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The following is the definition.

Let $\{s_n\}$ be a sequence of real numbers. Let $E$ be the set of numbers $x$ such that $s_{n_{k}}\rightarrow{x}$. This set $E$ contains all subsequential limits, plus possibly the numbers $+\infty$, $-\infty$. Now, putting $s^* = \sup (E)$ and $s_{*} = \inf (E)$. The numbers $s^{*}$ and $s_{*}$ are called upper and lower limits of ${s_n}$. $${\lim_{n \to \infty}} \sup s_n = s^{*}$$ $${\lim_{n \to \infty}}\inf s_n = s_{*}$$

I have no idea what this means.

From my understanding, if a sequence has a limit the limit is unique. Why is this definition implying that a sequence has multiple limits?

Or is it implying that a subsequence of a sequence can have a different limit ?

My book lacks in examples and I cannot figure out what's going on at all ...

Also, I learned that $\infty$ is not a number. Why is this definition treating it as if it is one ?

Theorem. Let ${s_n}$ be a sequence of real numbers. Then $s^*$ has the following properties.

  a) $s^* \in E$

  b) If $x>s^*$, $\exists N \in \Bbb Z$ such that $\forall n \ge N$, $s_n < x$.

Moreover, $s^*$ is unique.

I was able to understand the proof of a), and b), but I couldn't really understand the proof of the uniqueness. The book says,

Suppose $p<q$ where both $p$ and $q$ are upper limits. Choose $x$ such that $p < x < q$. Since $p$ satisfies b), we have $s_n<x$ for $n \ge N$. But then $q$ cannot satisfy a).

1) Why is it guaranteed that such $x$ between $p$ and $q$ exist?

2) Why will $q$ not satisfy a) ?

Zev Chonoles
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hyg17
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    A sequence may not have a limit, and yet some subsequences of it will have limits (perhaps in the extended real line). For example, if $s_n=(-1)^n$, then the sequence $s_n$ does not converge, but the subsequences $s_{2n}$ and $s_{2n+1}$ do: The first converges to $1$, the second to $-1$. In this example, $E={-1,1}$, $\limsup s_n=1$, $\liminf s_n=-1$. If $s_n$ is an enumeration of the rationals, then $E=[-\infty,+\infty]$. If $s_n$ converges to $t$, then $E={t}$. Other behaviors are possible, keeping in mind that, always, $E\ne\emptyset$. – Andrés E. Caicedo May 17 '13 at 04:58
  • (To see that $E\ne\emptyset$, note that any sequence, no matter how "chaotic", always contains a monotone subsequence.) – Andrés E. Caicedo May 17 '13 at 05:00
  • That's a perfect explanation and example. I wish Walter Rudin's book had this... Anyhow, thanks ! – hyg17 May 17 '13 at 06:12

1 Answers1

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Why is this definition implying that a sequence has multiple limits?

Andres has given a good answer to this in the comments. In short, even if a sequence has no unique limit, subsequences of the sequence can have a unique limit. The quoted definition is talking about limits of subsequences of a sequence.

Why is it guaranteed that such x between p and q exist?

We assumed that $p<q$, this means that $q-p=\varepsilon$ for some $\varepsilon>0$. You can take $x$ to be $p+\frac{\varepsilon}{2}$.

Why will q not satisfy a) ?

Since $s_n<x<q$ for $n \ge N$,

$${\lim_{n \to \infty}} \sup s_n \le x<q.$$

So $q$ can't be the supremum of $s_n$ as $n\rightarrow \infty$ because $x$ will always be a tighter bound.

Alex L
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  • Thank you for the perfect explanation. I wish my book didn't assume that all of these are trivial enough that no explanation is needed... – hyg17 May 17 '13 at 06:14