# $\liminf a_n = \sup_{n \in \mathbb{N}}(\inf_{k \ge n}a_k)$

How to prove these equalities? Here $a_n$ is a sequence in $\mathbb{R} \cup \{- \infty, + \infty \}$

$\liminf_{n\to\infty} a_n = \sup_{n \in \mathbb{N}}(\inf_{k \ge n}a_k)$

and

$\limsup_{n\to\infty} a_n = \inf_{n \in \mathbb{N}}(\sup_{k \ge n}a_k)$

I know that $\ \limsup_{n\to\infty} a_n = \sup\{ \text{ all limit points of } a_n \}$ and similarly $\liminf_{n\to\infty} a_n$ is the infimum of that set. I feel that it should be somehow used in proving the above equalities but I'm not sure how to do that exactly.

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Is that the definition of limsup and liminf that you are using? The supremum and infimum of the set of limit points? – T. Eskin Jan 24 '13 at 20:36
Well, that's the definition that I saw used to prove that $\lim \inf a_n \le \lim \sup a_n$. But if you think a different definition would be more useful here, please share it :) – Hagrid Jan 24 '13 at 20:41
Are you assuming that $(a_{n})$ is a bounded sequence? Because otherwise by taking $a_{n}=0$ for odd $n$ and $a_{n}=n$ for even $n$, you have a sequence with $0=\infty$ with your definition. – T. Eskin Jan 24 '13 at 20:54
I'm sorry. I forgot to include a vital assumption. I've already edited the post, so it should be all right now. – Hagrid Jan 24 '13 at 21:05
With this definition it is not true if $(a_{n})$ is unbounded. – T. Eskin Jan 24 '13 at 21:10

I will help you with the limsup part so you get the idea. I am also assuming that you have a bounded sequence $(a_{n})_{n=1}^{\infty}$, because otherwise the claim would be $0=\infty$ if we take $a_{n}=0$ for odd $n$ and $a_{n}=n$ for even $n$, using the provided definition for limsup.

Let $A$ denote the set of all accumulation points of the sequence $(a_{n})_{n=1}^{\infty}$. Your definition is that $\limsup a_{n}=\sup A$. You want to show that $a:=\inf_{n\in\mathbb{N}}(\sup_{k\geq n} a_{k})\in A$ and $c\leq a$ for all $c\in A$, as these two conditions imply the given result.

First fix $\varepsilon>0$. Since $(a_{n})_{n=1}^{\infty}$ is bounded and $(\sup_{k\geq n} a_{k})_{n=1}^{\infty}$ is nonincreasing, then the limit $\lim_{n\to\infty}(\sup_{k\geq n} a_{k})$ exists and it equals $a$. Hence by definition of convergence there exists $n_{\varepsilon}\in\mathbb{N}$ so that $\sup_{k\geq n} a_{k}\in B(a,\varepsilon)$ for all $n\geq n_{\varepsilon}$ $(*)$, where $B(a,\varepsilon)$ is the $\varepsilon$-radius ball around $a$. In particular we find $n_{1}\in\mathbb{N}$ so that $a_{n_{1}}\in B(a,\varepsilon)$, because if $a_{k}\notin B(a,\varepsilon)$ for all $k$, then $\sup_{n\geq k}a_{n}\notin B(a,\varepsilon)$ for all $k$ since $B(a,\varepsilon)^{c}$ is a closed set, which is a contradiction with $(*)$. Moreover by $(*)$ we find $n_{2}>n_{1}$ so that $a_{n_{2}}\in B(a,\varepsilon)$, and thus inductively we find an increasing sequence of indices $n_{1},n_{2},n_{3},...$ so that $a_{n_{k}}\in B(a,\varepsilon)$ for all $k\in\mathbb{N}$. Since $\varepsilon>0$ was arbitrary, then this shows that $a$ is an accumulation point of the sequence $(a_{n})_{n=1}^{\infty}$, i.e. $a\in A$.

So it remains to show that for any $c\in A$ we have $c\leq a$. Choose $c\in A$. For each $k\in\mathbb{N}$ take $a_{n_{k}}\in B(c,\frac{1}{k})$ with $n_{k}\geq k$ (such elements exists, since $c\in A$) to obtain a subsequence $(a_{n_{k}})_{k=1}^{\infty}$ converging to $c$. I will leave it for you to verify that now also $\inf_{j\in\mathbb{N}}\,(\sup_{k\geq j}a_{n_{k}})=c$. Hence, since $n_{k}\geq k$ for all $k$, then \begin{align*} c &=\lim_{k\to\infty}a_{n_{k}}=\inf_{j\in\mathbb{N}}\,(\sup_{k\geq j}a_{n_{k}})\leq \inf_{j\in\mathbb{N}}\,(\sup_{k\geq j}a_{k})=a, \end{align*} i.e. $c\leq a$. So this shows the claim, because we have shown that $a=\sup A$‚ and the right hand side is exactly your definition of $\limsup a_{n}$.

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$\inf\limits_{k\geq n} a_k$ is an increasing sequence, thus $\liminf\limits_{n\to\infty} := \lim\limits_{n\to\infty}\inf\limits_{k \geq n} a_k = \sup\limits_{n\in\mathbb N}\inf\limits_{k \geq n} a_k$ You can change the limit to a supremum, because the sequence $\inf\limits_{k\geq n} a_k$ is increasing.

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I don't think these are the definitions that he is using. – T. Eskin Jan 24 '13 at 20:47

Hint: you can use the fact that

$$\lim\; \inf\; a_n = \lim_{n\to\infty} \inf(\bigcup_{k=n}^\infty a_k)$$

and $\displaystyle A_n=\inf(\bigcup_{k=n}^\infty a_k)$ is a monotone sequence.

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that's why we need to do a $\lim$ before $A_n$ – mez Jan 24 '13 at 21:29
and I'm confused by what you meant with for all $n\in\mathbb{N}$, $$\lim_{n\to\infty} A_n=\lim\;\inf a_n$$ and both sides are just one number. – mez Jan 24 '13 at 21:32
you are right, let me change it – mez Jan 24 '13 at 21:35