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I've been working with proofs involving $\limsup$ and $\liminf$, and I'm a bit confused regarding their general methodology. More specifically, I'm unsure about whether my approach to the following problem makes sense.

Problem. Let $(s_n)$ and $(t_n$) be sequences and suppose that there exists $N_0$ such that $s_n \leq t_n$ for all $n > N_0$. Show that $\liminf s_n \leq \liminf t_n$ and that $\limsup s_n \leq \limsup t_n$.

The way I approached it was as follows:

Let $N > N_0$. Then $\limsup_{N \rightarrow \infty} \{ s_n : n > N \} \leq t_n$ as $s_n \leq t_n$, and $\limsup s_n $ is the largest possible limit of a subsequence of $s_n$. As $t_n : n > N$ is (by definition) less than $\limsup_{N \rightarrow \infty} \{ t_n : n > N \}$, the proof is complete.

I'm pretty sure this is incorrect, however, and I'm generally unclear about the method behind such a proof. Any help is appreciated!

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Let $x_n = t_n - s_n$. Your problem is equivalent to the following: if $x_n \ge 0 \ \forall n \ge N$, then $\liminf x_n \ge 0$.

Remember that, for sequences of real numbers, $\liminf$ is the $\inf$ of the set of limit points of $\{x_n\} _{n \ge N}$. I find working with this description of $\liminf$ very visual, therefore very productive - give it a try! Therefore, let us focus on this set of limit points, first.

Let $x$ be a limit point of the set $\{x_n\} _{n \ge N}$. This means that there exist a subsequence of $(x_n) _{n \ge N}$ which tends to $x$. Since all the $x_n$ are positive, so will be the terms of the subsequence, hence so will be its limit: $x \ge 0$. We have discovered that the set of limit points is made up of only positive numbers.

Next, you have to take the $\inf$ of this set. But the $\inf$ of a set of positive numbers is itself positive, which is exactly what you needed to show.

A similar argument is for $\limsup$.

Notice that the essential ingredients that we have used here are that the limit of a sequence of positive numbers is itself positive, and that the $\inf$ of a set of positive numbers is itself positive.

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  • $\begingroup$ Thanks! This definitely helps. I always thought of $\liminf$ as the limit of the set of infima (which we know exists as it is nondecreasing, and therefore monotone). Does this definition reconcile with the one you listed in your answer? I guess I'm still a bit confused regarding the meaning of the terms. $\endgroup$ – manan Jul 17 '16 at 15:41
  • $\begingroup$ @manan: Yes, the two are equivalent, and are in turn equivalent to several other definitions. The one that I chose to work with is the one that I have found most helpful in reasoning about problems. $\endgroup$ – Alex M. Jul 17 '16 at 15:54
  • $\begingroup$ Ok, great! One more thing -- would you mind telling me what was incorrect about my original idea? I understand your approach, but I also want to see where I went wrong the first time so I don't make the same mistake in the future. Thanks :) $\endgroup$ – manan Jul 17 '16 at 15:56
  • $\begingroup$ @manan: It is essentially correct, but with a caveat: it is not true that, in general, $x_n \le \limsup x_n$ - there might be a finite number of terms larger than $\limsup x_n$. Think of the sequence $1, 0, 0, \dots$ (there is a single $1$, on the first position): the $\limsup$ is not $1$ (because there is no sub-sequence tending to $1$), it is $0$ - so the first term of the sequence is greater than the $\limsup$. Similarly for $\liminf$. So your proof is correct, save for a finite number of terms, which are not important when talking about limits, anyway. $\endgroup$ – Alex M. Jul 17 '16 at 16:03
  • $\begingroup$ Awesome, thank you so much! $\endgroup$ – manan Jul 17 '16 at 16:13

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