Question

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

Answer 1

@abb0t

Answer 2

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

Answer 3

\[S \subseteq R\] 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

Answer 4

Give me time to read it. Thanks :)

Answer 5

Wow this is great :)

Answer 6

I didnt follow case 2 though

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

No like how did u show S contains the infimum

Answer 11

Case 2.

Answer 12

I showed4 supremum.

Answer 13

Ohhhhhh ok I thought i was going senile

Answer 14

:)
nah don't say that.

Answer 15

lol Thankssssssss. THIS WAS AWESOME. U R REALLY CLEARRRRRRRR