1

What exactly is the meaning of the following $$u_N = \inf\{ s_n : n > N\} \ \ \text{ and} \ \ v_N = \sup\{ s_n : n > N\}$$

This might seem a stupid question, but I am not understanding the meaning of these sets. We know that the infimum and the supremum should be just one, so what is the meaning of the sets $u_N$ and $v_N$?

This all came out studying monotone sequences...

Aaron Maroja
  • 17,571
  • Both are numbers, not sets. Eg, $\operatorname{inf} {\frac{1}{n}, n \in \mathbb{N} } = 0$. – gnometorule Apr 08 '15 at 14:36
  • The wikipedia description is pretty good. I don't know what I could say in the abstract that's better. :) There are also many questions on MSE with solved examples you could look for. It takes a bit time to get used to. – gnometorule Apr 08 '15 at 14:47
  • $x = \lim \inf a_n$ is the smallest number where, no matter how far you go, you'll find numbers in the sequence $(a_n)$ bigger than x – Tryss Apr 08 '15 at 14:57
  • Another way to define $\lim\sup$ and $\lim\inf$ is as the $\sup$ and $\inf$ of the set of all subsequential limits. This is how they are defined in baby Rudin. – Math1000 Apr 08 '15 at 15:06

2 Answers2

2

The definition of limit superior and limit inferior you may find here, and an interesting post here.

As an example to elucidate you could take the sequence $x_n = 1 - \frac{1}{n}$. Defining $u_N$ and $v_N$ we have

$$u_N = \inf\,\,\{ 1 - \frac{1}{n}; n \geq N\} = \inf\,\,\{1 - \frac{1}{N}, 1 - \frac{1}{N+1}, 1- \frac{1}{N+2}, \ldots\} = 1 - \frac{1}{N}$$

and

$$v_N = \sup \,\,\{1 - \frac{1}{n} ; n \geq N\} = \sup\,\,\{1 - \frac{1}{N}, 1 - \frac{1}{N+1}, 1 - \frac{1}{N+2}, \ldots\} = 1$$

then $$\lim \inf x_n = \lim_{N \to \infty} u_N = 1 \ \ \text{and} \ \ \lim \sup x_n = \lim_{N \to \infty} v_N = 1$$

Community
  • 1
Aaron Maroja
  • 17,571
  • That's it. For each $N$ the set ${1 - \frac{1}{n} ; n \geq N}$ has an infimum and supremum (real set property). You look at those on the limit, to see whether your sequence has a limit. There are cases where $\lim x_n$ does not exist though, because $$\lim \sup x_n \neq \lim \inf x_n $$, for intance $x_n = (-1) ^{n+1}$ (check it), on this case we say that $x_n$ does not converge. – Aaron Maroja Apr 08 '15 at 15:09
0

If $u_N$ and $v_N$ are finite, it mean that $$\forall \varepsilon>0, \exists n>N: u_N<s_n<u_N+\varepsilon$$ and $$\forall \varepsilon>0,\exists n>N: v_N-\varepsilon<s_n<v_N.$$

If for example $u_N$ is not finite, (and is for example $+\infty $), it mean that $$\forall M\in \mathbb N, \exists n>N: M< s_n.$$

Surb
  • 55,662