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Prove that if \(\left\\{x_{n}\right\\}\) and \(\left\\{y_{n}\right\\}\) are bounded in \(E^{1}\), then \(\overline{\lim } x_{n}+\overline{\lim } y_{n} \geq \overline{\lim }\left(x_{n}+y_{n}\right) \geq \overline{\lim } x_{n}+\underline{\lim } y_{n}\) \(\geq \underline{\lim }\left(x_{n}+y_{n}\right) \geq \underline{\lim } x_{n}+\underline{\lim } y_{n}\).

Short Answer

Expert verified
The inequalities follow from properties of limsup and liminf, and the fact that the sequences are bounded.

Step by step solution

01

Understanding Bounded Sequences

The sequences \( \{x_{n}\} \) and \( \{y_{n}\} \) are bounded. This means there exist constants \( M_x \) and \( M_y \) such that \(|x_{n}| \leq M_x\) and \(|y_{n}| \leq M_y\) for all \( n \). These sequences remain within fixed bounds without increasing indefinitely.
02

Evaluate Upper Limits of Sums

The inequality \( \overline{\lim } x_{n}+\overline{\lim } y_{n} \geq \overline{\lim }\left(x_{n}+y_{n}\right) \) must be shown. Use the fact that the limsup of a sum is less than or equal to the sum of the limsups: \( \limsup (a_n + b_n) \leq \limsup a_n + \limsup b_n \). Thus, it follows directly for any bounded sequences.
03

Evaluate Combined Upper and Lower Limits

Show \( \overline{\lim }\left(x_{n}+y_{n}\right) \geq \overline{\lim } x_{n}+\underline{\lim } y_{n} \). Use the subadditive property of limsup and liminf. Since \( \limsup (a_n + b_n) \geq \limsup a_n + \liminf b_n \), this inequality holds for any sequences.
04

Evaluate Lower Limits of Sums

For \( \underline{\lim }\left(x_{n}+y_{n}\right) \geq \underline{\lim } x_{n}+\underline{\lim } y_{n} \), note that the liminf of a sum is greater than or equal to the sum of the liminfs: \( \liminf (a_n + b_n) \geq \liminf a_n + \liminf b_n \). This is a property of bounded sequences and completes the last inequality.

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Key Concepts

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

Limsup and Liminf
The concepts of \( \limsup \) (limit superior) and \( \liminf \) (limit inferior) are crucial in understanding the behavior of sequences, especially when determining their convergence or divergence properties. The \( \limsup \) of a sequence \( \{a_n\} \) is the largest limit point that the sequence can approach as \( n \) tends to infinity. Formally, it is defined as:\[ \limsup_{n \to \infty} a_n = \lim_{n \to \infty} (\sup_{k \geq n} a_k) \]Similarly, the \( \liminf \) is the smallest limit point that the sequence approaches and is defined as:\[ \liminf_{n \to \infty} a_n = \lim_{n \to \infty} (\inf_{k \geq n} a_k) \]Understanding these definitions helps in comprehending the nature of the inequalities associated with sequences, such as how adding sequences impacts their \( \limsup \) and \( \liminf \). For instance, when you add two sequences, their combined behavior will be governed by the range between their \( \limsup \) and \( \liminf \). These concepts also allow us to compare the behavior of sums of sequences with the sum of their individual limit points. For example, if we have two sequences \( \{x_n\} \) and \( \{y_n\} \), and we want to find the limits of their sum, we observe the impact on their \( \limsup \) and \( \liminf \) values using subadditive and superadditive properties.
Bounded Sequences
A sequence \( \{a_n\} \) is considered bounded if there exists a real number \( M \) such that the absolute value of each term is less than or equal to \( M \). In other words, \[ |a_n| \leq M \quad \text{for all } n. \]Being bounded implies that a sequence neither diverges to infinity nor oscillates without limit.Bounded sequences are particularly nice to work with in calculus and analysis because we can apply various theorems and properties, such as the Bolzano-Weierstrass theorem, which states that every bounded sequence has a convergent subsequence. This is important when considering the \( \limsup \) and \( \liminf \) of a sequence, as these concepts often rely on capturing the behavior of subsequences. In our given problem, both \( \{x_n\} \) and \( \{y_n\} \) are bounded sequences, allowing us to apply limits to analyze their sum \( \{x_n + y_n\} \). A bounded sequence helps in ensuring that the calculated limits are finite and thus meaningful in practical sense.
Inequality of Limits
The inequality of limits, particularly involving \( \limsup \) and \( \liminf \), can be understood through their mathematical properties. One important property utilized in our exercise is that for two sequences \( \{a_n\} \) and \( \{b_n\} \), it holds that: - \[ \limsup (a_n + b_n) \leq \limsup a_n + \limsup b_n \] - \[ \liminf (a_n + b_n) \geq \liminf a_n + \liminf b_n \]. These properties are crucial because they effectively provide a range within which the limits of the sum of two sequences can be established. When proving inequalities involving sums of sequences, these inequalities help to ascertain how the upper bounds (supremums) and lower bounds (infimums) interact and lead to either larger or smaller bounds when sequences are added together. In this specific exercise, these concepts were combined to prove a sequence of inequalities starting with the least upper bounds and extending to the greatest lower bounds for the combination of sequences \( \{x_n + y_n\}\). This illustrates how bounds are preserved and manipulated within these mathematical operations.

