Introductory Analysis Watch

Rohan77642
Badges: 12
Rep:
?
#1
Report Thread starter 1 week ago
#1
I need help calculating the supremum and infimum of the following sets.
The questions asks to compute, without proofs. Can someone verify my answers and help me with the ones I am stuck.

a) {m/n: where m and n are natural with m < n}

I got sup = 1 because m < n rearranges to m/n < 1.
I am stuck on the infimum. Is it 0?

b) {m/m+n: where m and n are natural}

I don't think this has a supremum and infimum. But its a guess. Nothing concrete.

c) { ((-1)^m)/n: where m and n are natural}

Sup = 1 and Inf = 0. For Sup I just thought of it as fixing m = even and then we have the set {1/n: where n is a natural number} and here sup = 1 and inf = 0.

a) {n/(3n+1): where n is natural}
Sup = 1/3 because lim n tends to infinity is 1/3?
Inf I have no idea.

These questions are from Abott Understanding Analysis. I am to start uni in October so I am self-teaching analysis, and hence I am a completely rookie/beginner. Just wanted to put it out there if my answers look ridiculous and stupid
0
reply
ghostwalker
  • Study Helper
Badges: 15
#2
Report 1 week ago
#2
(Original post by Rohan77642)
I need help calculating the supremum and infimum of the following sets.
The questions asks to compute, without proofs. Can someone verify my answers and help me with the ones I am stuck.

a) {m/n: where m and n are natural with m < n}

I got sup = 1 because m < n rearranges to m/n < 1.
I am stuck on the infimum. Is it 0?
Agreed, 0 and 1.

We can see that m/n is always less than 1. So, 1 is an upper bound. Is it the least upper bound? Well yes, because we can think of the elments (n-1)/n for n>1, which belong to this set, and taking n as large as we like, we can get as close to 1 as we like, from below.

As to 0.

Well we know m/n is greater than 0. So, 0 is a lower bound. Is it the greatest lower bound? Again yes, since we can consider the elements 1/n which belong to the set, and in a similar fashion we can get as close to 0 as we like, from above.


b) {m/m+n: where m and n are natural}

I don't think this has a supremum and infimum. But its a guess. Nothing concrete.
It has both.

m,n are both positive integers. so, m/(n+n) is always positive. Also m < m+n, so m/(m+n) < 1. So we know the set is bounded below and above. What might the sup and inf be?


c) { ((-1)^m)/n: where m and n are natural}

Sup = 1 and Inf = 0. For Sup I just thought of it as fixing m = even and then we have the set {1/n: where n is a natural number} and here sup = 1 and inf = 0.
Sup is 1, since the largest positive element is 1.

However inf is not 0. If m is odd, then the element is negative. What's the lowest negative number it can be?

a) {n/(3n+1): where n is natural}
Sup = 1/3 because lim n tends to infinity is 1/3?
Inf I have no idea.
It might help to divide top and bottom by n, think about what you have and consider the first few values of n.

The sup is indeed 1/3, since all terms are less than 1/3, and making n as large as we like we can get as close to 1/3 as we like, from below.

Have a further think on the inf.
Last edited by ghostwalker; 1 week ago
reply
RDKGames
  • Community Assistant
Badges: 20
Rep:
?
#3
Report 1 week ago
#3
(Original post by Rohan77642)
I need help calculating the supremum and infimum of the following sets.
The questions asks to compute, without proofs. Can someone verify my answers and help me with the ones I am stuck.

a) {m/n: where m and n are natural with m < n}

I got sup = 1 because m < n rearranges to m/n < 1.
I am stuck on the infimum. Is it 0?
Sounds good. The infimum is indeed 0. You can make the denominator as large as you want and the fraction will tend to zero.

b) {m/m+n: where m and n are natural}

I don't think this has a supremum and infimum. But its a guess. Nothing concrete.
It does. What happens when you fix m and let n get very large? Or what happens when n=1 and you let m get very large?

c) { ((-1)^m)/n: where m and n are natural}

Sup = 1 and Inf = 0. For Sup I just thought of it as fixing m = even and then we have the set {1/n: where n is a natural number} and here sup = 1 and inf = 0.
Supremum looks good, but infimum not so much. I don't see why you set m being even here; we lose generality.

What happens when n=1 and m is odd??

a) {n/(3n+1): where n is natural}
Sup = 1/3 because lim n tends to infinity is 1/3?
Inf I have no idea.
Supremum is good.

