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Chapter 1: Problem 16

(a) Prove: If \(S\) is bounded above and \(\beta=\sup S,\) then \(\beta \in \partial S\). (b) State the analogous result for a set bounded below.

Short Answer

Expert verified
Question: Prove that if a set S is bounded above and has a supremum, then the supremum belongs to the boundary of the set. State the analogous result for sets bounded below. Answer: If a set S is bounded above and has a supremum \(\beta\), then \(\beta \in \partial S\), i.e., the supremum belongs to the boundary of the set. Analogously, if a set S is bounded below and has an infimum \(\alpha\), then \(\alpha \in \partial S\).

Step by step solution

01

Understand the definitions of supremum and boundary of a set

The supremum (or least upper bound) of a set S, denoted as \(\sup S\), is the smallest real number that is greater than or equal to every element in S. In other words, if \(\beta = \sup S\), then \(\beta\) is an upper bound of S and there is no upper bound of S smaller than \(\beta\). The boundary of a set, denoted as \(\partial S\), consists of all points \(x\) such that every open neighborhood of \(x\) contains points both from S and its complement \(S^c\).
02

Prove that the supremum belongs to the boundary of the set S

To show that \(\beta \in \partial S\), we need to prove that every open neighborhood of \(\beta\) contains points from both S and \(S^c\). Let's consider an open neighborhood \((\beta-\epsilon, \beta+\epsilon)\) of \(\beta\) for some \(\epsilon > 0\). Since \(\beta\) is the supremum of S and \(\beta - \epsilon < \beta\), there must exist an element \(s \in S\) such that \(\beta - \epsilon < s \leq \beta\). So, we have found a point \(s \in S\) inside the open neighborhood \((\beta-\epsilon, \beta+\epsilon)\). Now, because \(\beta\) is an upper bound of S, no element in S is greater than \(\beta\). Therefore, there exists a point \(t \in S^c\) inside the open neighborhood \((\beta-\epsilon, \beta+\epsilon)\) such that \(\beta \leq t < \beta + \epsilon\). Since we've found points in both S and \(S^c\) inside the open neighborhood \((\beta-\epsilon, \beta+\epsilon)\), we can conclude that \(\beta \in \partial S\).
03

State the analogous result for a set bounded below

Correspondingly, if a set S is bounded below and \(\alpha = \inf S\) (greatest lower bound or infimum), then \(\alpha \in \partial S\).

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Key Concepts

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

Supremum
In real analysis, the concept of a supremum is central to understanding the behavior of sets in the real number system. The supremum, often denoted as \( \sup S \), is described as the smallest real number that is greater than or equal to every element within a set \( S \). This means, for the set \( S \), if we have a number \( \beta = \sup S \), \( \beta \) acts as the least upper bound. Its significance lies in its comprehensiveness - \( \beta \) is not only an upper bound but also the least of such bounds for \( S \).

To visualize this better, imagine you have a collection of numbers and you're looking for a cap that just barely covers them all. That cap should be snug, not allowing any higher numbers from the set to poke through, but also can't be so tight that you could potentially find a smaller cap that fits just as well. That's what \( \beta \) is: the perfect fit. It's noteworthy that the supremum may or may not be an element of \( S \) itself, but is always considered a part of its closure.
Boundary of a Set
Moving on to the boundary of a set, which we denote as \( \partial S \), is akin to mapping the outline of a shape. It comprises all points \( x \) that have the peculiar property that every open interval around \( x \) manages to catch points from both the set \( S \) and the set of its outsiders, \( S^c \), which is the complement of \( S \). Think of it as standing on a beach where the water meets the sand; no matter how small a circle you draw around your feet, the circle will include both water and sand鈥攖his is the essence of a boundary point.

For the supremum to be part of this unique 'beach line', any tiny move around it must include elements below (from the set) and right above (from the outside) it. As you might notice, this characteristic is naturally possessed by the supremum of a set that is bounded above. This relationship is intrinsic in nature because, by definition, the supremum must be approached by elements of the set as closely as possible without being exceeded, which means right around the supremum, you will always witness a mingling of the set and its complement - a hallmark of boundary points.
Bounded Set
A bounded set refers to a set of numbers that has both a ceiling and a floor鈥攖here's a limit to how high and low the numbers can go. Formally, the set \( S \) is bounded above if there exists an upper bound \( M \) such that for every \( s \in S \: s \leq M \) and it is bounded below if there is a lower bound \( N \) such that for every \( s \in S \: s \geq N\). So, if you can confidently say, 'The numbers in this set will never go beyond these bounds,' then your set is a well-defined playground with clear fences around it.

In terms of practical applications, bounded sets are like knowing the maximum capacity of a venue or the height restrictions on a roller coaster鈥攊t gives us parameters within which we can operate. Properties of bounded sets are crucial in various fields, such as optimization, where certain formulas only work within known bounds, and in calculus, where the convergence of functions is significantly influenced by bounded intervals.

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Most popular questions from this chapter

$$ 1^{2}+3^{2}+\cdots+(2 n-1)^{2}=\frac{n\left(4 n^{2}-1\right)}{3} $$

Let \(S\) and \(T\) be nonempty sets of real numbers and define $$ S+T=\\{s+t \mid s \in S, t \in T\\} $$ (a) Show that $$ \sup (S+T)=\sup S+\sup T $$ if \(S\) and \(T\) are bounded above and $$ \inf (S+T)=\inf S+\inf T $$ if \(S\) and \(T\) are bounded below. (b) Show that if they are properly interpreted in the extended reals, then (A) and (B) hold if \(S\) and \(T\) are arbitrary nonempty sets of real numbers.

Prove or disprove: A set has no limit points if and only if each of its points is isolated.

If \(a_{i} \geq 0, i \geq 1,\) then $$ \left(1+a_{1}\right)\left(1+a_{2}\right) \cdots\left(1+a_{n}\right) \geq 1+a_{1}+a_{2}+\cdots+a_{n} $$

Suppose that \(m\) and \(n\) are integers, with \(0 \leq m \leq n .\) The binomial coefficient \(\left(\begin{array}{l}n \\ m\end{array}\right)\) is the coefficient of \(t^{m}\) in the expansion of \((1+t)^{n} ;\) that is, $$ (1+t)^{n}=\sum_{m=0}^{n}\left(\begin{array}{l} n \\ m \end{array}\right) t^{m} $$ From this definition it follows immediately that $$ \left(\begin{array}{l} n \\ 0 \end{array}\right)=\left(\begin{array}{l} n \\ n \end{array}\right)=1, \quad n \geq 0 $$ For convenience we define $$ \left(\begin{array}{r} n \\ -1 \end{array}\right)=\left(\begin{array}{c} n \\ n+1 \end{array}\right)=0, \quad n \geq 0 $$
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