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This is half of Theorem 3.19 from Baby Rudin. Rudin claims the proof is trivial. What I've come up with so far doesn't seem trivial, however, and is probably also wrong (my problem with it is pointed out below). This makes me wonder whether I'm overlooking some useful fact and/or using an unprofitable approach.

Theorem. If $s_n \leq t_n$ for $n \geq N$, where $N$ is fixed, then

$$ \liminf_{n \rightarrow \infty} s_n \leq \liminf_{n \rightarrow \infty} t_n. $$

Proof. Suppose that $s_n \leq t_n$ if $n \geq N$, but that

$$ \liminf_{n \rightarrow \infty} s_n > \liminf_{n \rightarrow \infty} t_n. $$

Let $E_s$ denote the set of all subsequential limits of $\{s_n\}$, and let $E_t$ denote the set of all subsequential limits of $\{t_n\}$. Then

$$ \inf E_s > \inf E_t. $$

This implies that there exists some $x \in E_t$ such that $\inf E_s > x > \inf E_t$, since otherwise $\inf E_t$ would not be the greatest lower bound of $E_t$. Hence some subsequence of $\{t_n\}$, say $\{t_{n_i}\}$, converges to $x < \inf E_s$.

Lemma. (from Rudin) If $x < \liminf_{n \rightarrow \infty} s_n$, then there exists an integer $N$ such that if $n \geq N$, then $s_n > x$.

By the lemma, there exists an integer $N_0$ such that if $n \geq N_0$, then $s_n > x$.

Now, let $\epsilon = \inf_{n \geq N_0} \{s_n - x\}$. This is where I think my proof breaks down. Can't $\epsilon$ be zero? Then, since $\{t_{n_i}\}$ converges to $x$, there exists an integer $N_1$ such that if $n_i \geq N_1$, then $|t_{n_i} - x| < \epsilon$. But this means that, if $n_i \geq \max\{N, N_0, N_1\}$, we have both

$$ s_{n_i} > t_{n_i}, $$

as well as $s_{n_i} \geq t_{n_i}$, a contradiction.

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    $\begingroup$ I've been there ; sometimes reading proofs in analysis that are considered "trivial" makes it a little harsh on yourself because you can't see why it is straight-forward when you don't have experience with some of the tools. Actually in this case it's straightforward, if you draw a little picture, you shouldn't see any problem believing the result, and the reason why it is so easy to believe is if you think about the proof given by copper.hat, which is pretty standard ; when limits exists, for inequalities it makes sense to "evaluate limits on both sides" ; the same remains true for liminfs. $\endgroup$ – Patrick Da Silva Jul 28 '12 at 7:01
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    $\begingroup$ The idea is just that "if I'm always smaller than this guy for every $n$, then I'm not gonna be bigger than him no matter how big $n$ grows". Letting $n$ "grow" is essentially taking limits. I wanted to give you an intuitive reasoning behind the proof ; see copper.hat's answer for details. $\endgroup$ – Patrick Da Silva Jul 28 '12 at 7:02
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    $\begingroup$ I interpret 'trivial' or 'obvious' as 'it has been proved'. $\endgroup$ – copper.hat Jul 28 '12 at 7:18
  • $\begingroup$ i don't quite get why $\inf E_s > \inf E_t$, i think it may not true in general. $\endgroup$ – Mathematics Jul 28 '12 at 8:22
  • $\begingroup$ Thanks for the encouragement and intuition, Patrick. $\endgroup$ – Jefferson Huang Jul 28 '12 at 13:12
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You are making it too hard for yourself.

By definition, $\liminf_{n \to \infty} a_n = \lim_{n \to \infty} \inf_{k\geq n} a_k$. Also, notice that the sequence $\inf_{k\geq n} a_k$ is non-decreasing.

So, if $s_n \leq t_n$, then clearly $\inf_{k\geq n} s_k \leq t_n$, for any $n$. From this it follows that $\inf_{k\geq n} s_k \leq \inf_{k\geq n} t_k$, for any $n$ (if not, then $\inf_{k\geq n} s_k > t_{k'}$, for some $k' \geq n$, which is an immediate contradiction).

Since both sides are non-decreasing, we have $\inf_{k\geq n} s_k \leq \lim_{n' \to \infty} \inf_{k\geq n'} t_k = \liminf_{n \to \infty} t_n$, and then we have $\lim_{n' \to \infty} \inf_{k\geq n'} s_k = \liminf_{n \to \infty} s_n \leq \liminf_{n \to \infty} t_n$, as desired.

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    $\begingroup$ If $a = \lim \inf_{n \to \infty} a_n$ and $b = \lim_{n \to \infty} \inf_{k \geq n} a_k$ then the fact that $a=b$ is not simply by definition, at least not by the definition given in Rudin, and needs to be proven. By Rudin's definition $a$ is the infimum of the set of all subequential limits whereas $b$ is the convergence point of the sequence $\{t_n\}$ where $t_n = \inf_{k \geq n} a_k$. Intuitively I can see why these two are the same but I am finding a rigorious proof to be nontrivial. $\endgroup$ – kyp4 Aug 5 '15 at 1:55
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Rudin is rather terse and sly in this section, even by his own standards. His definition of limsup and liminf is rather indirect because, for some reason, he didn't want to deal with sequences of extended reals. The answer from copper.hat is great, but if you want to work from Rudin's definition you'd need to formulate a definition for convergence of extended reals and prove a lemma that the Rudin's definition agrees with the result that copper.hat's equivalent definition gives. These are all slightly tedious and time-consuming. It does make you wonder what Rudin may have had in mind.

A fairly trivial result that Rudin doesn't mention even in passing is that if you have two subsets of extended reals $A$ and $B$ such that $A\subseteq B$, we can conclude $$\inf B\leq \inf A\quad\text{and}\quad\sup A\leq \sup B\,.$$

We might be able to adapt this result.

For each $n\in\mathbb{N}$, let us define $$S_n=[s_n, \infty),\quad T_n=[t_n, \infty),\quad S=\bigcup_{k=1}^\infty S_k,\quad T=\bigcup_{k=1}^\infty T_k\,.$$

Since $s_n\leq t_n$ for every $n$, we have that $T\subseteq S$. Which then implies that $$\inf S\leq\inf T\,.$$

Of course, this in turn demands that we would verify $\inf S=\liminf s_n$ and $\inf T=\liminf t_n$. This is slightly tedious but is fairly straightforward (dare I say trivial) from the definitions of subsequential limits.

A similar argument using unions of lower sets, instead of unions of upper sets, gives the limsup inequality.

I personally wonder if Rudin had an interleaving argument in mind---creating a new sequence from the two sequences that made these inequalities obvious. But I couldn't construct such a sequence.

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