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I came across the following problem about limit supremums and infimums of sequences:

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

  1. Prove that $b = \text{lim sup} \ a_n \implies (\forall \epsilon >0) \ a_n < b+\epsilon$ ultimately and $a_n > b-\epsilon$ frequently.

  2. Show that the condition first condition characterizes the limit superior, in the sense that if $b \in \mathbb{R}$ satisfies the condition then necessarily $b = \text{lim sup} \ a_n$.

Here is my attempted solution:

  1. Proof. Let $(a_n)$ is a bounded sequence in $\mathbb{R}$. Suppose $b = \text{lim sup} \ a_n$. By definition, $$b = \text{lim sup} \ a_n = \inf\limits_{n \geq 1} \left(\sup\limits_{k \geq n} \ a_k \right)$$ Let $A = \left\{\sup\limits_{k \geq n} \ a_k \right\}$. Then by definition of infimum, $b+ \epsilon >a_n$ for every $a_n \in A$. So $(\exists N) \ \ni n \geq N \implies a_n < b+ \epsilon$. Thus $a_n < b+ \epsilon$ frequently. Since $b$ is a lower bound of $A$, $b \leq a_n$ for all $n$. Thus $b- \epsilon < a_n$ for all $n$. Hence $a_n > b- \epsilon$ frequently.

  2. Proof. Suppose $(\forall \epsilon >0) \ a_n < b+ \epsilon$ ultimately and $a_n > b- \epsilon$ frequently. Then $b \leq \text{lim sup} \ a_n$ and $b \geq \text{lim sup} \ a_n$. Hence $b = \text{lim sup} \ a_n$ by antisymmetry.

Are these on the right track?

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In your attempted proof, I am confused between $a_n$ being a member of the original sequence and $a_n$ being an element of $A$; the suprema do not need to be in the original sequence. –  Henry Jun 26 '11 at 17:43
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What do you mean by "ultimately" and "frequently"? –  omar Jun 26 '11 at 18:42
    
@Omar: "Ultimately" means that there exists $N$ such that the condition holds for all $n\geq N$. "Frequently" means that for every $N$ there exists (at least one) $k\geq N$ such that the condition holds for $k$. That is, a condition on a sequence holds "ultimately" if it holds for a tail, and holds "frequently" if it holds for a subsequence. –  Arturo Magidin Jun 26 '11 at 18:55
    
For the first condition in part 1 you need to take a given $\epsilon$ and show that there is an N beyond which the condition holds - eventually the sequence stays below $b+ \epsilon$. For the second condition, given $\epsilon$ consider what would happen if there were only finitely many n for which $a_n > b- \epsilon$ - what would that do to your lim sup (hint - there would be an N beyond which the condition would never hold)? Frequently here should mean infinitely often, but not necessarily for all n or for all n beyond a certain point. –  Mark Bennet Jun 26 '11 at 19:02
    
Damien, I think it will serve you well if you come to understand these concepts (after which the proofs will become obvious). GH Hardy "Pure Mathematics" has a short section (no 82 in 10th edition, p156, also read sect 81) on the limits of indetermination of a bounded function. Amongst other things it has narrative definitions, and a short narrative proof of what you are asked and a number of examples which illustrate the various possibilities which may arise. Hardy may help you to understand why lim sup is a useful concept. –  Mark Bennet Jun 26 '11 at 20:05

1 Answer 1

up vote 6 down vote accepted

As soon as you write "Then by the definition of infimum", you are saying incorrect things.

First: if $b$ is the infimum of $A$, then for every $\epsilon\gt 0$ there exists at least one $a\in A$ such that $a\lt b+\epsilon$. However, you claim this holds for all elements of $A$, which is false. Take $A = (0,1)$. Then $\inf A = 0$; and it is indeed true that for every $\epsilon\gt0$ there exists at least one $a\in A$ with $a\lt\epsilon$; but if $\epsilon=\frac{1}{10}$, it is certainly false that every $a\in A$ is smaller than $\epsilon$.

Second: the elements of $A$ are not $a_n$s! They are suprema of infinite sequences of $a_n$s, and as such cannot be assumed to be $a_n$s. For example, if $a_n = 1-\frac{1}{n}$, then $A=\{1\}$, and no $a_n$ is equal to any element of $A$.

So that sentence is not just wrong, it's doubly wrong. The rest of course is now nonsense.

