Intro Analysis - Proof - max M

Can you use the fact that S has a largest member to extend that fact to T? Or, if you want to use your idea, which is correct, of using induction on the number of members in S, then you should say that, since S has 1 member, that member must be the largest member of S. Now, assume that, whenever a set has k members, it has a largest member. Then consider a set T with k+1 members. Choose one of those members, call it x, and let S be the set T- {x}. Then S has exactly k members. What can you say about the largest members of S and T? Can you use the fact
  • #1
brntspawn
12
0

Homework Statement


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.
 
Physics news on Phys.org
  • #2
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.
 
  • #3
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.
 
  • #4
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.
 
  • #5
brntspawn said:
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.
 
  • #6
Thanks Dick, I will work with that, I replied before seeing your post.
 
  • #7
Ok, let's 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.
 
  • #8
brntspawn said:
Ok, let's 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.
 
  • #9
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 anyone 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?
 

1. What is "Intro Analysis"?

"Intro Analysis" is a branch of mathematics that focuses on understanding and proving the properties of real numbers, functions, and sets. It is often a foundational course for students pursuing degrees in mathematics, engineering, and other related fields.

2. What is a "Proof" in mathematics?

A "Proof" in mathematics is a logical argument that demonstrates the validity of a mathematical statement or theorem. It is a step-by-step explanation of how a statement can be derived from previously proven statements or axioms, using accepted rules of logic.

3. What does "max M" mean in the context of Intro Analysis?

"max M" refers to the maximum or largest value of a given set of numbers or functions. In Intro Analysis, the concept of "max M" is often used in proofs to show that a certain value or function is the largest possible value within a given set of values or functions.

4. Why is "Intro Analysis" important in the field of science?

"Intro Analysis" is important in the field of science because it provides a rigorous framework for understanding and analyzing real-world problems. By learning how to prove mathematical statements and properties, scientists can apply these skills to develop and test theories, models, and experiments.

5. How can one improve their skills in "Intro Analysis" and proving mathematical statements?

One can improve their skills in "Intro Analysis" and proving mathematical statements by practicing regularly, seeking help from professors or tutors, and learning from example proofs. It is also helpful to have a solid understanding of algebra, calculus, and other mathematical concepts before diving into Intro Analysis.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
103
  • Calculus and Beyond Homework Help
Replies
4
Views
460
  • Calculus and Beyond Homework Help
Replies
1
Views
455
  • Calculus and Beyond Homework Help
Replies
1
Views
532
  • Calculus and Beyond Homework Help
Replies
1
Views
450
  • Calculus and Beyond Homework Help
Replies
4
Views
125
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
466
  • Calculus and Beyond Homework Help
Replies
34
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
761
Back
Top