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Let ${\left\{ {{c_n}} \right\}_{n \ge 1}}$, a bounded sequence of real numbers.

Let us define: $$\begin{array}{l} {a_n} = \inf \left\{ {{c_k}:k \ge n} \right\} \\ {b_n} = \sup \left\{ {{c_k}:k \ge n} \right\} \\ \end{array}$$

$$\begin{array}{l} \lim \inf c_n: = \mathop {\lim }\limits_{n \to \infty } {a_n} \\ \lim \sup c_n: = \mathop {\lim }\limits_{n \to \infty } {b_n} \\ \end{array}$$

I don't understand very well the above sets definitions, and why it implies that: $a_n$ is the smallest partial limit and $b_n$ is the greatest partial limit.

Martin Sleziak
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AndrePoole
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2 Answers2

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You might want to see this and this too.

Claim Let $(c_{n_k})_{k\geqslant 1})$ be a subsequence of $(c_k)_{k\geqslant 1}$. Then $$\liminf_{k\to\infty} c_k\leqslant \liminf_{k\to\infty} c_{n_k}\leqslant \limsup_{k\to\infty} c_{n_k}\leqslant \limsup_{k\to\infty} c_k$$

Proof We know that for any subsequence, $n_k\geqslant k$. This means that we have the containments

$$\{c_k:k\geqslant N\}\supseteq \{c_{n_k}:k\geqslant N\}$$

hold. Thus $$\sup_{k\geqslant N}c_k=\sup\{c_k:k\geqslant N\}\geqslant \sup\{c_{n_k}:k\geqslant N\}=\sup_{n_k\geqslant N}c_{n_k}$$ $$\inf_{k\geqslant N}c_k=\inf\{c_k:k\geqslant N\}\leqslant \inf\{c_{n_k}:k\geqslant N\}=\inf_{n_k\geqslant N}c_{n_k}$$

Taking the limits proves both claims. $\blacktriangleleft$

Corollary Let $(c_{n_k})_{k\geqslant 1})$ be a convergent subsequence of $(c_k)_{k\geqslant 1}$. Then $$\liminf_{k\to\infty} c_k\leqslant \lim_{k\to\infty} c_{n_k}\leqslant \limsup_{k\to\infty} c_k$$

Community
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Pedro
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For every sequence {${a_n}$}, Define a set of indexes (subset of Natural numbers) - $I={b_1,b_2,...}$, such that $b1<b2<..$. Than the sequence ${a_{b_n}}$ called sub-sequence of {${a_n}$}. a subsequence doesn't have to converge (!). Notice the subsequence that converges to the minimal possible value - this value is the lim inf. Similarly is the lim sup. So lim sup is the maximal partial limit of the sequence and lim inf is ithe minimal partial limit of the sequence.

Hope it helped.

Astro Nauft
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