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Show that if the sequence$(x_n)$ is bounded, then $(x_n)$ converges iff $\lim\sup_{x\to\infty}(x_n)=\lim\inf_{x\to\infty}(x_n)$.

The definitions that I’m using:

$$\begin{align*} &\liminf_{n\to\infty}x_n=\lim_{n\to\infty}\inf_{m\ge n}x_m\\ &\limsup_{n\to\infty}x_n=\lim_{n\to\infty}\sup_{m\ge n}x_m \end{align*}$$

This’s the first time I deal with $\lim\inf$ things, can someone give me help?

Thank you.

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  • $\begingroup$ There are at least two possible definitions of $\limsup$ and $\liminf$ that you might have been given; what definitions are you using? $\endgroup$ – Brian M. Scott Jun 29 '13 at 7:20
  • $\begingroup$ en.wikipedia.org/wiki/Limit_superior_and_limit_inferior The first one. $\endgroup$ – A. Chu Jun 29 '13 at 7:22
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    $\begingroup$ Here's a little help: give names to the sequences $\inf_{m\geq n} x_m$ and $\sup_{m\geq n} x_m$. How can you compare those to $x_n$ ? What does it mean that they have same limit ? $\endgroup$ – zozoens Jun 29 '13 at 7:46
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I’ll get you started. For one direction, suppose that $\lim\limits_{n\to\infty}x_n=x$; we want to show that $$\limsup_{n\to\infty}x_n=\liminf_{n\to\infty}x_n\;.$$ The most natural guess is that this is true because both are equal to $x$, so let’s try to prove that.

In order to show that $\limsup\limits_{n\to\infty}x_n=x$, we must show that $\lim\limits_{n\to\infty}\sup_{k\ge n}x_k=x$. To do this, we must show that for each $\epsilon>0$ there is an $m_\epsilon\in\Bbb N$ such that

$$\left|x-\sup_{k\ge n}x_k\right|<\epsilon\quad\text{whenever}\quad n\ge m_\epsilon\;.$$

Since $\lim\limits_{n\to\infty}x_n=x$, what we actually know is that for each $\epsilon>0$ there is an $m_\epsilon'\in\Bbb N$ such that $|x-x_n|<\epsilon$ whenever $n\ge m_\epsilon'$.

  • Show that if $|x-x_n|<\epsilon$ for all $n\ge m_\epsilon'$, then $\left|x-\sup\limits_{k\ge n}x_k\right|\le\epsilon$. Conclude that if we set $m_\epsilon=m_{\epsilon/2}'$, say, then $$\left|x-\sup_{k\ge n}x_k\right|<\epsilon\quad\text{whenever}\quad n\ge m_\epsilon$$ and hence $\limsup\limits_{n\to\infty}x_n=x$.

  • Modify the argument to show that $\liminf\limits_{n\to\infty}x_n=x$.

For the other direction, suppose that $$\limsup_{n\to\infty}x_n=\liminf_{n\to\infty}x_n=x\;;$$ we want to show that $\langle x_n:n\in\Bbb N\rangle$ converges. The natural candidate for the limit of the sequence is $x$, so we should try to prove that $\lim\limits_{n\to\infty}x_n=x$, i.e., that for each $\epsilon>0$ there is an $m_\epsilon\in\Bbb N$ such that $|x-x_n|<\epsilon$ whenever $n\ge m_\epsilon$. What we know is that

$$\lim_{n\to\infty}\sup_{k\ge n}x_k=x=\lim_{n\to\infty}\inf_{k\ge n}x_k\;,$$

i.e., that for each $\epsilon>0$ there is an $m_\epsilon'\in\Bbb N$ such that

$$\left|x-\sup_{k\ge n}x_k\right|<\epsilon\quad\text{and}\quad\left|x-\inf_{k\ge n}x_k\right|<\epsilon\quad\text{whenever}\quad n\ge m_\epsilon'\;.$$

(Why can I use a single $m_\epsilon'$ instead of requiring separate ones for each of the two limits?)

  • Show that if $\ell\ge n$, then $$|x-x_\ell|\le\max\left\{\left|x-\sup_{k\ge n}x_k\right|,\left|x-\inf_{k\ge n}x_k\right|\right\}\;,$$ and conclude that setting $m_\epsilon=m_\epsilon'$ will ensure that $|x-x_n|<\epsilon$ whenever $n\ge m_\epsilon$ and hence that the sequence converges to $x$.
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  • $\begingroup$ "Show that if $|x-x_n|<\epsilon$ for all $n\ge m_\epsilon'$, then $\left|x-\sup\limits_{k\ge n}x_k\right|\le\epsilon$." This seems trivial, do I really need to write something about it? $\endgroup$ – A. Chu Jun 30 '13 at 6:19
  • $\begingroup$ @ᴊᴀsᴏɴ: I grant that it’s intuitively obvious, but it’s not entirely trivial to write down a justification, so yes, you ought to do so. $\endgroup$ – Brian M. Scott Jun 30 '13 at 6:22
  • $\begingroup$ So actually for this part, we first say $|x_n-x|\le\epsilon/2,\forall n\geq N$, then conclude that $|\sup_{k\geq n}x_k-x|\le\epsilon,\forall n\geq N$, right? $\endgroup$ – A. Chu Jun 30 '13 at 6:31
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    $\begingroup$ @ᴊᴀsᴏɴ: In fact if $r\ge n$, you can use $$\inf_{k\ge n}x_k\le x_r\le\sup_{k\ge n}x_k$$ to conclude that $$x-\inf_{k\ge n}x_k\ge x-x_r\ge x-\sup_{k\ge n}x_k$$ and then (with just a little work) that $$|x-x_r|\le\max\{\left|x-\sup_{k\ge n}x_k\right|,\left|x-\inf_{k\ge n}x_k\right|\}\;.$$ And if $n\ge m_\epsilon'$, you can finally conclude that $|x-x_r|<\epsilon$. $\endgroup$ – Brian M. Scott Jun 30 '13 at 7:07
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    $\begingroup$ @ᴊᴀsᴏɴ: Excellent! You’re very welcome. $\endgroup$ – Brian M. Scott Jun 30 '13 at 7:33

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