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Intro Analysis - Proof - max M

by brntspawn
Tags: analysis, intro, proof
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brntspawn
#1
Jan10-10, 08:35 PM
P: 12
1. The problem statement, all variables and given/known data
If S[tex]\subset[/tex]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[tex]\in[/tex]R such that for all x[tex]\in[/tex]S, 1[tex]\geq[/tex]x
So 1 is an upper bound for S
1[tex]\in[/tex]S, so by definition 1=max S
Let S={m+1}
Then m+1>m for all m[tex]\in[/tex]S
So m+1 is an upper bound for S
Since m+1[tex]\in[/tex]S, then by definition m+1=max S
Therefore if S[tex]\subset[/tex]R is finite and non-empty, then S has a maximum.
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LCKurtz
#2
Jan10-10, 08:49 PM
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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.
Dick
#3
Jan10-10, 09:37 PM
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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.

brntspawn
#4
Jan10-10, 09:53 PM
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.
Dick
#5
Jan10-10, 09:57 PM
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Quote Quote by brntspawn View Post
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.
That's a good attitude. But don't start with S={1}. That doesn't lead anywhere. Start with assuming that the number of elements in S, call it |S|, is 1. If S has one element can you prove it? Now proceed by induction on the number of elements in S, not the contents of S.
brntspawn
#6
Jan10-10, 10:00 PM
P: 12
Thanks Dick, I will work with that, I replied before seeing your post.
brntspawn
#7
Jan11-10, 01:13 AM
P: 12
Ok, lets see if this is any better:

Assume |S|=1
so -1[tex]\leq[/tex]S[tex]\leq[/tex]1
since 1[tex]\in[/tex][tex]\Re[/tex] and [tex]\forall[/tex]s[tex]\in[/tex]S, 1[tex]\geq[/tex]s, then by definition 1 is an upper bound
since 1[tex]\in[/tex]S then 1=max(S)
Thus true for 1, now consider the case for m+1
Assume |S|=m+1
then -m-1[tex]\leq[/tex]S[tex]\leq[/tex]m+1
[tex]\forall[/tex]m[tex]\in[/tex]S, m+1>m and since m+1[tex]\in[/tex] S and therefore m+1[tex]\in[/tex] [tex]\Re[/tex] then by definition m+1=Max(S)
Therefore if S[tex]\subset[/tex]R is finite and non-empty, then S has a maximum.
Altabeh
#8
Jan11-10, 04:56 AM
P: 665
Quote Quote by brntspawn View Post
Ok, lets see if this is any better:

Assume |S|=1
so -1[tex]\leq[/tex]S[tex]\leq[/tex]1
since 1[tex]\in[/tex][tex]\Re[/tex] and [tex]\forall[/tex]s[tex]\in[/tex]S, 1[tex]\geq[/tex]s, then by definition 1 is an upper bound
since 1[tex]\in[/tex]S then 1=max(S)
Thus true for 1, now consider the case for m+1
Assume |S|=m+1
then -m-1[tex]\leq[/tex]S[tex]\leq[/tex]m+1
[tex]\forall[/tex]m[tex]\in[/tex]S, m+1>m and since m+1[tex]\in[/tex] S and therefore m+1[tex]\in[/tex] [tex]\Re[/tex] then by definition m+1=Max(S)
Therefore if S[tex]\subset[/tex]R is finite and non-empty, then S has a maximum.
What about when m+1 is not in S? I mean an open set would also have a maximum though not included in the set itself or let's speak technically and say it's supremum or the least upper bound which cannot be exceeded by any x in S. How can you prove that one with induction? Remember on the real line the supremum of any set is the same as that of its set closure.
HallsofIvy
#9
Jan11-10, 06:11 AM
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Your problem that you are completely misinterpreting "|S|= 1" which I assume you saw somewhere. "S" is a set, not a number. |S| is "the number of elements in set S" or "the size of S". It has nothing whatsoever to do with the size of the numbers in S. Saying |S|= 1 does NOT mean "S= 1" or "S= -1" or that those numbers are in S, not does it mean that members of S are between -1 and 1, it just means that there is exactly one number in S. And, if there is only one number in S, that number must be the "largest" number in S.

Now, assume that, whenever |S|= k (that is, whenever as set S has k members), it has a largest member, max(S). Let T be a set containing k+1 members. Let "x" be any one of those members and S be the set T- {x}, that is, S is T with the single member x removed. Then S has exactly k members. What can you say about the largest members of S and T?


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