Question

For each \(n \in \mathbb{N}\), let \(I_{n}=\left\\{x|| x-1 \mid\frac{1}{n}\right\\}\) Find \(\bigcup_{n} I_{n}\) and \(\bigcap_{n} I_{n}\).

Short answer

The union \(\bigcup_{n} I_{n}\) is the set of all real numbers. The intersection \(\bigcap_{n} I_{n}\) is the empty set.

Step-by-step solution

Understand the Definition of \(I_{n}\)

The set \(I_{n}\) is defined as the set of all \(x\) such that \(| x-1 | < n\) and \(| x+1 | > \frac{1}{n}\). This means that \(x\) needs to be closer to 1 than to \(n\) and farther from -1 than \(\frac{1}{n}\).

Find the Limiting Cases

As \(n\) gets bigger, the condition \(| x-1 | < n\) becomes less restrictive because \(n\) is the radius of the interval centered at 1, so it encompasses all real numbers for large \(n\). On the other hand, as \(n\) gets bigger, the condition \(| x+1 | > \frac{1}{n}\) becomes more restrictive because \(\frac{1}{n}\) approaches 0. This means, for large \(n\), \(x\) has to be farther from -1 than a small positive quantity. The smallest value of \(x\) that satisfies the equation \(| x+1 | > \frac{1}{n}\) is \(x = -1 + \frac{1}{n}\). Thus, for large \(n\), \(x\) must be greater than -1.

Find the Union \(\bigcup_{n} I_{n}\)

The union of the sets \(\bigcup_{n} I_{n}\) encompasses all values that are included in at least one of the sets \(I_{n}\). Because for each \(n\), the set \(I_{n}\) includes all \(x\) that are closer to 1 than to \(n\) and farther from -1 than \(\frac{1}{n}\), the union \(\bigcup_{n} I_{n}\) includes all real numbers.

Find the Intersection \(\bigcap_{n} I_{n}\)

The intersection of the sets \(\bigcap_{n} I_{n}\) encompasses all values that are included in all of the sets \(I_{n}\). Because for each \(n\), the set \(I_{n}\) includes all \(x\) which are closer to 1 than to \(n\) and farther from -1 than \(\frac{1}{n}\), and we know \(x\) must be greater than -1 for all \(n\), no value of \(x\) meets the requirement for all \(n\). Therefore, the intersection \(\bigcap_{n} I_{n}\) is the empty set.

Key concepts

Set Union and Intersection

When discussing sets in mathematics, two fundamental operations we perform are union and intersection. The union of sets, denoted by \( \bigcup \), is a set that contains all elements that are in at least one of the involved sets. It is a combination of elements from each contributing set.

For example, if we have sets \( A = \{1, 2, 3\} \) and \( B = \{3, 4, 5\} \), then the union \( A \cup B = \{1, 2, 3, 4, 5\} \). It includes every element from both sets, without repeating any.

The intersection of sets, on the other hand, is denoted by \( \bigcap \) and includes only those elements that are present in every one of the sets. Using the same sets \( A \) and \( B \), the intersection \( A \cap B = \{3\} \), which is the element common to both sets.

In the context of real analysis problems like the ones we are discussing, finding the union means gathering all values of \( x \) that belong to at least one \( I_n \). Conversely, finding the intersection means identifying those values of \( x \) that satisfy conditions for all \( I_n \).

Absolute Value Inequalities

Absolute value inequalities involve expressions like \(|x - c| < a\) or \(|x + c| > b\). These types of inequalities help determine the distance between a number and a point on the number line.

The expression \( |x - c| < a \) indicates that \( x \) is within a distance of \( a \) from the point \( c \). It can be rewritten as a double inequality: \( c-a < x < c+a \). This illustrates the range of values \( x \) can take.

Similarly, \( |x + c| > b \) implies that \( x \) is more than \( b \) units away from the point \( -c \). Hence, \( x \) cannot be a value within \( -c-b \) and \( -c+b \). These forms allow us to visually grasp the set of possible solutions for such inequalities.

In the given problem, we deal with these inequalities to define the sets \( I_n \). The \(|x - 1| < n\) condition ensures \( x \) is near 1, while \(|x + 1| > \frac{1}{n}\) restricts it from being near -1.

Limits and Convergence

The notions of limits and convergence are fundamental to understanding infinite processes in calculus and real analysis. A sequence or a set converges if its elements approach a finite value or a precisely defined state as an index or condition grows infinitely.

For instance, the term \( \frac{1}{n} \) converges to 0 as \( n \) approaches infinity. This concept of convergence is crucial for grasping how properties of sets like \( I_n \) change as \( n \) becomes very large.

In our problem, as \( n \) increases, the range dictated by \( |x+1| > \frac{1}{n} \) becomes a stricter condition. This means \( x \) must stay away from -1, highlighting how closely we analyze limits for accurate solutions. Similarly, understanding that \( |x-1| < n \) becomes less restrictive, assisting us to infer that any \( x \) closer to 1 is always valid in this limit as \( n \to \infty \).

Together, limits and the convergence behavior help decode how the union and intersection of infinitely many sets result in potentially different outcomes.