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]
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]