Monotone nature of sequences used to define 'lim sup' and 'lim inf'

Question

Hi guys, this is from the definition of lim sup and lim inf.

I don't quite understand the second paragraph which goes "As n increases...".

How exactly could (yₙ) be monotone decreasing? Doesn't it depend on the original sequence?

For example when the original sequence diverges to +infinity...since we are taking the sup the set of values for yₙ will always be +infinity and hence not monotone decreasing right?

Please help me get around this,is my intution correct or have I got the definitions wrong...really appreciate your help.

Thanks for your time.

Answer 1

Let $E_n=\{x_k: k\geq n\}$. Then $E_{n+1}\subset E_n$ for all $n$.

Claim: If $A \subset B$, then $\sup A \leq \sup B$.

Proof. For all $b\in B$, we know $b \leq \sup B$. Since $A\subset B$, we know that for all $a\in A,$ we have $a\leq \sup B$. Hence, $\sup B$ is an upper bound for $A$. By definition of supremum as the least upper bound, it follows that $\sup A \leq \sup B$.

From here, it should be obvious that $y_n$ is monotone decreasing. Your remark about blow up at infinity is valid, as in this case, we just have $y_n=+\infty$ for all $n$, which for the convenience of notation, we say approaches $+\infty$.

A similar argument holds for the lim inf.

One of the really nice things about lim inf and lim sup in analysis is that they always exist in some intuitive way, and this is helpful when one wants to work with subsequences (which gets more at the heart of compactness, a big topic in research).

The nice thing about monotone sequences is that even when the limit "doesn't exist," it always doesn't exist because of blow up, not because it just oscillates.

Answer 2

What you have to realize is that if $I\subseteq J\subseteq\Bbb N$, then

$$\sup\{x_n:n\in I\}\le\sup\{x_n:n\in J\}\,.\tag{1}$$

This is because $\{x_n:n\in J\}$ includes every $x_n$ with $n\in I$, so its supremum is at least as big as the supremum of $\{x_n:n\in I\}$, and it also contains $x_n$ for each $n\in J\setminus I$. These extra members cannot decrease the supremum and might increase it.

In the present instance we have $n_1,n_2\in\Bbb N$ with $n_1\le n_2$, and we’re looking at $I=\{k\in\Bbb N:k\ge n_2\}$ and $J=\{k\in\Bbb N:k\ge n_1\}$. Since $n_1\le n_2$, $I\subseteq J$, and $(1)$ holds:

$$y_{n_2}=\sup\{x_k:k\ge n_2\}\le\sup\{x_k:k\ge n_1\}=y_{n_1}\,,$$

and $\langle y_n:n\in\Bbb N\rangle$ is indeed non-increasing. The specific values $y_n$ certainly depend on the original sequence, but the fact that they are non-increasing does not.

If the original sequence increases without bound, then $y_n=\infty$ for each $n\in\Bbb N$, so $\langle y_n:n\in\Bbb N\rangle$ is a constant sequence (in the extended reals), and all constant sequences are non-increasing. (They are also non-decreasing.) Note that the claim is not that $\langle y_n:n\in\Bbb N\rangle$ is strictly decreasing: the term decreasing in this context generally means non-strictly decreasing, i.e., non-increasing.

Answer 3

Think about listing out the elements in $(x_n)$: $$ x_1, x_2, x_3, x_4, x_5, \dots $$ Then \begin{align*} y_1 & = \sup\{ x_1, x_2, \dots \},\\ y_2 &= \sup\{ x_2, x_3, x_4, \dots \},\\ \end{align*} and so on.

So $y_1$ is the "maximum" over all $x_1, x_2, \dots$ and $y_2$ is the "maximum" over all $x_2, x_3, x_4, \dots$, and so on. This means that $y_1$ can only be greater or equal to $y_2$: if $x_1 \leq y_2$, then $y_1 = y_2$, and if $x_1 > y_2$, then $y_1 = x_1 > y_2$.

You can draw this relationship like this: $$ \underbrace{x_1, \underbrace{x_2, x_3, x_4, \dots}_{\sup \text{of these elements} = y_2}}_{\sup \text{of these elements}= y_1} $$ This nesting effect is why we can say $(y_n)$ is monotone decreasing, without knowing anything about $(x_n)$.

Similar logic applies to general $y_k$. You can compare $y_{k}$ and $y_{k+1}$ by noting that $y_k = \max \{ y_{k+1}, x_k \} \geq y_{k+1}$. Thus, $(y_k)$ is monotone decreasing.