verify wording on supremum & infimum question

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bluefrog

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I am unsure whether to post this problem in the Calculus or Advanced Algebra forums.
I'd like to verify the a) logic and b) wording of the following problem:
Determine supremum and infimum of [MATH] X=\Bigl \lbrace \frac{n}{n+1} : n \in \Bbb{N} \Bigr \rbrace [/MATH]
For supremum: [MATH] \forall x \in X, x \lt 1 \text { and upper bound of } X=1 [/MATH]Need to show that 1 is the least upper bound. Assume 1 is not the least upper bound, hence
[MATH]\exists \; m>0 \text{ s.t. } 1-m [/MATH] is a lower upper bound [MATH]\forall x \in X [/MATH].
However, [MATH]\exists x \in X \text{ s.t. } 1>x>1-m[/MATH], since
[MATH]\exists n \in \Bbb{N} \text { s.t. } 1-m < \frac{n}{n+1} \Rightarrow \\ 1-\frac{n}{n+1}<m \Rightarrow \frac{n+1}{n+1} - \frac{n}{n+1} < m \Rightarrow \frac{1}{m} < n+ 1 \Rightarrow \frac{1}{m} -1 < n [/MATH]This shows that [MATH]\exists x \in X \text{ s.t. } x>1-m[/MATH], which contradicts assumption.
Hence, [MATH]x<1 \; \forall x \in X \text{ and } \sup{X}=1 [/MATH]
For infimum, for [MATH]n \in \Bbb{N} [/MATH] where [MATH]x=1 \text{, then lower bound } = \frac{1}{1+1}=\frac{1}{2}, \text{ since } \frac{1}{2} \in X \Rightarrow \inf{X}=\frac{1}{2}[/MATH]
 
bluefrog said:
I am unsure whether to post this problem in the Calculus or Advanced Algebra forums.
I'd like to verify the a) logic and b) wording of the following problem:
Determine supremum and infimum of [MATH] X=\Bigl \lbrace \frac{n}{n+1} : n \in \Bbb{N} \Bigr \rbrace [/MATH]
For supremum: [MATH] \forall x \in X, x \lt 1 \text { and upper bound of } X=1 [/MATH]Need to show that 1 is the least upper bound. Assume 1 is not the least upper bound, hence
[MATH]\exists \; m>0 \text{ s.t. } 1-m [/MATH] is a lower upper bound [MATH]\forall x \in X [/MATH].
However, [MATH]\exists x \in X \text{ s.t. } 1>x>1-m[/MATH], since
[MATH]\exists n \in \Bbb{N} \text { s.t. } 1-m < \frac{n}{n+1} \Rightarrow \\ 1-\frac{n}{n+1}<m \Rightarrow \frac{n+1}{n+1} - \frac{n}{n+1} < m \Rightarrow \frac{1}{m} < n+ 1 \Rightarrow \frac{1}{m} -1 < n [/MATH]This shows that [MATH]\exists x \in X \text{ s.t. } x>1-m[/MATH], which contradicts assumption.
Hence, [MATH]x<1 \; \forall x \in X \text{ and } \sup{X}=1 [/MATH]
For infimum, for [MATH]n \in \Bbb{N} [/MATH] where [MATH]x=1 \text{, then lower bound } = \frac{1}{1+1}=\frac{1}{2}, \text{ since } \frac{1}{2} \in X \Rightarrow \inf{X}=\frac{1}{2}[/MATH]
Click to expand...
bluefrog, you must follow your assigned textbook. That said, the book Naive Set Theory, by Paul Halmos is often called the Bible of working mathematicians, defines \(\mathbb{N}=\{0,1,2,\cdots\}\) i.e. as containing zero. For me that means that \(0\) is the infimum of X.
What does your textbook say?
 
bluefrog said:
I am unsure whether to post this problem in the Calculus or Advanced Algebra forums.
I'd like to verify the a) logic and b) wording of the following problem:
Determine supremum and infimum of [MATH] X=\Bigl \lbrace \frac{n}{n+1} : n \in \Bbb{N} \Bigr \rbrace [/MATH]
For supremum: [MATH] \forall x \in X, x \lt 1 \text { and upper bound of } X=1 [/MATH]Need to show that 1 is the least upper bound. Assume 1 is not the least upper bound, hence
[MATH]\exists \; m>0 \text{ s.t. } 1-m [/MATH] is a lower upper bound [MATH]\forall x \in X [/MATH].
However, [MATH]\exists x \in X \text{ s.t. } 1>x>1-m[/MATH], since
[MATH]\exists n \in \Bbb{N} \text { s.t. } 1-m < \frac{n}{n+1} \Rightarrow \\ 1-\frac{n}{n+1}<m \Rightarrow \frac{n+1}{n+1} - \frac{n}{n+1} < m \Rightarrow \frac{1}{m} < n+ 1 \Rightarrow \frac{1}{m} -1 < n [/MATH]This shows that [MATH]\exists x \in X \text{ s.t. } x>1-m[/MATH], which contradicts assumption.
Hence, [MATH]x<1 \; \forall x \in X \text{ and } \sup{X}=1 [/MATH]
For infimum, for [MATH]n \in \Bbb{N} [/MATH] where [MATH]x=1 \text{, then lower bound } = \frac{1}{1+1}=\frac{1}{2}, \text{ since } \frac{1}{2} \in X \Rightarrow \inf{X}=\frac{1}{2}[/MATH]
Click to expand...
I accept the part for supremum. The part for the infimum is totally garbage. Just because 1/2 is in the set does not mean that it is the infimum. After all following that logic 2/3 is the infimum as well as it too is in the set.
What exactly did you mean when you said x=1?? Your own definition stated that x is in X. But 1 is NOT in X.
To fix your proof you need to remove x=1 and you must show that X is an increasing sequence which you need to argue that the 1st term is the infimum. That was a sloppy sloppy mistake on your part. Please fix and post back.
 
pka said:
bluefrog, you must follow your assigned textbook. That said, the book Naive Set Theory, by Paul Halmos is often called the Bible of working mathematicians, defines \(\mathbb{N}=\{0,1,2,\cdots\}\) i.e. as containing zero. For me that means that \(0\) is the infimum of X.
What does your textbook say?
Click to expand...
Halmos' book kept me up at nights, ok for many many nights, but it was/is a classic book to study from.
 
Thanks pka & jomo

In answer to your questions
The definition for natural numbers from my LSE study guide is
"natural numbers are the positive integers", where zero is not considered positive or negative.

[MATH]x=1[/MATH] was a typo for infimum, should have been [MATH]n=1[/MATH].

My understanding of maximum, minimum, supremum and infimum in regards to bounds is that if an element is within the set and is clearly the max or min, then, since it is an element it is also the supremum and infimum respectively.
Hence, I stated [MATH]\frac{1}{2} \in X[/MATH].
Perhaps if I illustrated a sequence at the beginning, like for example
[MATH]X=\Bigl \lbrace \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, ... \Bigr \rbrace[/MATH], that might make it more clear?
 
But you never stated that 1/2 was the smallest number in X. Being clearly the max or min is nonsense! It must be shown. Do you see how important that is? In the writing of a proof can I just write that it is obviously true or do I have to prove the obvious?
 
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