1. more proofs in analysis

Problem 1:

Let S be a non-empty set bounded by the subset of real numbers. Prove that sup S is unique.

Problem 2:

Let S and T be non-empty bounded subsets of real numbers were S is a subset of T. Prove that
inf T <= inf S <= sup S <= sup T

Problem 3:

A) prove: if x and y are real numbers with x < y, then there are infitely many rational numbers in the interval [x,y]

B) repeat part A for irrational numbers

2. Here is some help. On #1.
Suppose that A=sup(S) and B=sup(S) is A is not B then one is less than the other: say that B<A. The by definition of supremum, there is an element, x, in S such that B<x<=A. Do you see the contradiction?

On #3(a). We know that between any two real numbers there is a rational number. Then there is a rational number between x & y, r_1. There is a rational number between x & r-1, r_2. How do we know that r_1 & r_2 are distinct? For each positive integer n>=3, there is a rational number r_n between x & r_{n-1}. How does this prove the statement?

3. Thank you for your help! I will look in to the insight you gave me. These problems were from a test, these are the problems that I got wrong. The professor is willing to give us a fourth of the problems missed back if we can come up with the right solution by monday. So how would you complete the problem?

4. Originally Posted by luckyc1423
Problem 1:

Let S be a non-empty set bounded by the subset of real numbers. Prove that sup S is unique.

Problem 2:

Let S and T be non-empty bounded subsets of real numbers were S is a subset of T. Prove that
inf T <= inf S <= sup S <= sup T

Problem 3:

A) prove: if x and y are real numbers with x < y, then there are infitely many rational numbers in the interval [x,y]

B) repeat part A for irrational numbers

Problem 1: I think Plato handled that one well.

Problem 2:

Let S and T be non-empty bounded subsets of real numbers were S is a subset of T. We have two cases: (1) S=T, and (2) S is a proper subset of T. We show that in either case, inf T <= inf S <= sup S <= sup T.

case 1: If S=T, then it follows immediately that inf T <= inf S <= sup S <= sup T.

case 2: S is a proper subset of T.

Then for all s in S, we have s in T.

By definition: infT<= supT and infS<=supS. Thus it is sufficient to show that (i) infT<=infS and (ii) supS<=supT.

(i) By definition, infT<= t, for all t in T, and so infT<=s for all s in S (since for all s in S, we have s in T). Let s0 = min{infS, infT}, so by definition, infT<=s0. So we have infT<=infT (which is trivial) or infT<=infS. Thus we have infT<=infS.

(ii) The proof of supS<=supT is similar to the part (i), we just change "inf" to "sup", "<=" to ">=", and "min" to "max"

Problem 3:
(A) Let x and y are real numbers with x < y. We will show that there are infinately many rationals between x and y, that is, x<(m/n)<y, for m,n integers and n>0.

x<(m/n)<y so we have nx<m<ny. Since x<y, we have y - x>0. By the Archimedean property, we have n(y - x)>1 for some n in N. Since ny - nx>1, clearly there is at least one integer between them, so nx<m<ny holds. Now we show that such an m exists. By the Archimedean property we have, for some k>max{|nx|,|ny|},

-k < nx < ny < k

So the set {a in Z: -k<a<=k and nx<a} is finite and nonempty. S we can let our m = min{a in Z: -k<a<=k and nx<a}.

Then xn< m, and m-1 <= an

so m = (m - 1) + 1 <= xn + 1 < xn + (yn - xn) = yn

so nx < m < ny holds.

(B) Let I be the set of irrational numbers. We show that for x,y in R, x<y, we have i in I, such that x < i < y

lemma: the set {i : i = r + sqrt(2), r in Q} is a subset of I.
Assume to the contrary that r + sqrt(2) is not a member of I. Then r + sqrt(2) = m/n for some m,n integers, n>0. So sqrt(2) = m/n - r = (m - nr)/n. Since m-nr and n are in Z. We have sqrt(2) being a rational number, which is clearly a contradiction. Thus, it must be the case that r + sqrt(2) are in I.

Now we show x < i < y.
Let x, y be real numbers. Then x-sqrt(2) and y-sqrt(2) are also real. By the denseness of Q property, proven in (A) above, we have some r in Q, such that x-sqrt(2) < r < y-sqrt(2). Adding sqrt(2) throughout the system, we obtain:

x < r + sqrt(2) < y. Be r + sqrt(2) represents an irrational number i, so x < i < y.

