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Chapter 1: Problem 4

Find the largest \(\epsilon\) such that \(S\) contains an \(\epsilon\) -neighborhood of \(x_{0}\). (a) \(x_{0}=\frac{3}{4}, S=\left[\frac{1}{2}, 1\right)\) (b) \(x_{0}=\frac{2}{3}, S=\left[\frac{1}{2}, \frac{3}{2}\right]\) (c) \(x_{0}=5, S=(-1, \infty)\) (d) \(x_{0}=1, S=(0,2)\)

Short Answer

Expert verified
Question: What is the largest 饾湒 such that 饾憜 contains an 饾湒-neighborhood of 饾懃鈧� for the following cases: (a) \(x_{0}=\frac{3}{4}\) and \(S=\left[\frac{1}{2}, 1\right)\) (b) \(x_{0}=\frac{2}{3}\) and \(S=\left[\frac{1}{2}, \frac{3}{2}\right]\) (c) \(x_{0}=5\) and \(S=(-1, \infty)\) (d) \(x_{0}=1\) and \(S=(0,2)\) Answer: (a) The largest 饾湒 is \(\frac{1}{4}\). (b) The largest 饾湒 is \(\frac{1}{6}\). (c) The largest 饾湒 is \(6\). (d) The largest 饾湒 is \(1\).

Step by step solution

01

Identify the interval and point \(x_{0}\).

For this case, we have \(x_{0}=\frac{3}{4}\) and \(S=\left[\frac{1}{2}, 1\right)\).
02

Calculate the distance to the nearest boundary point.

There are two boundary points in consideration: \(\frac{1}{2}\) and \(1\). Since \(x_{0}\) is closer to \(1\), we find the distance: \(\epsilon = 1-\frac{3}{4}=\frac{1}{4}\).
03

Largest epsilon value.

The largest \(\epsilon\) is \(\frac{1}{4}\). (b)
04

Identify the interval and point \(x_{0}\).

For this case, we have \(x_{0}=\frac{2}{3}\) and \(S=\left[\frac{1}{2}, \frac{3}{2}\right]\).
05

Calculate the distance to the nearest boundary point.

There are two boundary points in consideration: \(\frac{1}{2}\) and \(\frac{3}{2}\). Since \(x_{0}\) is closer to \(\frac{1}{2}\), we find the distance: \(\epsilon = \frac{2}{3}-\frac{1}{2}=\frac{1}{6}\).
06

Largest epsilon value.

The largest \(\epsilon\) is \(\frac{1}{6}\). (c)
07

Identify the interval and point \(x_{0}\).

For this case, we have \(x_{0}=5\) and \(S=(-1, \infty)\).
08

Calculate the distance to the nearest boundary point.

There is only one boundary point in consideration: \(-1\). We find the distance: \(\epsilon = 5-(-1)=6\).
09

Largest epsilon value.

The largest \(\epsilon\) is \(6\). (d)
10

Identify the interval and point \(x_{0}\).

For this case, we have \(x_{0}=1\) and \(S=(0,2)\).
11

Calculate the distance to the nearest boundary point.

There are two boundary points in consideration: \(0\) and \(2\). Since \(x_{0}\) is equidistant from both, we choose either one of them. We find the distance: \(\epsilon = 1-0=1\).
12

Largest epsilon value.

The largest \(\epsilon\) is \(1\).

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Key Concepts

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

Epsilon-Neighborhood
An epsilon-neighborhood is a fundamental concept in real analysis that denotes a set of points within a certain distance from a central point. Specifically, if we take any point, say x0, the epsilon-neighborhood of x0 is the set of points that lie within an interval (x0 - 蔚, x0 + 蔚) where 蔚 is a positive number.

