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?

  • 1
    $\begingroup$ 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. $\endgroup$ – Henry Jun 26 '11 at 17:43
  • 1
    $\begingroup$ What do you mean by "ultimately" and "frequently"? $\endgroup$ – omar Jun 26 '11 at 18:42
  • $\begingroup$ @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. $\endgroup$ – Arturo Magidin Jun 26 '11 at 18:55
  • $\begingroup$ 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. $\endgroup$ – Mark Bennet Jun 26 '11 at 19:02
  • $\begingroup$ 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. $\endgroup$ – Mark Bennet Jun 26 '11 at 20:05

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 $\limsup a_n\lt b+\epsilon$; this holds for all $\epsilon\gt 0$, so $\limsup a_n \leq b$.

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

  • $\begingroup$ @Arturo: I think it should be for all $n \geq N$ we have that $b \leq a_n \leq A_{N} < b+ \epsilon$. $\endgroup$ – Damien Jun 26 '11 at 19:25
  • $\begingroup$ @Damien: Where? Which line of which paragraph? I have a lot of inequalities, I don't know which one you are refering to. $\endgroup$ – Arturo Magidin Jun 26 '11 at 19:27
  • $\begingroup$ @Arturo: It would be the third line of the seventh "paragraph." Also in the 9th paragraph, fourth line: $\text{lim inf} \ a_n = \sup \{A_n: n \in \mathbb{N} \} \geq b- \epsilon$. $\endgroup$ – Damien Jun 26 '11 at 19:31
  • $\begingroup$ @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 $\mathop{\inf}\limits_{n\in\mathbb{N}}A_n$. $\endgroup$ – Arturo Magidin Jun 26 '11 at 19:35
  • $\begingroup$ @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 $\text{lim inf} \ a_n = \inf\{A_n: n \in \mathbb{N} \}$? $\endgroup$ – Damien Jun 26 '11 at 19:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.