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Most popular questions from this chapter

For \(n>0\) define $$ \left(\begin{array}{l} n \\ k \end{array}\right)=\left\\{\begin{array}{ll} \frac{n !}{k !(n-k) !}, & 0 \leq k \leq n \\ 0, & \text { otherwise } \end{array}\right. $$ Verify Pascal's law, $$ \left(\begin{array}{l} n+1 \\ k+1 \end{array}\right)=\left(\begin{array}{l} n \\ k \end{array}\right)+\left(\begin{array}{c} n \\ k+1 \end{array}\right) $$ Then prove by induction on \(n\) that (i) \((\forall k \mid 0 \leq k \leq n)\left(\begin{array}{l}n \\\ k\end{array}\right) \in N ;\) and (ii) for any field elements \(a\) and \(b\), \((a+b)^{n}=\sum_{k=0}^{n}\left(\begin{array}{l}n \\ k\end{array}\right) a^{k} b^{n-k}, \quad n \in N\) (the binomial theorem). What value must \(0^{0}\) take for (ii) to hold for all \(a\) and \(b ?\)

Prove that in any field, $$ a^{n+1}-b^{n+1}=(a-b) \sum_{k=0}^{n} a^{k} b^{n-k}, \quad n=1,2,3, \ldots . $$ Hence for \(r \neq 1\) $$ \sum_{k=0}^{n} a r^{k}=a \frac{1-r^{n+1}}{1-r} $$ (sum of \(n\) terms of a geometric series).

Prove that (i) if \(\lim x_{n}=+\infty\) and \((\forall n) x_{n} \leq y_{n},\) then also \(\lim y_{n}=+\infty,\) and (ii) if \(\lim x_{n}=-\infty\) and \((\forall n) y_{n} \leq x_{n},\) then also \(\lim y_{n}=-\infty\).

Prove that (i) if \((\forall \varepsilon>0) a \geq b-\varepsilon,\) then \(a \geq b ;\) (ii) if \((\forall \varepsilon>0) a \leq b+\varepsilon,\) then \(a \leq b\).

Let \(A\) and \(B\) be subsets of an ordered field \(F\). Assuming that the required lub and glb exist in \(F\), prove that (i) if \((\forall x \in A)(\forall y \in B) x \leq y,\) then \(\sup A \leq \inf B ;\) (ii) if \((\forall x \in A)(\exists y \in B) x \leq y,\) then sup \(A \leq \sup B\) (iii) if \((\forall y \in B)(\exists x \in A) x \leq y,\) then \(\inf A \leq \inf B\).

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