When solving exercises involving epsilon-neighborhoods, understanding this concept helps you to visualize the interval around x0 which contains all points that are 'close' to x0. In the given exercise, for instance, the goal was to find the largest epsilon such that the neighborhood is completely contained within the given set S. It's crucial to pick the right epsilon that ensures all points in the neighborhood are inside S without touching or crossing the interval's boundaries.
Boundary Points of an Interval
In the context of interval calculations in real analysis, the term boundary points refers to the 'edges' of intervals. An interval is a contiguous block of numbers within the real number line. The boundary points are those at the very end or very beginning of these blocks.

An interval can be closed or open at its boundaries; a closed interval includes its boundary points, denoted by square brackets [ ], whereas an open interval does not, denoted by parentheses ( ). It's important to identify whether the boundary points belong to the set because it influences the selection of epsilon when seeking an epsilon-neighborhood. For example, in the exercise, the largest allowable epsilon depends on whether the interval is open or closed at the boundary nearest to x0.
Real Analysis Interval Calculations
Calculations involving intervals are a key aspect of real analysis, often requiring you to determine relations between points and intervals, the length of intervals, or the maximum allowed epsilon for an epsilon-neighborhood.

When performing these calculations, precision is key. You may need to determine whether a point lies within an interval, how far it is from the interval's boundaries, and if those boundaries are included in the set. Use inequalities to express these relationships and check the calculations carefully. In terms of the given exercise, calculating the distance to the nearest boundary point determines the maximum size of the epsilon-neighborhood that can fit within the interval without exceeding it.
Nearest Boundary Distance
The nearest boundary distance is a measure of how far a point is from the nearest end of an interval. It's vital for determining the size of an epsilon-neighborhood in real analysis. The nearest boundary distance is calculated simply by finding the absolute difference between the point in question and each of the interval's boundary points, then selecting the shortest distance.

For example, when working on problems like the ones given, always keep in mind to look at both boundaries, even though an open interval might suggest that the point has 'infinite distance' to that end. You must compare the distances to both ends and choose the smaller one, as this distance will give you the largest possible epsilon for your epsilon-neighborhood, ensuring the entire neighborhood stays within the interval's bounds.

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Most popular questions from this chapter

Prove: (a) \(\left(S_{1} \cap S_{2}\right)^{0}=S_{1}^{0} \cap S_{2}^{0}\) (b) \(S_{1}^{0} \cup S_{2}^{0} \subset\left(S_{1} \cup S_{2}\right)^{0}\)

A set \(S\) is. in a set \(T\) if \(S \subset T \subset \bar{S}\). (a) Prove: If \(S\) and \(T\) are sets of real numbers and \(S \subset T,\) then \(S\) is dense in \(T\) if and only if every neighborhood of each point in \(T\) contains a point from \(S\). (b) State how (a) shows that the definition given here is consistent with the restricted definition of a dense subset of the reals given in Section \(1.1 .\)

Take the following statement as given: If \(p\) is a prime and \(a\) and \(b\) are integers such that \(p\) divides the product \(a b\), then \(p\) divides \(a\) or \(b\). (a) Prove: If \(p, p_{1}, \ldots, p_{k}\) are positive primes and \(p\) divides the product \(p_{1} \cdots p_{k}\), then \(p=p_{i}\) for some \(i\) in \(\\{1, \ldots, k\\}\). (b) Let \(n\) be an integer \(>1\). Show that the prime factorization of \(n\) found in Example 1.2 .7 is unique in the following sense: If $$ n=p_{1} \cdots p_{r} \quad \text { and } \quad n=q_{1} q_{2} \cdots q_{s} $$ where \(p_{1}, \ldots, p_{r}, q_{1}, \ldots, q_{s}\) are positive primes, then \(r=s\) and \(\left\\{q_{1}, \ldots, q_{r}\right\\}\) is a permutation of \(\left\\{p_{1}, \ldots, p_{r}\right\\}\)

(a) Show that the union of finitely many closed sets is closed. (b) Give an example showing that the union of infinitely many closed sets may fail to be closed.

Prove: An isolated point of \(S\) is a boundary point of \(S^{c}\).
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