Exercise 6.4.3 (Basic properties of limit superior and limit inferior)

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extbf{Proposition 6.4.12.} Let $\left(a_{n} ight)_{n=m}^{\infty}$ be a sequence of real numbers, let $L^{+}$ be the limit superior of this sequence, and let $L^{-}$be the limit inferior of this sequence (thus both $L^{+}$and $L^{-}$are extended real numbers). \begin{enumerate}[label=(\alph*)] \item For every $x>L^{+}$, there exists an $N \geq m$ such that $a_{n}L^{+}$, the elements of the sequence $\left(a_{n} ight)_{n=m}^{\infty}$ are eventually less than $x$.) Similarly, for every $yy$ for all $n \geq N$. \item For every $xx$. (In other words, for every $xL^{-}$and every $N \geq m$, there exists an $n \geq N$ such that $a_{n}

Prakash Anand · Accepted Answer

When doing these types of problems (multiple parts) usually one can use previously proved results to build up to new results. \par
(c) \begin{proof} Let $A_{k}=\inf \left\{a_{n}ight\}_{n \geq k}$ and $B_{k}=\sup \left\{a_{n}ight\}_{n \geq k}$. Then $A_{k} \leq B_{k}, A_{k}$ are increasing (why?) and $B_{k}$ are decreasing in $k$. Clearly, $A_{m}=i n f\left\{a_{n}ight\} \leq$ $A_{k} \leq \sup A_{k}=\liminf \left\{a_{n}ight\}$ and similarly for the $B_{k}$ 's, so we must show $\liminf \left\{a_{n}ight\} \leq \limsup \left\{a_{n}ight\} .$ Now for all $n \geq k$ we have $A_{k} \leq A_{n} \leq B_{n}$.
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Thus $A_{k} \leq B_{n} \forall n, k$ (Why?). So first taking the supremum over $k$ in this inequality, $\liminf a_{n} \leq B_{m} \forall m$ and then taking the infemum over m gives $\liminf a_{n} \leq \lim \sup a_{n}$.\end{proof}

(d) \begin{proof} If $c$ is a limit point of the sequence, then $\forall \epsilon \gtrless 0$ and $\forall N$ there exists $n \geq N$ such that $A_{n}-\epsilon \leq c \leq B_{n}+\epsilon$ (Why?). This means $\lim \inf a_{n}-\epsilon \leq c \leq$ $\limsup a_{n}+\epsilon$ (Again, I have skipped a small step, How?). As $\epsilon$ is arbitrary, we are done.\end{proof}

(e) \begin{proof} We use (b) for this one. Given $\epsilon \geqslant 0 \forall N$, there exists $n \geq N$ such that $L^{+}-\epsilon \leq a_{n} \leq L^{+}$. Note that since $L^{+}$is finite, $L^{+}-\epsilon 
eq L^{x}$. But this says that $\forall \epsilon \gtrless 0$ the sequence $\left\{a_{n}ight\}$ is continually $\epsilon$ steady. Similarly for $L^{-} . \diamond$\end{proof}

(f) \begin{proof} If $a_{n}$ converges to $c$ then we know $a_{n}$ is Cauchy and hence bounded. Therefore, $L^{-}$and $L^{+}$are finite, and hence limit points. But $\left\{a_{n}ight\}$ has only one limit point. So $L^{-}=c=L^{+}$.
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Conversely, $L^{-}=c=L^{+}$, then given $\epsilon \geqslant 0$ we have the existence of $N$ large enough that $L^{-}-\epsilon \leq a_{n} \leq L^{+}+\epsilon$. So $\forall \epsilon \geqslant 0,\left\{a_{n}ight\}$ is eventually $\epsilon$ close to $c$.\end{proof}

Exercise 6.4.3 (Basic properties of limit superior and limit inferior)

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Exercise 6.4.3 (Basic properties of limit superior and limit inferior)

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