Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Prove that \(\overline{\lim } x_{n}<+\infty\left(\underline{\lim } x_{n}>-\infty\right)\) iff \(\left\\{x_{n}\right\\}\) is bounded above (below) in \(E^{1}\).

Short Answer

Expert verified
The sequence is bounded above (below) if and only if its limit superior (inferior) is finite (not infinite).

Step by step solution

01

Understanding Limits

First, recall the definitions of the limits involved. The upper limit, or limit superior, of a sequence \( \{x_n\} \) is the largest limit point of the sequence, while the lower limit, or limit inferior, is the smallest limit point. These can be denoted as \( \overline{\lim} \, x_n \) and \( \underline{\lim} \, x_n \), respectively.
02

Forward Implication: Limit Superior Condition

To show \( \overline{\lim} \, x_n < +\infty \) implies \( \{x_n\} \) is bounded above, assume \( \overline{\lim} \, x_n = M < +\infty \). This means for any \( \epsilon > 0 \), there exists \( N \) such that \( x_n < M + \epsilon \) for all \( n > N \). Thus, \( \{x_n\} \) is bounded above by \( M + \epsilon \).
03

Forward Implication: Limit Inferior Condition

Similarly, to show \( \underline{\lim} \, x_n > -\infty \) implies \( \{x_n\} \) is bounded below, assume \( \underline{\lim} \, x_n = m > -\infty \). This implies for any \( \epsilon > 0 \), there exists \( N \) such that \( x_n > m - \epsilon \) for all \( n > N \). Thus, \( \{x_n\} \) is bounded below by \( m - \epsilon \).
04

Reverse Implication: Bounded Above

Suppose \( \{x_n\} \) is bounded above by some constant \( M \). This means for all \( n \), \( x_n \leq M \). Consequently, \( \overline{\lim} \, x_n \leq M < +\infty \) since it cannot exceed any point beyond this bound.
05

Reverse Implication: Bounded Below

Similarly, if \( \{x_n\} \) is bounded below by some constant \( m \), then for all \( n \), \( x_n \geq m \). Consequently, \( \underline{\lim} \, x_n \geq m > -\infty \) since it cannot be less than any point above this bound.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Bounded Sequences
A bounded sequence is a sequence of numbers where all elements are confined within a specific range. The concept of boundedness is vital for understanding sequences because it helps determine the behavior of a sequence over time. A sequence that is bounded above means there is an upper limit beyond which no elements of the sequence can go.
A sequence bounded below means there is a lower limit. This concept works similarly. Understanding bounded sequences in both upper and lower bounds involves recognizing how they operate:
  • In a sequence bounded above, there exists a number that is greater than or equal to every number in the sequence.
  • In a sequence bounded below, there exists a number that is less than or equal to every number in the sequence.
Bounded sequences are essential in determining other sequence characteristics like convergence or divergence. They offer a framework to manage and predict the behavior of complex sequences in advanced mathematics. When a sequence is both bounded above and below, it is simply called a "bounded sequence."
Upper Limit
The upper limit, or the limit superior (\( \overline{\lim} \)), is a critical characteristic of a sequence. It helps in understanding the sequence's long-term behavior. Specifically, the upper limit is the largest value that a sequence converges to, or hovers around, as it progresses.
Finding a sequence’s upper limit involves considering its largest subsequential limits.The concept might get clearer with some keywords:
  • "Largest Limit Point": The greatest limit that subsequences of a given sequence can achieve.
  • "Convergence": The upper limit is the value to which parts of the sequence approach as it continues indefinitely.
When \( \overline{\lim} x_n < +\infty \), it indicates there is an upper bound to the sequence, showing the values do not grow beyond a certain point indefinitely. Analyzing and determining this upper limit is crucial for understanding whether the sequence itself is bounded above or not.
Lower Limit
The lower limit, or the limit inferior (\( \underline{\lim} \)), acts as the flip side of the upper limit. It represents the lowest value that a sequence converges to, or closely approaches, as it progresses over time.This concept also revolves around the smallest subsequential limits within the sequence, making it equally important as the upper limit in creating a complete picture of a sequence's behavior.Some clarifications regarding the lower limit:
  • "Smallest Limit Point": Refers to the minimal subsequent boundary that parts of the sequence can achieve.
  • "Approach": Indicates how closely sequences get to this lower point as they expand indefinitely.
When \( \underline{\lim} x_n > -\infty \), it implies that the sequence does not trend towards negative infinity, establishing a lower bound.Therefore, knowing the lower limit can help determine if a sequence is bounded below. This concept is critical for analyses where lower bounds are pivotal to the understanding of sequence behavior.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

For any field elements \(a, b\) and natural numbers \(m, n,\) prove that (i) \(a^{m} a^{n}=a^{m+n}\) (ii) \(\quad\left(a^{m}\right)^{n}=a^{m n} ;\) (iii) \(\quad(a b)^{n}=a^{n} b^{n} ;\) (iv) \(\quad(m+n) a=m a+n a ;\) (v) \(n(m a)=(n m) \cdot a ;\) (vi) \(\quad n(a+b)=n a+n b\).

Prove the Bernoulli inequalities: For any element \(\varepsilon\) of an ordered field, (i) \((1+\varepsilon)^{n} \geq 1+n \varepsilon\) if \(\varepsilon>-1 ;\) (ii) \((1-\varepsilon)^{n} \geq 1-n \varepsilon\) if \(\varepsilon<1 ; n=1,2,3, \ldots\)

Prove that the rational subfield \(R\) of any ordered field is Archimedean. [Hint: If $$ x=\frac{k}{m} \text { and } y=\frac{p}{q} \quad(k, m, p, q \in N), $$ then \(n x>y\) for \(n=m p+1]\)

Prove in \(E^{1}\) that (i) \(\sum_{k=1}^{n} k=\frac{1}{2} n(n+1) ;\) (ii) \(\sum_{k=1}^{n} k^{2}=\frac{1}{6} n(n+1)(2 n+1) ;\) (iii) \(\sum_{k=1}^{n} k^{3}=\frac{1}{4} n^{2}(n+1)^{2}\); (iv) \(\sum_{k=1}^{n} k^{4}=\frac{1}{30} n(n+1)(2 n+1)\left(3 n^{2}+3 n-1\right)\).

For any two subsets \(A\) and \(B\) of an ordered field \(F,\) let \(A+B\) denote the set of all sums \(x+y\) with \(x \in A\) and \(y \in B ;\) i.e., $$ A+B=\\{x+y \mid x \in A, y \in B\\} $$ Prove that if \(\sup A=p\) and \(\sup B=q\) exist in \(F,\) then $$ p+q=\sup (A+B) $$ similarly for infima. [Hint for sup: By Theorem 2 , we must show that (i) \((\forall x \in A)(\forall y \in B) x+y \leq p+q\) (which is easy) and \(\left(\mathrm{ii}^{\prime}\right)(\forall \varepsilon>0)(\exists x \in A)(\exists y \in B) x+y>(p+q)-\varepsilon\) Fix any \(\varepsilon>0 .\) By Theorem 2 , $$ (\exists x \in A)(\exists y \in B) \quad p-\frac{\varepsilon}{2}\left(p-\frac{\varepsilon}{2}\right)+\left(q-\frac{\varepsilon}{2}\right)=(p+q)-\varepsilon $$ as required.]

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free