1
$\begingroup$

I have just started reading set theory for a course I am taking on probability and random processes. There is this rather elementary problem for which I would appreciate an input. Find $\lim_{n\mapsto\infty} \ E_n \ where \ E_n = (0,\frac1n), n\in N$.

My Solution: To find the limit, I have to know $lim_{n\mapsto\infty} \ inf \ E_n$ and $lim_{n\mapsto\infty} \ sup \ E_n$. Now I have read that the limit infimum of a sequence of sets {$E_n$}, $n\in N$ is the set {$x\in\Omega:x \ belongs \ to \ E_n \ for \ all \ but \ finitely \ many \ values \ of \ n$}. By this definition, $\lim_{n\mapsto\infty} \ inf\ E_n = (0,\frac1n)=\emptyset $ since from the sequence, $E_1=(0,1)$; $E_2=(0,\frac12)$; $E_3=(0,\frac13)$; $E_4=(0,\frac14)$; $E_5=(0,\frac15)$;......

I can see that for any number $x$ (of the form $\frac1 k, k \in N$) in $(0,1)$, there exist infinitely many $n$ for which $x\notin E_n$.

Similarly, the limit supremum of a sequence of sets {$E_n$}, $n\in N$ is the set {$x\in\Omega:x \ belongs \ to \ E_n \ for \ infinitely \ many \ values \ of \ n$}. While I can guess the limit supremum is also $\emptyset $, I am not really able to explain it. Would appreciate if someone can help me understand this concept more clearly.

$\endgroup$
3
  • 2
    $\begingroup$ What is the limit for sets? $\endgroup$
    – Hanul Jeon
    May 21, 2017 at 10:12
  • $\begingroup$ Intuitively, the limit of a sequence of sets is the set to which that sequence converges. Thank you for the question; it complimented the understanding. $\endgroup$
    – GA-Student
    May 21, 2017 at 11:16
  • $\begingroup$ Definitions are your friends. Sometimes coming up with a rigorous definition that captures the intuition of a concept is a big accomplishment. $\endgroup$
    – hardmath
    May 21, 2017 at 13:18

1 Answer 1

0
$\begingroup$

What you mean by limit here is $$E =^{\text{def}} \bigcap_{i=1}^\infty E_i$$ since for all $n\in \Bbb N^*, E_{n}\supset E_{n+1}$.

Now if you take $x\in E$, then $x\in E_1$ therefore $x\gt 0$. However then there exist infinitely many such $n\in\Bbb N^*$ such that ${1\over n}\le x$ and thus $x\notin E_n$ for infinitely many $n\in\Bbb N^*$. So $x\notin E$.

This means $$E = \emptyset$$

$\endgroup$
1

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .