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Let $s\in \mathbb{Z}^+$ and $q=2s+1$.

Define $T_{s,i}(x)=\frac{x}{q} + \frac{2i}{q}$ where $x\in\mathbb{R}$ and $i=0,1,\ldots,s$.

Define recursively $I_{s,0} = [0,1]$ and $I_{s,n+1} =\bigcup_{i=0}^s T_{s,i} (I_{s,n})$

Let $\Gamma (s) = \bigcap_{n\in\omega} I_{s,n}$

The text i'm reading says $\lim_{s\to \infty} \Gamma (s)$ does not exist but $\lim\inf_{s\to \infty} \Gamma (s)$ exists.

I have no idea what this limit refers to..

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2 Answers 2

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Given a family of sets $(X_i)_{i\in\Bbb N}$, you can define its upper and lower limits as follows : $$\mathrm{lim~inf}~X_i=\bigcup_{n=0}^{\infty}\bigcap_{k=n}^{\infty}X_k$$ and $$\mathrm{lim~sup}~X_i=\bigcap_{n=0}^{\infty}\bigcup_{k=n}^{\infty}X_k.$$ The first set is the set of all elements that appear in all $X_n$ provided $n$ is large enough, and the second is the set of all elements that never disappear, that is, for all $n$ there is $N\geq n$ such that $X_N$ contains it. Obviously $$\mathrm{lim~inf}~X_i\subset \mathrm{lim~sup}~X_i$$ and I've seen some older probability texts define the sequence $(X_i)$ to have a limit whenever the two sets are equal. One then sets $$\lim X_i=\mathrm{liminf}~X_i= \mathrm{limsup}~X_i.$$

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The relevant definitions can be found here:

$$\liminf_{s\to\infty}\Gamma(s)=\bigcup_{s\ge 1}\left(\bigcap_{k\ge s}\Gamma(k)\right)\;,\tag{1}$$

$$\limsup_{s\to\infty}\Gamma(s)=\bigcap_{s\ge 1}\left(\bigcup_{k\ge s}\Gamma(k)\right)\;,\tag{2}$$

and $\displaystyle\lim_{s\to\infty}\Gamma(s)$ exists iff $\displaystyle\liminf_{s\to\infty}\Gamma(s)=\limsup_{s\to\infty}\Gamma(s)$, in which case $\displaystyle\lim_{s\to\infty}\Gamma(s)=\liminf_{s\to\infty}\Gamma(s)$.

It follows from $(1)$ that $x\in\liminf\limits_{s\to\infty}\Gamma(s)$ iff $x$ is in all of the sets $\Gamma(s)$ from some $s$ on:

$$x\in\liminf_{s\to\infty}\Gamma(s)\quad\text{iff}\quad\exists m\in\Bbb Z^+\text{ such that }x\in\Gamma(s)\text{ for all }s\ge m\;.$$

It follows from $(2)$ that $x\in\limsup\limits_{s\to\infty}\Gamma(s)$ iff $x$ belongs to infinitely many of the sets $\Gamma(s)$:

$$x\in\limsup_{s\to\infty}\Gamma(s)\quad\text{iff}\quad\forall m\in\Bbb Z^+~\exists s\ge m\text{ such that }x\in\Gamma(s)\;.$$

It follows that $\lim\limits_{s\to\infty}\Gamma(s)$ exists iff each point that belongs to infinitely many of the $\Gamma(s)$ belongs to all of them from some point on; apparently this is not the case, however.

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