The second part does not seem to be proving anything; you are just asserting things. Why do the conditions imply the inequalities? What properties of the limit superior are you using? It's a mystery.

Rather: let $A_n = \mathop{\sup}\limits_{k\geq n}(a_k)$. Prove that $A_n$ is a decreasing sequence: $A_{n+1}\leq A_n$ for all $n\in \mathbb{N}$. Your set $A$ is precisely the set of $A_n$s.

Now let $b= \inf A = \inf\{ A_n\}$. By the definition of infimum, for every $\epsilon\gt 0$ there exists $N$ such that $b\leq A_N\lt b+\epsilon$. Since the sequence of $A_n$s is decreasing, then for all $n\geq N$ we have $b\leq A_n\leq A_N\lt b+\epsilon$, so in fact we have that $A_n\lt b+\epsilon$ ultimately. Moreover, since $a_n\leq A_m$ for all $n\geq m$, this implies that $a_n\lt b+\epsilon$ ultimately, as required.

For the second clause of the first part, let $\epsilon\gt 0$. Then $b-\epsilon\lt A_n$ for all $n$. Now remember what $A_n$ is. $A_n = \mathop{\sup}\limits_{k\geq n}(a_k)$; since $b-\epsilon\lt A_n$, there exists $k\geq n$ such that $b-\epsilon\lt a_k\leq A_n$. That is: for all $n$, there exists $k\geq n$ such that $a_k\gt b-\epsilon$, so $a_k\gt b-\epsilon$ frequently.

For part (2), let $b$ be a real number that satisfies the given properties. Since for every $\epsilon\gt 0$ we have that $b-\epsilon \lt a_n$ frequently, that means that $b-\epsilon$ is not an upper bound for $\{a_k\mid k\geq n\}$ for any $n$. Therefore, $\sup\{a_k\mid k\geq n\} = A_n\gt b-\epsilon$. This holds for all $A_n$, so $\liminf a_n = \inf\{A_n\mid n\in\mathbb{N}\} \geq b-\epsilon$. This holds for all $\epsilon\gt 0$, so $\liminf a_n \geq b$.

Now, since $b+\epsilon\gt a_n$ ultimately, then $b+\epsilon$ is an upper bound for $A_m$ all sufficiently large $m$; since the $A_m$ are decreasing, that means that $\inf A_m \lt b+\epsilon$, hence $\liminf a_n\lt b+\epsilon$; this holds for all $\epsilon\gt 0$, so $\liminf a_n \leq b$.

Now that we have established the inequalities (rather than merely asserting them), we have $b\leq \liminf a_n \leq b$, hence $b=\liminf a_n$, as claimed. QED

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@Arturo: I think it should be for all $n \geq N$ we have that $b \leq a_n \leq A_{N} < b+ \epsilon$. –  Damien Jun 26 '11 at 19:25
    
@Damien: Where? Which line of which paragraph? I have a lot of inequalities, I don't know which one you are refering to. –  Arturo Magidin Jun 26 '11 at 19:27
    
@Arturo: It would be the third line of the seventh "paragraph." Also in the 9th paragraph, fourth line: $\text{lim inf} \ a_n = \sup \{A_n: n \in \mathbb{N} \} \geq b- \epsilon$. –  Damien Jun 26 '11 at 19:31
    
@Damien: No, in the third line of the seventh paragraph, I wrote what I meant to write. Since the $A_n$s form a decreasing sequence, and $b\leq A_n$ for all $n$, you have that if $A_N\lt b+\epsilon$, then for all $n\geq N$ you have $b\leq A_n\leq A_N\lt b+\epsilon$. There is also no error in the 9th paragraph fourth line. What makes you think those $A_n$'s are supposed to be $a_n$? $\inf\{A_n\mid n\in\mathbb{N}\}$ is the same as $\mathop{\inf}\limits_{n\in\mathbb{N}}A_n$. –  Arturo Magidin Jun 26 '11 at 19:35
    
@Arturo: In the third line of the seventh paragraph you have $b_{\epsilon}$ instead of $b+\epsilon$. And in the 9th paragraph fourth line, shouldnt $\text{lim inf} \ a_n = \sup \{A_n: n \in \mathbb{N} \}$ instead of $\text{lim inf} \ a_n = \inf\{A_n: n \in \mathbb{N} \}$? –  Damien Jun 26 '11 at 19:38

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