# Prove $\liminf$ definition

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

• Shouldn't $x_n \geq B - \varepsilon$ be $x_n > B - \varepsilon$ ? I haven't read further yet. – flowian Oct 27 '18 at 13:56
• I think it doesn’t change anything in the argument assuming the strict inequality. – gimusi Oct 27 '18 at 14:00
• True, having read further, it does not. – flowian Oct 27 '18 at 14:01
• The key point is that if $B>L$ we can find $\epsilon$ such that eventually $x_n<B-\epsilon$ that is $\epsilon<(B-L)/2$. – gimusi Oct 27 '18 at 14:07
• 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 ? – 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 $$\lim\inf x_n -\epsilon for all $$k$$ sufficiently large. If you denote $$\lim\inf x_n$$ by $$B$$ then $$B-\epsilon 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 $$B then $$B 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 $$(1)$$.