(f(x)=\frac{x^2}{1+x^2}\), if you input values from a closed set such as \([-1, 1]\), the resulting outputs form what is termed the "image" of this function over that input range.

• For instance, when evaluating our given function over the closed interval \([-1, 1]\), we find that the image is \([0, 1]\).
• This implies that although our inputs were from a closed set, the resulting image does not necessarily form a closed set because it may not include all possible boundary or limit points.

Therefore, the image of a function is a critical concept when assessing whether maps from various sets retain certain properties such as closeness.

Limit Points

Limit points, also known as accumulation points, are essential when discussing closed sets. A point is considered a limit point of a set if every neighborhood of that point contains at least one other point from the set. In simpler terms, you can get as close as you want to the limit point without ever losing contact with the elements of the set.

• For the set \([0, 1]\), all points within this range, as well as the point \'1\' itself, are limit points.
• However, if we consider the behavior of our function \(f(x)=\frac{x^2}{1+x^2}\) evaluated over \([-1, 1]\), we observe that the point \'1\' is not included in the image, which is why \([0, 1)\) is not closed.

Recognizing which points are limit points is crucial as it tells us whether a resulting range (or image) retains properties such as closeness and helps in understanding the boundaries of a mapped set.