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

HippoCampus Calculus & Advanced Math
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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
\[S \subseteq R\] 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

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Give me time to read it. Thanks :)
Wow this is great :)
I didnt follow case 2 though
Ummm Also werent we supposed to prove that it contains its supremum and infimum
Like I followed what u did but i am just confused why we needed to show case 2
Well supremem is the upper bound, and infimum is the lower bound. I just used the definitions.
No like how did u show S contains the infimum
Case 2.
I showed4 supremum.
Ohhhhhh ok I thought i was going senile
:) nah don't say that.
lol Thankssssssss. THIS WAS AWESOME. U R REALLY CLEARRRRRRRR

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