This is homework assignment.

Let $(x_n)$ be bounded sequence. Prove following equation

$$\liminf_{n \rightarrow \infty}\, x_n = \max \{ B \in \mathbb{R} : \forall \varepsilon > 0 \{n \in \mathbb{N}: x_n \leq B - \varepsilon\} \; \text{is finite set} \}$$

I don't understand right side of equation, where it says that $\{ n \in \mathbb{N}: x_n \leq B - \varepsilon \}$ is finite set.

I understand concept of $\liminf_{n\rightarrow\infty}$ though.

$\liminf_{n\rightarrow\infty}\; x_n = \lim_{n\rightarrow\infty}\inf\{x_k: k \geq n\}$


I will not give a full proof but some insight in order to understand what it is going on and then proceed by your self to find or follow the formal proof.

Let denote as

$$L=\liminf_{n \rightarrow \infty}\, x_n\in \mathbb{R}$$

then consider

$$ B \in \mathbb{R} : \forall \varepsilon > 0 \{n \in \mathbb{N}: x_n \leq B - \varepsilon\}\; \text{is a finite set} $$

it means that $\forall \varepsilon > 0\, \exists \bar n$ such that $\forall n\ge \bar n$, that is eventually

$$x_n \geq B - \varepsilon$$

Now recall that by the definition we have also that $\forall \varepsilon > 0$

  • $x_n\ge L-\varepsilon$ eventually

  • $x_n\le L+\varepsilon$ frequently

Here is a sketch of what is going on

enter image description here

we can see that when $B>L$ there is some problem with the condition $x_n \geq B - \varepsilon$.

  • $\begingroup$ Shouldn't $x_n \geq B - \varepsilon$ be $x_n > B - \varepsilon$ ? I haven't read further yet. $\endgroup$ – flowian Oct 27 '18 at 13:56
  • $\begingroup$ I think it doesn’t change anything in the argument assuming the strict inequality. $\endgroup$ – gimusi Oct 27 '18 at 14:00
  • $\begingroup$ True, having read further, it does not. $\endgroup$ – flowian Oct 27 '18 at 14:01
  • $\begingroup$ The key point is that if $B>L$ we can find $\epsilon$ such that eventually $x_n<B-\epsilon$ that is $\epsilon<(B-L)/2$. $\endgroup$ – gimusi Oct 27 '18 at 14:07
  • $\begingroup$ In the last part, in definition we have. $\liminf_{n\rightarrow\infty}x_n = L \Leftrightarrow \; \forall \varepsilon \exists N\in \mathbb{N}: n \geq N \implies |\inf_{k \geq n}x_k - L| \leq \varepsilon$. isn't it ? $\endgroup$ – flowian Oct 27 '18 at 17:02

Suppose $\{n:x_n\leq B-\epsilon\}$ is a finite set for each $\epsilon.$ Then $x_n >B-\epsilon$ for all $n$ sufficiently large which implies $\lim\inf x_n \geq B-\epsilon$ . This is true for all $\epsilon$, so $\lim\inf x_n \geq B$. This proves LHS $\geq$ RHS. On the other hand $\lim\inf x_n -\epsilon <x_k$ for all $k$ sufficiently large. If you denote $\lim\inf x_n $ by $B$ then $B-\epsilon <x_k$ for all $k$ sufficiently large. So $B$ belongs to the set in the RHS so RHS $\geq B=\lim\inf x_n $. This completes the proof. I have used two properties of $\lim\inf$:

1) if $B <lim\inf x_n$ then $B<x_n$ for all $n$ sufficiently large

2) if $x_n \geq c$ for all $n$ sufficiently large then $\lim \inf x_n \geq c$.


On base of the definition that you are familiar with we can find:

  • If $y>\liminf_{n\to\infty} x_n$ then the set $\{n\mid x_n\leq y\}$ is infinite.

  • If $y<\liminf_{n\to\infty} x_n$ then the set $\{n\mid x_n\leq y\}$ is finite.

Now have a look at the condition:$$\{n\mid x_n\leq B-\epsilon\}\text{ is finite for every }\epsilon>0\tag1$$

The second bullet guarantees that $(1)$ is satisfied for $B=\liminf_{n\to\infty} x_n$.

If $B>\liminf_{n\to\infty} x_n$ then we can find an $\epsilon>0$ such that also $B-\epsilon>\liminf_{n\to\infty} x_n$ and then the first bullet guarantees that $(1)$ is not satisfied.

This means exactly that $\liminf_{n\to\infty} x_n$ is maximal element of the set of elements $B\in\mathbb R$ that satisfy $(1)$.


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.