1
$\begingroup$

My textbook defines the limit supremum and limit infimum of a sequence $(a_i)$ as

$$ b_k = \sup\{a_i : i \ge k\} \\ c_k = \inf\{a_i : i \ge k\} $$

The book also states that $b_k \le b_{k+1}$ and $c_k \ge c_{k+1}$, but I don't understand why these inequalities are true.

Take the supremum. $b_1$ is the supremum of $a_1, a_2, a_3, \cdots$. If we look at $b_2$, it's the supremum of $a_2, a_3, \cdots$. How is that $b_2 \ge b_1$? For example, if the sequence is strictly decreasing, so that $a_1 > a_2 > a_3 > \cdots$, then removing $a_1$ from the set we're taking the supremum (least upper bound) of should only make the supremum smaller, right?

The image on Wikipedia seems to bear this out, unless I'm reading it wrong. Can someone clarify this for me?

$\endgroup$
2
$\begingroup$

You are correct and the book has made some mistake: $b_k$ is a decreasing sequence and $c_k$ is an increasing sequence. Your logic is fine, but we can make things precise. I'll demonstrate this for $b_k$ (the $c_k$ case is completely analogous).

$$b_k = \sup\{a_i : i \ge k\}=\max\{a_k,\sup\{a_i : i \ge k+1\}\}\ge\sup\{a_i : i \ge k+1\}=b_{k+1}$$

$\endgroup$
2
$\begingroup$

Your argument is right. $b_k$ is the supremum of more numbers than $b_{k+1}$, hence it can hardly be smaller. So indeed we have $b_1\ge b_2\ge b_3\ge \ldots$. And likewise $c_1\le c_2\le c_3\le \ldots$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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