Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Here's an example.

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..

share|cite|improve this question
up vote 3 down vote accepted

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 $$\lim X_i=\mathrm{liminf}~X_i= \mathrm{limsup}~X_i.$$

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.