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I am trying to understand the proof to Theorem 10.7(i) in Ross's Elementary Analysis. I've found other proofs for the theorem, but my issue is in understanding the specific details in Ross's proof, and these haven't helped. Additionally, I haven't gotten to subspaces yet. My questions are within brackets within the proof. Thanks!

Here are the definitions of lim sup and lim inf that I'm given: $$\limsup s_n=\lim_{N\to\infty}\sup\{s_n:n\gt N\}$$ $$\liminf s_n=\lim_{N\to\infty}\inf\{s_n:n\gt N\}$$

Theorem 10.7

Let $(s_n)$ be a sequence in $\mathbb{R}$.

(i) If $\lim s_n$ is defined [as a real number, $+\infty$, or $-\infty$], then $\liminf s_n=\lim s_n=\limsup s_n$.

Proof

We use the notation $u_{N}=\inf\{s_n:n\gt N\}$, $v_{N}=\sup\{s_n:n\gt N\}$, $u=\lim u_{N}=\liminf s_n$ and $v=\lim v_{N}=\limsup s_n$.

Suppose $\lim s_n=+\infty$. Let M be a positive real number. Then there is a positive integer N so that $$n\gt N \text { implies} \,s_n\gt M. $$

Then $u_{N}=\inf\{s_n:n\gt N\} \ge M.$ [Why is it $\ge M?$ Why isn't it just $\gt M?$] It follows that $m\gt N$ implies $u_m \ge M.$ [Where does m come from? Is it arbitrary? What is $u_m$? Is it just another item in the sequence $(u_{N})$?] In other words, the sequence $(u_{N})$ satisfies the condition defining $\lim u_{N}=+\infty$, i.e., $\liminf s_n=+\infty.$ Likewise $\limsup s_n=+\infty.$

The case of $\lim s_n=-\infty$ is handled in a similar manner.

Now suppose $\lim s_n=s$ where s is a real number. Consider $\epsilon>0.$ There exists a positive integer $N$ such that $|s_n-s|<\epsilon$ for $n>N$. Thus $s_n < s + \epsilon$ [How does this follow? I read it is from the triangle inequality, but I don't see how it follows.] for $n>N$, so

$$v_{N}=\sup\{s_n:n\gt N\}\le s+\epsilon$$.

Also, $m>N$ implies $v_m \le s+\epsilon$ [Why is it necessary to state this? It seems like it was just shown with $v_{N}$.], so $\limsup s_n = \lim v_m \le s+ \epsilon.$ Since $\limsup s_n \le s + \epsilon$ for all $\epsilon > 0$, no matter how small, we conclude $\limsup s_n \le \lim s_n$. A similar argument shows $\lim s_n \le \liminf s_n.$ Since $\liminf s_n \le \limsup s_n$, we infer all three numbers are equal:

$$\liminf s_n = \lim s_n = \limsup s_n.$$

Mark
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1 Answers1

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  1. Because the infimum of a set of numbers greater than $M$ doesn't have to be greater than $M$. All you can say is that it is greater than or equal to $M$.
  2. Before this, the sequence $(u_N)_{N\in\Bbb N}$ was defined and $u_m$ is a term of this sequence, with $m>N$.
  3. It follows from$$s_n-s\leqslant|s_n-s|<\varepsilon\implies s_n<s+\varepsilon.$$
  4. Yes, it was shown just for $v_N$, but the same argument proves that $m>N\implies v_n<s+\varepsilon$.
José Carlos Santos
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  • Thanks Jose! For 2, it seems like the definition of $\lim s_n=+\infty$ has been met by $u_{N}$. Why does it again need to be shown for $u_m$? – Mark Apr 02 '21 at 17:44
  • For 4, I understand how $v_m \le s+\epsilon$ follows in the proof. I just don't see why it's necessary, since it was just shown that $v_{N} \le s+\epsilon$. Why show it again, why can't you use $v_{N}$? – Mark Apr 02 '21 at 17:47
  • Asserting that “the definition of $\lim s_n=+\infty$ has been met by $u_N$” makes no sense. The number $N$ is just one natural number. Asserting that $\lim s_n=+\infty$ means that, for every $M>0$, you have $s_n>M$ for every $n$ large enough. – José Carlos Santos Apr 02 '21 at 17:57
  • For 2, what I mean is that it seems like the definition for a limit of $+\infty$ has already been satisfied with $u_{N}$. Why do you need to show that $u_m$ satisfies it too? – Mark Apr 03 '21 at 21:10
  • I have already provided an answer to that question in my previous comment. – José Carlos Santos Apr 03 '21 at 21:25