## anonymous 3 years ago Show that any non-empty finite set S ⊂ ℝ contains both its supremum and infimum. (Hint: use induction)

1. anonymous

@abb0t

2. anonymous

To be honest I really didnt study the chapter fully yet I was just looking for a tutor so hopefully I wont be lost lol

3. abb0t

$S \subseteq R$ such that |S| = n Proceed by induction on n. If n = 1, then S = {a} for some $a \in R$. Then, sup S = a. Assume that all subsets of R of order k contain their respective suprema. Let S be a subset of R in order of k+1. Label the elements of S as: $\left\{ a_1, a_2...a_{k+1} \right\}$ Let $T = \left\{ a_1, a_2...a_k \right\}$ By the inductive assumption: $\sup T= a_{i_o}$ for some $1 \le i_o \le k$ and $\sup(S-T) = a_{k+1}$ Thus clamining case 1: $a_{i_o} \le a_{k+1}$ since $a_{i_o} = \sup T$ and $a_{i_o} \le a_{k+1}$, you know that for all $1 \le i_o \le k, a_i \le a_{i_o} \le a_{k+1}$ thus $\sup S = a_{k+1}$ Case 2: $a_{i_o} \ge a_{k+1}$ since $a_{i_o} = \sup T$ and $a_{i_o} \ge a_{k+1}$ you know that for all $1 \le i \le k+1, a_i \le a_{i_o}$ thus $\sup S = a_{i_o}$ Thus, S contains its supremum

4. anonymous

Give me time to read it. Thanks :)

5. anonymous

Wow this is great :)

6. anonymous

I didnt follow case 2 though

7. anonymous

Ummm Also werent we supposed to prove that it contains its supremum and infimum

8. anonymous

Like I followed what u did but i am just confused why we needed to show case 2

9. abb0t

Well supremem is the upper bound, and infimum is the lower bound. I just used the definitions.

10. anonymous

No like how did u show S contains the infimum

11. abb0t

Case 2.

12. abb0t

I showed4 supremum.

13. anonymous

Ohhhhhh ok I thought i was going senile

14. abb0t

:) nah don't say that.

15. anonymous

lol Thankssssssss. THIS WAS AWESOME. U R REALLY CLEARRRRRRRR