# Intro Analysis - Proof - max M

by brntspawn
Tags: analysis, intro, proof
 P: 12 1. The problem statement, all variables and given/known data If S$$\subset$$R is finite and non-empty, then S has a maximum. Can someone look over this? I struggled a bit in my first proof class, which is why I am asking for help, so I really am unsure if this is right at all. Let S={1} So 1$$\in$$R such that for all x$$\in$$S, 1$$\geq$$x So 1 is an upper bound for S 1$$\in$$S, so by definition 1=max S Let S={m+1} Then m+1>m for all m$$\in$$S So m+1 is an upper bound for S Since m+1$$\in$$S, then by definition m+1=max S Therefore if S$$\subset$$R is finite and non-empty, then S has a maximum.
 PF Patron HW Helper Thanks P: 6,774 Does R represent the real numbers, so S is a finite subset of R? If so, why would you suppose 1 is in S? And why would 1 be an upper bound for S. Your "proof" looks very confused. And why would you be using induction? Perhaps if you gave a careful complete statement of the problem, we could give helpful suggestions.
 HW Helper Sci Advisor Thanks P: 24,510 Your 'proof' is complete gibberish. On the other hand, induction isn't a bad idea, as LCKurtz suggests. But do induction on the number of elements in S.
P: 12

## Intro Analysis - Proof - max M

What I posted was the entire problem. Yes R was for the set of Reals. It had a hint with it: Use induction, which is why I was trying to use induction.
Like I said I struggled in the intro class, so it is not surprising to me that my proof looks confusing.
I assume starting with S={1} is where my problems start, I was thinking that if S is finite in the reals, I could choose any number and go from there. The way we were taught induction was to show it was true for 1 first which is why I chose 1.
I am not looking for the problem to be done, as I very much need to be able to do this on my own, so hints in the right direction would be great.
HW Helper
 P: 12 Ok, lets see if this is any better: Assume |S|=1 so -1$$\leq$$S$$\leq$$1 since 1$$\in$$$$\Re$$ and $$\forall$$s$$\in$$S, 1$$\geq$$s, then by definition 1 is an upper bound since 1$$\in$$S then 1=max(S) Thus true for 1, now consider the case for m+1 Assume |S|=m+1 then -m-1$$\leq$$S$$\leq$$m+1 $$\forall$$m$$\in$$S, m+1>m and since m+1$$\in$$ S and therefore m+1$$\in$$ $$\Re$$ then by definition m+1=Max(S) Therefore if S$$\subset$$R is finite and non-empty, then S has a maximum.
 Quote by brntspawn Ok, lets see if this is any better: Assume |S|=1 so -1$$\leq$$S$$\leq$$1 since 1$$\in$$$$\Re$$ and $$\forall$$s$$\in$$S, 1$$\geq$$s, then by definition 1 is an upper bound since 1$$\in$$S then 1=max(S) Thus true for 1, now consider the case for m+1 Assume |S|=m+1 then -m-1$$\leq$$S$$\leq$$m+1 $$\forall$$m$$\in$$S, m+1>m and since m+1$$\in$$ S and therefore m+1$$\in$$ $$\Re$$ then by definition m+1=Max(S) Therefore if S$$\subset$$R is finite and non-empty, then S has a maximum.