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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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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. –  Patrick Da Silva Jul 28 '12 at 7:01
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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. –  Patrick Da Silva Jul 28 '12 at 7:02
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I interpret 'trivial' or 'obvious' as 'it has been proved'. –  copper.hat Jul 28 '12 at 7:18
    
i don't quite get why $\inf E_s > \inf E_t$, i think it may not true in general. –  Mathematics Jul 28 '12 at 8:22
    
Thanks for the encouragement and intuition, Patrick. –  Jefferson Huang Jul 28 '12 at 13:12

2 Answers 2

up vote 4 down vote accepted

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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Thanks, copper.hat! –  Jefferson Huang Jul 28 '12 at 13:08

I really liked your idea as it avoids definitions outside of Rudin. Here's my shot at a proof using your idea.

(This is the $\sup$ case, and $s^* = \limsup_{n \rightarrow \infty} s_n = \sup{E_s}$ just like in Rudin.)

Theorem

$$\forall n\ge N,s_n\le t_n \implies s^* \le t^*$$ Proof

If $s^* = -\infty$ or $t^* = +\infty$, the result follows. Hence assume $s^* \neq -\infty $ and $ t^* \neq +\infty.$

Lemma: $s^*,t^* \in \mathbb{R}$. (As opposed to $\mathbb{R} \cup\{-\infty,+\infty\}.)$

Proof: Assume $t^* = -\infty$. By Rudin Theorem 3.17 (a), $t^*\in E_t$ so that $E_t = \{-\infty\}.$
Then for any subsequence $\{t_{n_k}\} $,$\{t_{n_k}\} $ must be bounded above (otherwise it would contain a subsequence that approached $+\infty$). By Definition 3.15 if $\{t_{n_k}\} $ does not approach $-\infty$ then $\{t_{n_k}\} $ contains a subsequence that is bounded below. Then this subsequence is both bounded above and below. Then $\{t_{n_k}\} $ contains a convergent subsequence. Hence $\forall \{t_{n_k}\},t_{n_k}\rightarrow t^*=-\infty.$Then for any real $M$,$$\exists K, k\ge K \implies t_{n_k} \le M.$$ For any $\{s_{n_k}\}$, let $n_p =\min\{n_k|n_k\ge N\}$. Then for any real $M$,$$k\ge \max(p,K) \implies s_{n_k}\le t_{n_k} \le M.$$ Hence $s_{n_k} \rightarrow -\infty$ for all $\{s_{n_k}\}$ so that $E_s = \{-\infty\}$, contradicting our assumption that $s^* \neq -\infty $. A similar argument shows $s^*\neq +\infty.$

Suppose $s^* > t^*$. Then by 3.17 (b) $$\exists G, n\ge G \implies t_{n} \lt s^*$$ Then by 3.17(a) $s^*\in E_s$ so that there exists $\{s_{n_k}\} $ such that $s_{n_k}\rightarrow s^*$. Let $n_q =\min\{n_k|n_k\ge G\}$. Then$$k\ge \max(p,q) \implies s_{n_k}\le t_{n_k} \lt s^*$$ Since $s_{n_k}\rightarrow s^*$ we have $$\forall \epsilon > 0,\exists r, k\ge r \implies |s^*-s_{n_k}| < \epsilon$$ Then $\forall \epsilon > 0$ $$k\ge \max(p,q,r) \implies |s^*-t_{n_k}|\le|s^*-s_{n_k}| \lt \epsilon$$ so that $$ t_{n_k}\rightarrow s^* > t^* = \sup{E_t}$$ which is a contradiction.

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