5. Originally Posted by luckyc1423
Problem 1:

Let S be a non-empty set bounded by the subset of real numbers. Prove that sup S is unique.

Problem 2:

Let S and T be non-empty bounded subsets of real numbers were S is a subset of T. Prove that
inf T <= inf S <= sup S <= sup T

Problem 3:

A) prove: if x and y are real numbers with x < y, then there are infitely many rational numbers in the interval [x,y]

B) repeat part A for irrational numbers
Here is, perhaps, a better proof for problem 2:

Let S and T be non-empty bounded subsets of real numbers were S is a subset of T.

By definition, infT<=supT and infS<=supS. Thus we have only to prove infT<=infS and supS<=supT.

Let s = infS, and t = infT. Then, by definition of infimum, if m is a lower bound for S, then s>=m. Now let y be in S. Then y is in T, since S is a subset of T. Thus, t <= y for every y in T, since t = infT. So t is a lower bound for S. But s>= t by the definition of infimum. so we have infS>=infT.

Showing supS<=supT is similar to the proof above.

6. Thanks alot for that insight, this helps me out a ton! Have a good day!

7. Originally Posted by luckyc1423
Thanks alot for that insight, this helps me out a ton! Have a good day!
So you followed everything right? What i did for parts 2 and 3 were not just insights, they were actual proofs. You should be able to hand them in verbatim and get full credit. I'd probably hand in the last proof i gave for problem 2. Of course where i wrote "the proof of <blank> is similar to <blank>", i'd actually write out the proof.

8. Okay, is the first problem a complete proof?
I am only asking this stuff is I can manage to get a fourth of the points back on the problems I missed. I didnt do well on the test, matter of fact I got a D, but I talked to the prof and he said considering how everyone else sucked on the test to, that would be a low b or a C.
I am usually very good in math but proofs are not my thing. I am praying and hoping I can pull a C out of the class because I am scheduled to graduate this coming May, and I already have a job lined up starting June first, so I have alot riding on this class.
I usually have homework once a week that is somewhat as complicated as this, so I am always needing help in this course.
Did you see my other post about the binomial coefficient?

9. Jhevon, how does what you did in part 3(a) prove that there are infinitely many rationals between x & y?

10. Originally Posted by Plato
Jhevon, how does what you did in part 3(a) prove that there are infinitely many rationals between x & y?
in 3(a) i showed that i can find integers m,n so that x< m/n < y. the set of integers is infinite, so there are infinite numbers of the form m/n. and of course, anything of the form m/n is rational

in 3 (b) since r is any element of Q, and Q is infinite, it follows there are infinite numbers of the form r + srqt(2). I showed that this number was irrational and that it was between x and y, so there are infinite irrationals of this form between x and y

11. Originally Posted by Jhevon
in 3(a) i showed that i can find integers m,n so that x< m/n < y. the set of integers is infinite, so there are infinite numbers of the form m/n
No that is no proof! You have shown that the is at least one rational.
You have not shown that there are infinitely many.

12. For any a<b real numbers.

Define now new numbers
a-sqrt(2)<b-sqrt(2)

We know there exists a rational "r" thus,

a-sqrt(2)<r<b-sqrt(2)

Thus,
a<r+sqrt(2)<b

Q.E.D.

13. Originally Posted by Plato
No that is no proof! You have shown that the is at least one rational.
You have not shown that there are infinitely many.
Read the proof again, i showed there is at least one m that satisfies nx < m < ny. But there are infinitely many n's that can work. Thus, for infinitely many n's i can find at least one m for each of them, so i end up with infinitely many m/n numbers

14. Originally Posted by Jhevon
Read the proof again, i showed there is at least one m that satisfies nx < m < ny. But there are infinitely many n's that can work. Thus, for infinitely many n's i can find at least one m for each of them, so i end up with infinitely many m/n numbers
The way I proved it for homework, is by contradiction. I assumed that there are only finitely many rational numbers. And I let the set S represent all the rationals, which is finite. Then I show that leads to a contradiction.

15. Originally Posted by luckyc1423
Problem 1:

Let S be a non-empty set bounded by the subset of real numbers. Prove that sup S is unique.
I do not see how that is a problem. Since "sup S" is used, does it not mean the definition is "well-defined", i.e. it is unique. Hence there is nothing to prove.

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