But notice that this is clearly an increasing sequence. Therefore you can find the smallest element of this set.
2
reply
Rohan77642
Badges: 12
Rep:
?
#4
Report Thread starter 1 week ago
#4
(Original post by ghostwalker)
Agreed, 0 and 1.

We can see that m/n is always less than 1. So, 1 is an upper bound. Is it the least upper bound? Well yes, because we can think of the elments (n-1)/n for n>1, which belong to this set, and taking n as large as wel like, be can get as close to 1 as we like.

As to 0.

Well we know m/n is greater than 0. So, it's a lower bound. Is it the greatest lower bound? Again yes, since we can consider the elements 1/n which belong to the set, and in a similar fashion we can get as close to 0 as we like.




It has both.

m,n are both positive integers. so, m/(n+n) is always positive. Also m < m+n, so m/(m+n) < 1. So we know the set is bounded below and above. What might the sup and inf be?




Sup is 1, since the largest positive element is 1.

However inf is not 0. If m is odd, then the element is negative. What's the lowest negative number it can be?



It might help to divide top and bottom by n, think about what you have and consider the first few values of n.

The sup is indeed 1/3, since all terms are less than 1/3, and making n as large as we like we can get as close to 1/3 as we like.

Have a further think on the inf.
Thank you for helping me .

So I have gathered some thoughts based on your advice and updated my solutions below.

b) we have m < m+n, therefore m/m+n < 1. Therefore sup = 1. As for infimum, I think its 0. Because if we say m is 1, then we have 1/1+n where now if we say n tends to infinity then we get the limit tends to 0. So I am guessing its 0. It obviously cant be negative cause m, n are naturals.

c) Yes. Inf is not 0. My mistake. I think its -1.

d) for inf by using your advice if I divide up and down by n then I have 1/(3+1/n). Now, 1/n is maximized when n is 1. So inf is 1/4?

Thanks again for helping. Really appreciate it.
0
reply
Rohan77642
Badges: 12
Rep:
?
#5
Report Thread starter 1 week ago
#5
(Original post by RDKGames)
Sounds good. The infimum is indeed 0. You can make the denominator as large as you want and the fraction will tend to zero.



It does. What happens when you fix m and let n get very large? Or what happens when n=1 and you let m get very large?



Supremum looks good, but infimum not so much. I don't see why you set m being even here; we lose generality.

What happens when n=1 and m is odd??



Supremum is good.

But notice that this is clearly an increasing sequence. Therefore you can find the smallest element of this set.
Thanks. I have updated some of my answers in a recent post.

Thanks for the help. Appreciate it
0
reply
ghostwalker
  • Study Helper
Badges: 15
#6
Report 1 week ago
#6
(Original post by Rohan77642)
b) we have m < m+n, therefore m/m+n < 1. Therefore sup = 1.
The second "therefore" needs a little more explanation. 1 is clearly an upper bound. If we consider the subset formed by m/(m+1) we can see that this gets as close as we like to 1 as m is increased. And therefore the sup is 1.

As for infimum, I think its 0. Because if we say m is 1, then we have 1/1+n where now if we say n tends to infinity then we get the limit tends to 0. So I am guessing its 0. It obviously cant be negative cause m, n are naturals.
Yes.

c) Yes. Inf is not 0. My mistake. I think its -1.
Yes.

d) for inf by using your advice if I divide up and down by n then I have 1/(3+1/n). Now, 1/n is maximized when n is 1. So inf is 1/4?
Yes.
Last edited by ghostwalker; 1 week ago
reply
Rohan77642
Badges: 12
Rep:
?
#7
Report Thread starter 1 week ago
#7
(Original post by ghostwalker)
The "therefore" needs a little more explanation. 1 is clearly an upper bound. If we consider the subset formed by m/(m+1) we can see that this gets as close as we like to 1 as m is increased. And therefore the sup is 1.



Yes.



Yes.



Yes.
Thanks a lot.
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • Cardiff University
    Undergraduate Open Day Undergraduate
    Wed, 27 Mar '19
  • University of Portsmouth
    Postgraduate and Part-Time Open Evenings Postgraduate
    Wed, 27 Mar '19
  • Middlesex University London
    Postgraduate Open Evening Postgraduate
    Wed, 27 Mar '19

How old were you when you first saw porn?

I've never seen it (221)
22.97%
Before I was 12 (346)
35.97%
13 (153)
15.9%
14 (111)
11.54%
15 (56)
5.82%
16 (36)
3.74%
17 (11)
1.14%
18 (8)
0.83%
Between the ages of 19 - 25 (15)
1.56%
Over 25 (5)
0.52%

Watched Threads

View All